Derivatives

Derivative of a function f(x) at any real number a is given by –

{where h is a very small positive number}

∴ derivative of f(x) = 3x at x = 2 is given as –

Hence,

Derivative of f(x) = 3x at x = 2 is 3

Derivative of a function f(x) at any real number a is given by –

{where h is a very small positive number}

∴ derivative of x² – 2 at x = 10 is given as –

⇒ f’(10) = 0 + 20 = 20

Hence,

Derivative of f(x) = x² – 2 at x = 10 is 20

Derivative of a function f(x) at any real number a is given by –

{where h is a very small positive number}

∴ derivative of 99x at x = 100 is given as –

Hence,

Derivative of f(x) = 99x at x = 100 is 99

Derivative of a function f(x) at any real number a is given by –

{where h is a very small positive number}

∴ derivative of x at x = 1 is given as –

Hence,

Derivative of f(x) = x at x = 1 is 1

Derivative of a function f(x) at any real number a is given by –

{where h is a very small positive number}

∴ derivative of cos x at x = 0 is given as –

∵ we can’t find the limit by direct substitution as it gives 0/0 (indeterminate form)

So we need to do few simplifications to evaluate the limit.

As we know that 1 – cos x = 2 sin²(x/2)

Dividing the numerator and denominator by 2 to get the form (sin x)/x to apply sandwich theorem, also multiplying h in numerator and denominator to get the required form.

Using algebra of limits we have –

Use the formula:

∴ f’(0) = – 1×0 = 0

Hence,

Derivative of f(x) = cos x at x = 0 is 0

Derivative of a function f(x) at any real number a is given by –

{where h is a very small positive number}

∴ derivative of cos x at x = 0 is given as –

∵ we can’t find the limit by direct substitution as it gives 0/0 (indeterminate form)

∴ Use the formula: {sandwich theorem}

∴ f’(0) = 1

Hence,

Derivative of f(x) = tan x at x = 0 is 1

Derivative of a function f(x) at any real number a is given by –

{where h is a very small positive number}

∴ derivative of sin x at x = π/2 is given as –

{∵ sin (π/2 + x) = cos x }

∵ we can’t find the limit by direct substitution as it gives 0/0 (indeterminate form)

So we need to do few simplifications to evaluate the limit.

As we know that 1 – cos x = 2 sin²(x/2)

Dividing the numerator and denominator by 2 to get the form (sin x)/x to apply sandwich theorem, also multiplying h in numerator and denominator to get the required form.

Using algebra of limits we have –

Use the formula:

∴ f’(π/2) = – 1×0 = 0

Hence,

Derivative of f(x) = sin x at x = π/2 is 0

Derivative of a function f(x) at any real number a is given by –

{where h is a very small positive number}

∴ derivative of x at x = 1 is given as –

Hence,

Derivative of f(x) = x at x = 1 is 1

Derivative of a function f(x) at any real number a is given by –

{where h is a very small positive number}

∴ derivative of 2cos x at x = π/2 is given as –

{∵ cos (π/2 + x) = – sin x }

∵ we can’t find the limit by direct substitution as it gives 0/0 (indeterminate form)

⇒ f’(π/2) =

Use the formula:

∴ f’(π/2) = – 2×1 = – 2

Hence,

Derivative of f(x) = 2cos x at x = π/2 is – 2

Derivative of a function f(x) at any real number a is given by –

{where h is a very small positive number}

∴ derivative of sin 2x at x = π/2 is given as –

{∵ sin (π + x) = – sin x & sin π = 0}

∵ we can’t find the limit by direct substitution as it gives 0/0 (indeterminate form)

We need to use sandwich theorem to evaluate the limit.

Multiplying 2 in numerator and denominator to apply the formula.

⇒ f’(π/2) = –

Use the formula:

∴ f’(π/2) = – 2×1 = – 2

Hence,

Derivative of f(x) = sin 2x at x = π/2 is – 2

We need to find derivative of f(x) = 2/x from first principle

Derivative of a function f(x) from first principle is given by –

{where h is a very small positive number}

∴ derivative of f(x) = 2/x is given as –

As h is cancelled and by putting h = 0 we are not getting any indeterminate form so we can evaluate the limit directly.

Hence,

Derivative of f(x) = 2/x

We need to find derivative of f(x) = 1/√x

Derivative of a function f(x) from first principle is given by –

{where h is a very small positive number}

∴ derivative of f(x) = 1/√x is given as –

Using algebra of limits –

Multiplying numerator and denominator by √x + √(x + h) to rationalise the expression so that we don’t get any indeterminate form after putting value of h

Using (a + b)(a – b) = a² – b²

Using algebra of limits –

Hence,

Derivative of f(x) = 1/√x

We need to find derivative of f(x) = 1/x³

Derivative of a function f(x) from first principle is given by –

{where h is a very small positive number}

∴ derivative of f(x) = 1/x³ is given as –

Using algebra of limits –

Using a³ – b³ = (a – b)(a² + ab + b²)

We have:

As h is cancelled and by putting h = 0 we are not getting any indeterminate form so we can evaluate the limit directly.

Hence,

Derivative of f(x) = 1/x³

We need to find derivative of

Derivative of a function f(x) from first principle is given by –

{where h is a very small positive number}

∴ derivative ofis given as –

Using algebra of limits –

As h is cancelled and by putting h = 0 we are not getting any indeterminate form so we can evaluate the limit directly.

Hence,

Derivative of f(x) =

We need to find derivative of f(x) =

Derivative of a function f(x) from first principle is given by –

{where h is a very small positive number}

∴ derivative of f(x) = is given as –

Using algebra of limits –

As h is cancelled and by putting h = 0 we are not getting any indeterminate form so we can evaluate the limit directly.

Hence,

Derivative of f(x) =

We need to find derivative of

Derivative of a function f(x) from first principle is given by –

{where h is a very small positive number}

∴ derivative of is given as –

Using algebra of limits –

Hence,

Derivative of f(x) =

We need to find derivative of

Derivative of a function f(x) is given by –

{where h is a very small positive number}

∴ derivative of is given as –

Using algebra of limits –

Hence,

Derivative of f(x) =

We need to find the derivative of f(x) = kxⁿ

Derivative of a function f(x) from first principle is given by –

{where h is a very small positive number}

∴ derivative of f(x) = kxⁿ is given as –

Using binomial expansion we have –

(x + h)ⁿ = ⁿC₀ xⁿ + ⁿC₁ x^{n – 1}h + ⁿC₂ x^{n – 2}h² + …… + ⁿC_n hⁿ

Take h common –

As there is no more indeterminate, so put value of h to get the limit.

⇒ f’(x) = k ⁿC₁ x^{n – 1} = k nx^{n – 1}

Hence,

Derivative of f(x) = kxⁿ is k nx^{n – 1}

We need to find derivative of f(x) = 1/√(3 – x)

Derivative of a function f(x) from first principle is given by –

{where h is a very small positive number}

∴ derivative of f(x) = 1/√(3 – x) is given as –

Using algebra of limits –

Multiplying numerator and denominator by √(3 – x) + √(3 – x – h) to rationalise the expression so that we don’t get any indeterminate form after putting value of h

Using (a + b)(a – b) = a² – b²

Using algebra of limits –

Hence,

We need to find the derivative of f(x) = x² + x + 3

Derivative of a function f(x) from first principle is given by –

{where h is a very small positive number}

∴ derivative of f(x) = x² + x + 3 is given as –

f’(x) =

⇒ f’(x) =

Using (a + b)² = a² + 2ab + b²

⇒ f’(x) =

⇒ f’(x) =

Take h common –

⇒ f’(x) =

⇒ f’(x) =

As there is no more indeterminate, so put value of h to get the limit.

⇒ f’(x) = (2x + 0 + 1)

⇒ f’(x) = 2x + 1 = 2x + 1

Hence,

Derivative of f(x) = x² + x + 3 is (2x + 1)

We need to find the derivative of f(x) = (x + 2)³

Derivative of a function f(x) from first principle is given by –

f’(x) = {where h is a very small positive number}

∴ derivative of f(x) = (x + 2)³ is given as –

f’(x) =

⇒ f’(x) =

Using a³ – b³ = (a – b)(a² + ab + b²)

⇒ f’(x) =

⇒ f’(x) =

As h is cancelled, so there is no more indeterminate form possible if we put value of h = 0.

So, evaluate the limit by putting h = 0

⇒ f’(x) =

⇒ f’(x) = (x + 0 + 2)² + (x + 2)(x + 2) + (x + 2)²

⇒ f’(x) = 3 (x + 2)²

⇒ f’(x) = 3 (x + 2)²

Hence,

Derivative of f(x) = (x + 2)³ is 3(x + 2)²

We need to find the derivative of f(x) = x³ + 4x² + 3x + 2

Derivative of a function f(x) from first principle is given by –

f’(x) = {where h is a very small positive number}

∴ derivative of f(x) = x³ + 4x² + 3x + 2is given as –

f’(x) =

⇒ f’(x) =

⇒ f’(x) =

Using (a + b)² = a² + 2ab + b² , we have –

⇒ f’(x) =

⇒ f’(x) =

⇒ f’(x) =

Take h common –

⇒ f’(x) =

As h is cancelled, so there is no more indeterminate form possible if we put value of h = 0.

So, evaluate the limit by putting h = 0

⇒ f’(x) =

⇒ f’(x) = 3x² + 3x(0) + 8x + 3 + 0² + 4(0)

⇒ f’(x) = 3x² + 8x + 3

Hence,

Derivative of f(x) = x³ + 4x² + 3x + 2 is 3x² + 8x + 3

We need to find the derivative of f(x) = (x² + 1)(x – 5)

Derivative of a function f(x) from first principle is given by –

f’(x) = {where h is a very small positive number}

∴ derivative of f(x) = (x² + 1)(x – 5) is given as –

f’(x) =

⇒ f’(x) =

⇒ f’(x) =

Using (a + b)² = a² + 2ab + b² and (a + b)³ = a³ + 3ab(a + b) + b³ we have –

⇒ f’(x) =

⇒ f’(x) =

Take h common –

⇒ f’(x) =

As h is cancelled, so there is no more indeterminate form possible if we put value of h = 0

∴ f’(x) =

So, evaluate the limit by putting h = 0

⇒ f’(x) = 3x² + 3(0)x + 0² + 1 – 10x – 5(0)

⇒ f’(x) = 3x² – 10x + 1

Hence,

Derivative of f(x) = (x² + 1)(x – 5) is 3x² – 10x + 1

We need to find derivative of f(x) = √(2x² + 1)

Derivative of a function f(x) from first principle is given by –

f’(x) = {where h is a very small positive number}

∴ derivative of f(x) = √(2x² + 1) is given as –

f’(x) =

⇒ f’(x) =

As the above limit can’t be evaluated by putting the value of h because it takes 0/0 (indeterminate form)

∴ multiplying denominator and numerator by to eliminate the indeterminate form.

⇒ f’(x) =

Using algebra of limits & a² – b² = (a + b)(a – b),we have –

⇒ f’(x) =

⇒ f’(x) =

⇒ f’(x) =

Using a² – b² = (a + b)(a – b), we have –

⇒ f’(x) =

⇒ f’(x) =

⇒ f’(x) =

Evaluating the limit by putting h = 0

∴ f’(x) =

∴ f’(x) =

Hence,

Derivative of f(x) = √(2x² + 1)

We need to find derivative of f(x) =

Derivative of a function f(x) from first principle is given by –

f’(x) = {where h is a very small positive number}

∴ derivative of f(x) = is given as –

f’(x) =

⇒ f’(x) =

⇒ f’(x) =

⇒ f’(x) =

Using algebra of limits –

⇒ f’(x) =

⇒ f’(x) =

⇒ f’(x) =

⇒ f’(x) =

⇒ f’(x) =

∴ f’(x) =

Hence,

Derivative of f(x) =

We need to find derivative of f(x) = e ^{– x}

Derivative of a function f(x) is given by –

f’(x) = {where h is a very small positive number}

∴ derivative of f(x) = e ^{– x} is given as –

f’(x) =

⇒ f’(x) =

⇒ f’(x) =

Taking e ^{– x} common, we have –

⇒ f’(x) =

Using algebra of limits –

⇒ f’(x) =

As one of the limits can’t be evaluated by directly putting the value of h as it will take 0/0 form.

So we need to take steps to find its value.

⇒ f’(x) =

Use the formula:

⇒ f’(x) = e ^{– x} × ( – 1)

⇒ f’(x) = – e ^{– x}

Hence,

Derivative of f(x) = e ^{– x} = – e ^{– x}

We need to find derivative of f(x) = e^3x

Derivative of a function f(x) is given by –

f’(x) = {where h is a very small positive number}

∴ derivative of f(x) = e^3x is given as –

f’(x) =

⇒ f’(x) =

⇒ f’(x) =

Taking e ^{– x} common, we have –

⇒ f’(x) =

Using algebra of limits –

⇒ f’(x) =

As one of the limits can’t be evaluated by directly putting the value of h as it will take 0/0 form.

So we need to take steps to find its value.

⇒ f’(x) =

Use the formula:

⇒ f’(x) = e^3x × (3)

⇒ f’(x) = 3e^3x

Hence,

Derivative of f(x) = e^3x = 3e^3x

We need to find derivative of f(x) = e^{ax + b}

Derivative of a function f(x) is given by –

f’(x) = {where h is a very small positive number}

∴ derivative of f(x) = e^{ax + b} is given as –

f’(x) =

⇒ f’(x) =

⇒ f’(x) =

Taking e^{ax + b} common, we have –

⇒ f’(x) =

Using algebra of limits –

⇒ f’(x) =

As one of the limits can’t be evaluated by directly putting the value of h as it will take 0/0 form.

So we need to take steps to find its value.

⇒ f’(x) =

Use the formula:

⇒ f’(x) = e^{ax + b} × (a)

⇒ f’(x) = ae^{ax + b}

Hence,

Derivative of f(x) = e^{ax + b} = ae^{ax + b}

We need to find derivative of f(x) = xe^x

Derivative of a function f(x) is given by –

f’(x) = {where h is a very small positive number}

∴ derivative of f(x) = xe^x is given as –

f’(x) =

⇒ f’(x) =

⇒ f’(x) =

Using algebra of limits, we have –

⇒ f’(x) =

⇒ f’(x) =

Again Using algebra of limits, we have –

⇒ f’(x) =

Use the formula:

⇒ f’(x) = e^x + xe^x

⇒ f’(x) = e^x(x + 1)

Hence,

Derivative of f(x) = xe^x = e^x(x + 1)

We need to find derivative of f(x) = – x

Derivative of a function f(x) is given by –

f’(x) = {where h is a very small positive number}

∴ derivative of f(x) = – xis given as –

f’(x) =

⇒ f’(x) =

⇒ f’(x) =

⇒ f’(x) =

∴ f’(x) = – 1

Hence,

Derivative of f(x) = – x = – 1

We need to find derivative of f(x) = ( – x) ^{– 1} = – 1/x

Derivative of a function f(x) is given by –

f’(x) = {where h is a very small positive number}

∴ derivative of f(x) = – 1/x is given as –

f’(x) =

⇒ f’(x) =

⇒ f’(x) =

⇒ f’(x) =

As h is cancelled and by putting h = 0 we are not getting any indeterminate form so we can evaluate the limit directly.

∴ f’(x) =

Hence,

Derivative of f(x) = ( – x) ^{– 1}

We need to find derivative of f(x) = sin (x + 1)

Derivative of a function f(x) is given by –

f’(x) = {where h is a very small positive number}

∴ derivative of f(x) = sin (x + 1) is given as –

f’(x) =

⇒ f’(x) =

We can’t evaluate the limits at this stage only as on putting value it will take 0/0 form. So, we need to do little modifications.

Use: sin A – sin B = 2 cos ((A + B)/2) sin ((A – B)/2)

∴ f’(x) =

⇒ f’(x) =

Using algebra of limits –

⇒ f’(x) =

Use the formula –

∴ f’(x) =

Put the value of h to evaluate the limit –

∴ f’(x) = cos (x + 1 + 0) = cos (x + 1)

Hence,

Derivative of f(x) = sin (x + 1) = cos (x + 1)

We need to find derivative of f(x) = cos (x – π/8)

Derivative of a function f(x) is given by –

f’(x) = {where h is a very small positive number}

∴ derivative of f(x) = cos (x – π/8) is given as –

f’(x) =

⇒ f’(x) =

We can’t evaluate the limits at this stage only as on putting value it will take 0/0 form. So, we need to do little modifications.

Use: cos A – cos B = – 2 sin ((A + B)/2) sin ((A – B)/2)

∴ f’(x) =

⇒ f’(x) =

Using algebra of limits –

⇒ f’(x) =

Use the formula –

∴ f’(x) =

Put the value of h to evaluate the limit –

∴ f’(x) = – sin (x – π/8 + 0) = – sin (x – π/8)

Hence,

Derivative of f(x) = cos (x – π/8) = – sin (x – π/8)

We need to find derivative of f(x) = xsin x

Derivative of a function f(x) is given by –

f’(x) = {where h is a very small positive number}

∴ derivative of f(x) = x sin x is given as –

f’(x) =

⇒ f’(x) =

⇒ f’(x) =

Using algebra of limits, we have –

⇒ f’(x) =

⇒ f’(x) =

Using algebra of limits we have –

∴ f’(x) = sin x +

We can’t evaluate the limits at this stage only as on putting value it will take 0/0 form. So, we need to do little modifications.

Use: sin A – sin B = 2 cos ((A + B)/2) sin ((A – B)/2)

∴ f’(x) =

⇒ f’(x) =

Using algebra of limits –

⇒ f’(x) =

Use the formula –

∴ f’(x) =

Put the value of h to evaluate the limit –

∴ f’(x) = sin x + x cos(x + 0) = sin x + x cos x

Hence,

Derivative of f(x) = (x sin x) is (sin x + x cos x)

We need to find derivative of f(x) = x cos x

Derivative of a function f(x) is given by –

f’(x) = {where h is a very small positive number}

∴ derivative of f(x) = x cos x is given as –

f’(x) =

⇒ f’(x) =

⇒ f’(x) =

Using algebra of limits, we have –

⇒ f’(x) =

⇒ f’(x) =

Using algebra of limits we have –

∴ f’(x) = cos x +

We can’t evaluate the limits at this stage only as on putting value it will take 0/0 form. So, we need to do little modifications.

Use: cos A – cos B = – 2 sin ((A + B)/2) sin ((A – B)/2)

∴ f’(x) = cos x +

⇒ f’(x) = cos x –

Using algebra of limits –

⇒ f’(x) =

Use the formula –

∴ f’(x) =

Put the value of h to evaluate the limit –

∴ f’(x) = cos x – x sin x

Hence,

Derivative of f(x) = x cos x is cos x – x sin x

We need to find derivative of f(x) = sin (2x – 3)

Derivative of a function f(x) is given by –

f’(x) = {where h is a very small positive number}

∴ derivative of f(x) = sin (2x – 3) is given as –

f’(x) =

⇒ f’(x) =

We can’t evaluate the limits at this stage only as on putting value it will take 0/0 form. So, we need to do little modifications.

Use: sin A – sin B = 2 cos ((A + B)/2) sin ((A – B)/2)

∴ f’(x) =

⇒ f’(x) =

Using algebra of limits –

⇒ f’(x) =

Use the formula –

∴ f’(x) =

Put the value of h to evaluate the limit –

∴ f’(x) = 2 cos (2x – 3 + 0) = 2cos (2x – 3)

Hence,

Derivative of f(x) = sin (2x – 3) = 2cos (2x – 3)

We need to find derivative of f(x) = √(sin 2x)

Derivative of a function f(x) is given by –

f’(x) = {where h is a very small positive number}

∴ derivative of f(x) = √(sin 2x) is given as –

f’(x) =

⇒ f’(x) =

We can’t evaluate the limits at this stage only as on putting value it will take 0/0 form. So, we need to do little modifications.

Multiplying numerator and denominator by √(sin 2(x + h)) + √(sin 2x), we have –

⇒ f’(x) =

Using a² – b² = (a + b)(a – b), we have –

⇒ f’(x) =

Again using algebra of limits, we get –

⇒ f’(x) =

Use: sin A – sin B = 2 cos ((A + B)/2) sin ((A – B)/2)

∴ f’(x) =

⇒ f’(x) =

Using algebra of limits –

⇒ f’(x) =

Use the formula –

∴ f’(x) =

Put the value of h to evaluate the limit –

∴ f’(x) =

Hence,

Derivative of f(x) = √(sin 2x) =

We need to find derivative of f(x) = (sin x)/x

Derivative of a function f(x) is given by –

f’(x) = {where h is a very small positive number}

∴ derivative of f(x) = (sin x)/x is given as –

f’(x) =

⇒ f’(x) =

⇒ f’(x) =

Using algebra of limits we have –

∴ f’(x) =

⇒ f’(x) =

⇒ f’(x) =

⇒ f’(x) =

Using algebra of limits, we have –

⇒ f’(x) =

⇒ f’(x) =

Using algebra of limits we have –

∴ f’(x) =

We can’t evaluate the limits at this stage only as on putting value it will take 0/0 form. So, we need to do little modifications.

Use: sin A – sin B = 2 cos ((A + B)/2) sin ((A – B)/2)

∴ f’(x) =

⇒ f’(x) =

Using algebra of limits –

⇒ f’(x) =

Use the formula –

∴ f’(x) =

Put the value of h to evaluate the limit –

∴ f’(x) =

Hence,

Derivative of f(x) = (sin x)/x is

We need to find derivative of f(x) = (cos x)/x

Derivative of a function f(x) is given by –

f’(x) = {where h is a very small positive number}

∴ derivative of f(x) = (cos x)/x is given as –

f’(x) =

⇒ f’(x) =

⇒ f’(x) =

Using algebra of limits we have –

∴ f’(x) =

⇒ f’(x) =

⇒ f’(x) =

⇒ f’(x) =

Using algebra of limits, we have –

⇒ f’(x) =

⇒ f’(x) =

Using algebra of limits we have –

∴ f’(x) =

We can’t evaluate the limits at this stage only as on putting value it will take 0/0 form. So, we need to do little modifications.

Use: cos A – cos B = – 2 sin ((A + B)/2) sin ((A – B)/2)

∴ f’(x) =

⇒ f’(x) =

Using algebra of limits –

⇒ f’(x) =

Use the formula –

∴ f’(x) =

Put the value of h to evaluate the limit –

∴ f’(x) =

Hence,

Derivative of f(x) = (cos x)/x is

We need to find derivative of f(x) = x² sin x

Derivative of a function f(x) is given by –

f’(x) = {where h is a very small positive number}

∴ derivative of f(x) = x² sin x is given as –

f’(x) =

⇒ f’(x) =

Using (a + b)² = a² + 2ab + b² ,we have –

⇒ f’(x) =

Using algebra of limits, we have –

⇒ f’(x) =

⇒ f’(x) =

⇒ f’(x) = 0×sin (x + 0) + 2x sin (x + 0) +

⇒ f’(x) =

Using algebra of limits we have –

∴ f’(x) = 2x sin x +

We can’t evaluate the limits at this stage only as on putting value it will take 0/0 form. So, we need to do little modifications.

Use: sin A – sin B = 2 cos ((A + B)/2) sin ((A – B)/2)

∴ f’(x) =

⇒ f’(x) =

Using algebra of limits –

⇒ f’(x) =

Use the formula –

∴ f’(x) =

Put the value of h to evaluate the limit –

∴ f’(x) = 2x sin x + x² cos(x + 0) = 2x sin x + x² cos x

Hence,

Derivative of f(x) = (x² sin x) is (2x sin x + x² cos x)

We need to find derivative of f(x) = √(sin (3x + 1))

Derivative of a function f(x) is given by –

f’(x) = {where h is a very small positive number}

∴ derivative of f(x) = √(sin (3x + 1)) is given as –

f’(x) =

⇒ f’(x) =

We can’t evaluate the limits at this stage only as on putting value it will take 0/0 form. So, we need to do little modifications.

Multiplying numerator and denominator by , we have –

⇒ f’(x) =

Using a² – b² = (a + b)(a – b), we have –

⇒ f’(x) =

Again using algebra of limits, we get –

⇒ f’(x) =

Use: sin A – sin B = 2 cos ((A + B)/2) sin ((A – B)/2)

∴ f’(x) =

⇒ f’(x) =

Using algebra of limits –

⇒ f’(x) =

Use the formula –

∴ f’(x) =

Put the value of h to evaluate the limit –

∴ f’(x) =

Hence,

Derivative of f(x) = √(sin (3x + 1)) =

We need to find derivative of f(x) = sin x + cos x

Derivative of a function f(x) is given by –

f’(x) = {where h is a very small positive number}

∴ derivative of f(x) = sin x + cos x is given as –

f’(x) =

⇒ f’(x) =

Using algebra of limits we have –

⇒ f’(x) =

We can’t evaluate the limits at this stage only as on putting value it will take 0/0 form. So, we need to do little modifications.

Use: sin A – sin B = 2 cos ((A + B)/2) sin ((A – B)/2) and

cos A – cos B = – 2 sin ((A + B)/2) sin ((A – B)/2)

∴ f’(x) =

Dividing numerator and denominator by 2 in both the terms –

⇒ f’(x) =

Using algebra of limits –

⇒ f’(x) =

Use the formula –

∴ f’(x) =

Put the value of h to evaluate the limit –

∴ f’(x) = cos (x + 0) – sin (x + 0) = cos x – sin x

Hence,

Derivative of f(x) = sin x + cos x = cos x – sin x

We need to find derivative of f(x) = x² e^x

Derivative of a function f(x) is given by –

f’(x) = {where h is a very small positive number}

∴ derivative of f(x) = x² e^x is given as –

f’(x) =

⇒ f’(x) =

⇒ f’(x) =

Using algebra of limits, we have –

⇒ f’(x) =

⇒

As 2 of the terms will not take indeterminate form if we put value of h = 0, so obtained their limiting value as follows –

∴ f’(x) = 0×e^{x + 0} + 2x e^{x + 0} +

Use the formula:

⇒ f’(x) = 2x e^x + x² e^x

⇒ f’(x) = 2x e^x + x² e^x

Hence,

Derivative of f(x) = x² e^x = 2x e^x + x² e^x

We need to find derivative of f(x) =

Derivative of a function f(x) is given by –

f’(x) = {where h is a very small positive number}

∴ derivative of f(x) = is given as –

f’(x) =

⇒ f’(x) =

⇒ f’(x) =

Taking common, we have –

⇒ f’(x) =

Using algebra of limits –

⇒ f’(x) =

⇒ f’(x) =

As one of the limits can’t be evaluated by directly putting the value of h as it will take 0/0 form.

So we need to take steps to find its value.

As h → 0 so, (2hx + h²) → 0

∴ multiplying numerator and denominator by (2hx + h²) in order to apply the formula –

∴ f’(x) =

Again using algebra of limits, we have –

⇒ f’(x) =

Use the formula:

⇒ f’(x) =

⇒ f’(x) =

∴ f’(x) =

Hence,

Derivative of f(x) =

We need to find derivative of f(x) = e^√2x

Derivative of a function f(x) is given by –

f’(x) = {where h is a very small positive number}

∴ derivative of f(x) = e^√2x is given as –

f’(x) =

⇒ f’(x) =

⇒ f’(x) =

Taking common, we have –

⇒ f’(x) =

Using algebra of limits –

⇒ f’(x) =

⇒ f’(x) =

As one of the limits can’t be evaluated by directly putting the value of h as it will take 0/0 form.

So we need to take steps to find its value.

As h → 0 so, () → 0

∴ multiplying numerator and denominator by in order to apply the formula –

∴ f’(x) =

Again using algebra of limits, we have –

⇒ f’(x) =

Use the formula:

⇒ f’(x) =

Again we get an indeterminate form, so multiplying and dividing √(2x + 2h) + √(2x) to get rid of indeterminate form.

∴ f’(x) =

Using a² – b² = (a + b)(a – b), we have –

⇒ f’(x) =

Using algebra of limits we have –

⇒ f’(x) =

⇒ f’(x) =

⇒ f’(x) =

∴ f’(x) =

Hence,

Derivative of f(x) = e^√2x =

We need to find derivative of f(x) = e^{√(ax + b)}

Derivative of a function f(x) is given by –

f’(x) = {where h is a very small positive number}

∴ derivative of f(x) = e^{√(ax + b)} is given as –

f’(x) =

⇒ f’(x) =

⇒ f’(x) =

Taking common, we have –

⇒ f’(x) =

Using algebra of limits –

⇒ f’(x) =

⇒ f’(x) =

As one of the limits can’t be evaluated by directly putting the value of h as it will take 0/0 form.

So we need to take steps to find its value.

As h → 0 so, () → 0

∴ multiplying numerator and denominator by in order to apply the formula –

∴ f’(x) =

Again using algebra of limits, we have –

⇒ f’(x) =

Use the formula:

⇒ f’(x) =

Again we get an indeterminate form, so multiplying and dividing √(ax + ah + b) + √(ax + b) to get rid of indeterminate form.

∴ f’(x) =

Using a² – b² = (a + b)(a – b), we have –

⇒ f’(x) =

Using algebra of limits we have –

⇒ f’(x) =

⇒ f’(x) =

⇒ f’(x) =

∴ f’(x) =

Hence,

Derivative of f(x) = e^{√(ax + b)} =

We need to find derivative of f(x) = a^√x

Derivative of a function f(x) is given by –

f’(x) = {where h is a very small positive number}

∴ derivative of f(x) = a^√x is given as –

f’(x) =

⇒ f’(x) =

⇒ f’(x) =

Taking common, we have –

⇒ f’(x) =

Using algebra of limits –

⇒ f’(x) =

⇒ f’(x) =

As one of the limits can’t be evaluated by directly putting the value of h as it will take 0/0 form.

So we need to take steps to find its value.

As h → 0 so, () → 0

∴ multiplying numerator and denominator by in order to apply the formula –

∴ f’(x) =

Again using algebra of limits, we have –

⇒ f’(x) =

Use the formula:

⇒ f’(x) =

Again we get an indeterminate form, so multiplying and dividing

√(x + h) + √(x) to get rid of indeterminate form.

∴ f’(x) =

Using a² – b² = (a + b)(a – b), we have –

⇒ f’(x) =

Using algebra of limits we have –

⇒ f’(x) =

⇒ f’(x) =

⇒ f’(x) =

∴ f’(x) =

Hence,

Derivative of f(x) = a^√x =

We need to find derivative of f(x) =

Derivative of a function f(x) is given by –

f’(x) = {where h is a very small positive number}

∴ derivative of f(x) = is given as –

f’(x) =

⇒ f’(x) =

⇒ f’(x) =

Taking common, we have –

⇒ f’(x) =

Using algebra of limits –

⇒ f’(x) =

⇒ f’(x) =

As one of the limits can’t be evaluated by directly putting the value of h as it will take 0/0 form.

So we need to take steps to find its value.

As h → 0 so, (2hx + h²) → 0

∴ multiplying numerator and denominator by (2hx + h²) in order to apply the formula –

∴ f’(x) =

Again using algebra of limits, we have –

⇒ f’(x) =

Use the formula:

⇒ f’(x) =

⇒ f’(x) =

∴ f’(x) =

Hence,

Derivative of f(x) =

We need to find derivative of f(x) = tan² x

Derivative of a function f(x) is given by –

f’(x) = {where h is a very small positive number}

∴ derivative of f(x) = tan² x is given as –

f’(x) =

⇒ f’(x) =

Using (a + b)(a – b) = a² – b² ,we have –

⇒ f’(x) =

Using algebra of limits, we have –

⇒ f’(x) =

⇒ f’(x) =

⇒ f’(x) = 2tan x

⇒ f’(x) = 2tan x

⇒ f’(x) =

Using: sin A cos B – cos A sin B = sin (A – B)

⇒ f’(x) =

Using algebra of limits we have –

∴ f’(x) =

Use the formula –

∴ f’(x) =

∴ f’(x) =

Hence,

Derivative of f(x) = (tan² x) is (2 tan x sec² x)

We need to find derivative of f(x) = tan (2x + 1)

Derivative of a function f(x) is given by –

f’(x) = {where h is a very small positive number}

∴ derivative of f(x) = tan (2x + 1) is given as –

f’(x) =

⇒ f’(x) =

⇒ f’(x) =

⇒ f’(x) =

⇒ f’(x) =

Using: sin A cos B – cos A sin B = sin (A – B)

⇒ f’(x) =

Using algebra of limits we have –

∴ f’(x) =

To apply sandwich theorem ,we need 2h in denominator, So multiplying by 2 in numerator and denominator by 2.

∴ f’(x) =

Use the formula –

⇒ f’(x) = 2×

∴ f’(x) =

Hence,

Derivative of f(x) = tan(2x + 1) is 2 sec² (2x + 1)

We need to find derivative of f(x) = tan (2x)

Derivative of a function f(x) is given by –

f’(x) = {where h is a very small positive number}

∴ derivative of f(x) = tan (2x) is given as –

f’(x) =

⇒ f’(x) =

⇒ f’(x) =

⇒ f’(x) =

⇒ f’(x) =

Using: sin A cos B – cos A sin B = sin (A – B)

⇒ f’(x) =

Using algebra of limits we have –

∴ f’(x) =

To apply sandwich theorem ,we need 2h in denominator, So multiplying by 2 in numerator and denominator by 2.

∴ f’(x) =

Use the formula –

⇒ f’(x) = 2×

∴ f’(x) =

Hence,

Derivative of f(x) = tan(2x) is 2 sec² (2x)

We need to find derivative of f(x) = √tan x

Derivative of a function f(x) is given by –

f’(x) = {where h is a very small positive number}

∴ derivative of f(x) = √tan x is given as –

f’(x) =

⇒ f’(x) =

As the limit takes 0/0 form on putting h = 0. So we need to remove the indeterminate form. As the numerator expression has square root terms so we need to multiply numerator and denominator by √tan (x + h) + √tan x.

⇒ f’(x) =

Using (a + b)(a – b) = a² – b² ,we have –

⇒ f’(x) =

Using algebra of limits, we have –

⇒ f’(x) =

⇒ f’(x) =

⇒ f’(x) =

⇒ f’(x) =

Using: sin A cos B – cos A sin B = sin (A – B)

⇒ f’(x) =

Using algebra of limits we have –

∴ f’(x) =

Use the formula –

⇒ f’(x) = ×

∴ f’(x) =

Hence,

Derivative of f(x) = √tan(x) is

We need to find derivative of f(x) =

Derivative of a function f(x) from first principle is given by –

f’(x) = {where h is a very small positive number}

∴ derivative of f(x) = is given as –

f’(x) =

⇒ f’(x) =

Use: sin A – sin B = 2 cos ((A + B)/2) sin ((A – B)/2)

⇒ f’(x) =

Using algebra of limits, we have –

⇒ f’(x) = 2

⇒ f’(x) = 2

⇒ f’(x) =

As, h → 0 ⇒ → 0

∴ To use the sandwich theorem to evaluate the limit, we need in denominator. So multiplying this in numerator and denominator.

⇒ f’(x) =

Using algebra of limits –

⇒ f’(x) =

Use the formula:

∴ f’(x) = × 1 ×

⇒ f’(x) =

⇒ f’(x) =

Again we get an indeterminate form, so multiplying and dividing √(2x + 2h) + √(2x) to get rid of indeterminate form.

∴ f’(x) =

Using a² – b² = (a + b)(a – b), we have –

⇒ f’(x) =

Using algebra of limits we have –

⇒ f’(x) =

⇒ f’(x) =

⇒ f’(x) =

∴ f’(x) =

Hence,

Derivative of f(x) = sin √2x =

We need to find derivative of f(x) =

Derivative of a function f(x) from first principle is given by –

f’(x) = {where h is a very small positive number}

∴ derivative of f(x) = is given as –

f’(x) =

⇒ f’(x) =

Use: cos A – cos B = – 2 sin ((A + B)/2) sin ((A – B)/2)

⇒ f’(x) =

Using algebra of limits, we have –

⇒ f’(x) = – 2

⇒ f’(x) = – 2

⇒ f’(x) =

As, h → 0 ⇒ → 0

∴ To use the sandwich theorem to evaluate the limit, we need in denominator. So multiplying this in numerator and denominator.

⇒ f’(x) =

Using algebra of limits –

⇒ f’(x) =

Use the formula:

∴ f’(x) = × 1 ×

⇒ f’(x) =

Again, we get an indeterminate form, so multiplying and dividing √(x + h) + √(x) to get rid of indeterminate form.

∴ f’(x) =

Using a² – b² = (a + b)(a – b), we have –

⇒ f’(x) =

Using algebra of limits we have –

⇒ f’(x) =

⇒ f’(x) =

⇒ f’(x) =

∴ f’(x) =

Hence,

Derivative of f(x) = cos √x =

We need to find derivative of f(x) =

Derivative of a function f(x) from first principle is given by –

f’(x) = {where h is a very small positive number}

∴ derivative of f(x) = is given as –

f’(x) =

⇒ f’(x) =

⇒ f’(x) =

⇒ f’(x) =

Use the formula: sin (A – B) = sin A cos B – cos A sin B

⇒ f’(x) =

Using algebra of limits, we have –

⇒ f’(x) =

⇒ f’(x) =

⇒ f’(x) =

⇒ f’(x) =

As, h → 0 ⇒ → 0

∴ To use the sandwich theorem to evaluate the limit, we need in denominator. So multiplying this in numerator and denominator.

⇒ f’(x) =

Using algebra of limits –

⇒ f’(x) =

Use the formula:

∴ f’(x) = × 1 ×

⇒ f’(x) =

Again, we get an indeterminate form, so multiplying and dividing √(x + h) + √(x) to get rid of indeterminate form.

∴ f’(x) =

Using a² – b² = (a + b)(a – b), we have –

⇒ f’(x) =

Using algebra of limits we have –

⇒ f’(x) =

⇒ f’(x) =

⇒ f’(x) =

∴ f’(x) =

Hence,

Derivative of f(x) = tan √x =

We need to find derivative of f(x) = tan x²

Derivative of a function f(x) from first principle is given by –

f’(x) = {where h is a very small positive number}

∴ derivative of f(x) = tan x²is given as –

f’(x) =

⇒ f’(x) =

⇒ f’(x) =

⇒ f’(x) =

Use the formula: sin (A – B) = sin A cos B – cos A sin B

⇒ f’(x) =

Using algebra of limits, we have –

⇒ f’(x) =

⇒ f’(x) =

⇒ f’(x) =

As, h → 0 ⇒ 2hx + h² → 0

∴ To use the sandwich theorem to evaluate the limit, we need 2hx + h² in denominator. So multiplying this in numerator and denominator.

⇒ f’(x) =

Using algebra of limits –

⇒ f’(x) =

⇒ f’(x) =

Use the formula:

∴ f’(x) = sec² x² × 1 × (2x + 0)

∴ f’(x) = 2x sec² x²

Hence,

Derivative of f(x) = tan x² = 2x sec² x²

Given,

f(x) = x⁴ – 2sin x + 3 cos x

we need to find f’(x), so differentiating both sides with respect to x –

∴

Using algebra of derivatives –

⇒ f’(x) =

Use the formula:

∴ f’(x) = 4x^{4 – 1} – 2 cos x + 3 ( – sin x)

⇒ f’(x) = 4x³ – 2 cos x – 3 sin x

Given,

f(x) = 3^x + x³ + 3³

we need to find f’(x), so differentiating both sides with respect to x –

∴

Using algebra of derivatives –

⇒ f’(x) =

Use the formula:

∴ f’(x) = 3^x log_e 3 – 3x^{3 – 1} + 0

⇒ f’(x) = 3^x log_e 3 – 3x²

Given,

f(x) =

we need to find f’(x), so differentiating both sides with respect to x –

∴

Using algebra of derivatives –

⇒ f’(x) =

⇒ f’(x) =

Use the formula:

∴ f’(x) =

⇒ f’(x) =

∴ f’(x) = x² – x^{( – 1/2)} – 10x ^{– 3}

Given,

f(x) = e^{x log a} + e^{a log x} + e^{a log a}

⇒ f(x) =

We know that, e^{log f(x) =} f(x)

∴ f(x) = a^x + x^a + a^a

we need to find f’(x), so differentiating both sides with respect to x –

∴

Using algebra of derivatives –

⇒ f’(x) =

Use the formula:

∴ f’(x) = a^x log_e a – ax^{a – 1} + 0

⇒ f’(x) = a^x log_e a – ax^{a – 1}

Given,

f(x) = (2x² + 1)(3x + 2)

⇒ f(x) = 6x³ + 4x² + 3x + 2

we need to find f’(x), so differentiating both sides with respect to x –

∴

Using algebra of derivatives –

⇒ f’(x) =

Use the formula:

∴ f’(x) = 6(3x^{3 – 1}) + 4(2x^{2 – 1}) + 3(x^{1 – 1}) + 0

⇒ f’(x) = 18x² + 8x + 3 + 0

∴ f’(x) = 18x² + 8x + 3

Given,

f(x) = log₃x + 3log_ex + 2 tan x

we need to find f’(x), so differentiating both sides with respect to x –

∴

Using algebra of derivatives –

⇒ f’(x) =

Use the formula:

∴ f’(x) =

⇒ f’(x) =

∴ f’(x) =

Given,

f(x) =

⇒ f(x) =

⇒ f(x) = x^3/2 + x^1/2 + x ^{– 1/2 +} x ^{– 3/2}

we need to find f’(x), so differentiating both sides with respect to x –

∴

Using algebra of derivatives –

⇒ f’(x) =

Use the formula:

∴ f’(x) =

⇒ f’(x) =

⇒ f’(x) =

∴ f’(x) =

Given,

f(x) =

Using (a + b)³ = a³ + 3a²b + 3ab² + b³

⇒ f(x) =

⇒ f(x) = x^3/2 + x^1/2 + x ^{– 1/2 +} x ^{– 3/2}

we need to find f’(x), so differentiating both sides with respect to x –

∴

Using algebra of derivatives –

⇒ f’(x) =

Use the formula:

∴ f’(x) =

⇒ f’(x) =

⇒ f’(x) =

∴ f’(x) =

Given,

f(x) =

⇒ f(x) =

⇒ f(x) = 2x + 3 + 4x ^{– 1}

we need to find f’(x), so differentiating both sides with respect to x –

∴ 2x + 3 + 4x ^{– 1}

Using algebra of derivatives –

⇒ f’(x) =

Use the formula:

∴ f’(x) = 2 + 0 + 4( – 1)x ^{– 1 – 1}

⇒ f’(x) = 2 – 4x ^{– 2}

∴ f’(x) = 2 – 4x ^{– 2}

Given,

f(x) =

⇒ f(x) =

⇒ f(x) = x² – 2x + x ^{– 1} – 2x ^{– 2}

we need to find f’(x), so differentiating both sides with respect to x –

∴

Using algebra of derivatives –

⇒ f’(x) =

Use the formula:

∴ f’(x) = 2x^{2 – 1} + 2x^{1 – 1} + ( – 1)x ^{– 1 – 1} – 2( – 2)x ^{– 2 – 1}

⇒ f’(x) = 2x + 2x⁰ – 1x ^{– 2} + 4x ^{– 3}

∴ f’(x) = 2x + 2 – x ^{– 2} + 4x ^{– 3}

Given,

f(x) =

⇒ f(x) =

⇒ f(x) = a cot x + b + c cosec x

we need to find f’(x), so differentiating both sides with respect to x –

∴ a cot x + b + c cosec x)

Using algebra of derivatives –

⇒ f’(x) =

Use the formula:

∴ f’(x) = a( – cosec² x) + 0 + c( – cosec x cot x)

⇒ f’(x) = – a cosec² x – c cosec x cot x

∴ f’(x) = – a cosec² x – c cosec x cot x

Given,

f(x) = 2 sec x + 3 cot x – 4 tan x

we need to find f’(x), so differentiating both sides with respect to x –

∴ 2 sec x + 3 cot x – 4 tan x)

Using algebra of derivatives –

⇒ f’(x) =

Use the formula:

∴ f’(x) = 2(sec x tan x) + 3( – cosec² x) – 4(sec² x)

⇒ f’(x) = 2sec x tan x – 3cosec² x – 4sec² x

∴ f’(x) = 2sec x tan x – 3cosec² x – 4sec² x

Given,

f(x) = a₀ xⁿ + a₁ x^{n – 1} + a₂ x^{n – 2} + ……. + a_{n – 1} x + a_n

we need to find f’(x), so differentiating both sides with respect to x –

∴ (a₀ xⁿ + a₁ x^{n – 1} + a₂ x^{n – 2} + ……. + a_{n – 1} x + a_n)

Using algebra of derivatives –

⇒ f’(x) =

Use the formula:

∴ f’(x) = a₀ n x^{n – 1} + a₁ (n – 1) x^{n – 1 – 1} + a₂(n – 2) x^{n – 2 – 1} + ……. + a_{n – 1} + 0

⇒ f’(x) = a₀ n x^{n – 1} + a₁ (n – 1) x^{n – 2} + a₂(n – 2) x^{n – 3} + ……. + a_{n – 1}

∴ f’(x) = a₀ n x^{n – 1} + a₁ (n – 1) x^{n – 2} + a₂(n – 2) x^{n – 3} + ……. + a_{n – 1}

Given,

f(x) =

using change of base formula for log, we can write –

log_x 3 = (log_e 3)/(log_e x)

∴ f(x) =

⇒ f(x) =

we need to find f’(x), so differentiating both sides with respect to x –

∴ )

Using algebra of derivatives –

⇒ f’(x) =

Use the formula:

∴ f’(x) = 2( – cosec x cot x) + 8(2^x log_e2) – 4/(log_e3) (1/x)

⇒ f’(x) =

∴ f’(x) =

Given,

f(x) =

⇒ f(x) =

⇒ f(x) = 2x² + 10x – 1 – 5x ^{– 1}

we need to find f’(x), so differentiating both sides with respect to x –

∴

Using algebra of derivatives –

⇒ f’(x) =

Use the formula:

∴ f’(x) = 2(2x^{2 – 1}) + 10(1) – ( – 1)(0)– 5( – 1)x ^{– 1 – 1}

⇒ f’(x) = 4x + 10 + 0 + 5x ^{– 2}

∴ f’(x) = 4x + 10 + 5x ^{– 2}

Given,

f(x) =

⇒ f(x) =

⇒ f(x) = – 0.5 log x + 5x^a – 3a^x + x^2/3 + 6x ^{– 3/4}

we need to find f’(x), so differentiating both sides with respect to x –

∴

Using algebra of derivatives –

⇒ f’(x) =

Use the formula:

∴ f’(x) =

⇒ f’(x) =

∴ f’(x) =

Given,

f(x) = cos (x + a)

Using cos (A + B) = cos A cos B – sin A sin B, we get –

∴ f(x) = cos x cos a – sin x sin a

we need to find f’(x), so differentiating both sides with respect to x –

∴ )

Using algebra of derivatives –

⇒ f’(x) =

As cos a and sin a are constants, so using algebra of derivatives we have –

⇒ f’(x) =

Use the formula:

∴ f’(x) = – sin x cos a – sin a cos x

⇒ f’(x) = – (sin x cos a + sin a cos x)

Using sin (A + B) = sin A cos B + cos A sin B, we get –

∴ f’(x) = – sin (x + a)

Given,

f(x) =

Using cos (A + B) = cos A cos B – sin A sin B, we get –

∴ f(x) =

⇒ f(x) = cos 2 cot x – sin 2

we need to find f’(x), so differentiating both sides with respect to x –

∴ )

Using algebra of derivatives –

⇒ f’(x) =

As cos a and sin a are constants, so using algebra of derivatives we have –

⇒ f’(x) =

Use the formula:

∴ f’(x) = – cosec² x cos 2 – sin 2 (0)

⇒ f’(x) = – cosec² x cos 2 – 0

∴ f’(x) = – cosec² x cos 2

Given,

y =

Using (a + b)² = a² + 2ab + b²

y =

⇒ y = {∵ sin²A + cos²A = 1 & 2sin A cos A = sin 2A}

⇒ y = 1 + sin x

Now, differentiating both sides w.r.t x –

∴ sin x)

Using algebra of derivatives –

⇒

Use:

∴

Hence, dy/dx at x = π/6 is

Given,

y =

y =

⇒ y = 2 cosec x – 3 cot x

Now, differentiating both sides w.r.t x –

∴ )

Using algebra of derivatives –

⇒

Use:

∴

Hence, dy/dx at x = π/4 is

Given,

y = 2x⁶ + x⁴ – 1

We need to find slope of tangent of f(x) at x = 1.

Slope of the tangent is given by value of derivative at that point. So we need to find dy/dx first.

As, y = 2x⁶ + x⁴ – 1

Now, differentiating both sides w.r.t x –

∴ )

Using algebra of derivatives –

⇒

Use:

∴

⇒

As, slope of tangent at x = 1 will be given by the value of dy/dx at x = 1

∴

Given,

y =

We need to prove:

As, y = ….equation 1

Now, differentiating both sides w.r.t x –

∴ )

Using algebra of derivatives –

⇒

Use:

∴

⇒

⇒

Multiplying x both sides –

⇒

⇒

Now, multiplying y both sides –

⇒

⇒ {from equation 1}

Using (a + b)(a – b) = a² – b²

⇒

⇒

Given,

y = x⁴ – 2x³ + 3x² + x + 5

We need to rate of change of f(x) w.r.t x.

Rate of change of a function w.r.t a given variable is obtained by differentiating the function w.r.t that variable only.

So in this case we will be finding dy/dx

As, y = x⁴ – 2x³ + 3x² + x + 5

Now, differentiating both sides w.r.t x –

∴ )

Using algebra of derivatives –

⇒

Use:

∴

⇒

∴ Rate of change of y w.r.t x is given by –

Given,

y =

We need to find dy/dx at x = 1

As, y =

Now, differentiating both sides w.r.t x –

∴ )

Using algebra of derivatives –

⇒

Use:

∴

⇒

∴

Given,

y = λ x² + μx + 12

Now, differentiating both sides w.r.t x –

∴ )

Using algebra of derivatives –

⇒

Use:

∴

Now, we have –

f’(x) = 2λx + μ

Given,

f’(4) = 15

⇒ 2λ (4) + μ = 15

⇒ 8λ + μ = 15 ……equation 1

Also f’(2) = 11

⇒ 2λ(2) + μ = 11

⇒ 4λ + μ = 11 …..equation 2

Subtracting equation 2 from equation 1, we have –

4λ = 15 – 11 = 4

∴ λ = 1

Putting λ = 1 in equation 2

4 + μ = 11

∴ μ = 7

Hence,

λ = 1 & μ = 7

Given,

Now, differentiating both sides w.r.t x –

∴ )

Using algebra of derivatives –

⇒

Use:

∴

⇒ f’(x) = x^{99 +} x⁹⁸ + ….. + x + 1

∴ f’(1) = 1⁹⁹ + 1⁹⁸ + … + 1 + 1 (sum of total 100 ones) = 100

∴ f’(1) = 100

As, f’(0) = 0 + 0 + ….. + 0 + 1 = 1

∴ we can write as

f’(1) = 100×1 = 100× f’(0)

Hence,

f’(1) = 100 f’(0) ….proved

Let, y = x³ sin x

We have to find dy/dx

As we can observe that y is a product of two functions say u and v where,

u = x³ and v = sin x

∴ y = uv

As we know that to find the derivative of product of two function we apply product rule of differentiation.

By product rule, we have –

…equation 1

As, u = x³

∴ …equation 2 {∵ }

As, v = sin x

…equation 3 {∵ }

∴ from equation 1, we can find dy/dx

∴

⇒ {using equation 2 & 3}

Hence,

Let, y = x³ e^x

We have to find dy/dx

As we can observe that y is a product of two functions say u and v where,

u = x³ and v = e^x

∴ y = uv

As we know that to find the derivative of product of two function we apply product rule of differentiation.

By product rule, we have –

…equation 1

As, u = x³

∴ …equation 2 {∵ }

As, v = e^x

…equation 3 {∵ }

∴ from equation 1, we can find dy/dx

∴

⇒ {using equation 2 & 3}

Hence,

Let, y = x² e^x log x

We have to find dy/dx

As we can observe that y is a product of three functions say u, v & w where,

u = x²

v = e^x

w = log x

∴ y = uvw

As we know that to find the derivative of product of three function we apply product rule of differentiation.

By product rule, we have –

…equation 1

As, u = x²

∴ …equation 2 {∵ }

As, v = e^x

…equation 3 {∵ }

As, w = log x

∴ …equation 4 {∵ }

∴ from equation 1, we can find dy/dx

∴

⇒ {using equation 2, 3 & 4}

Hence,

Let, y = xⁿ tan x

We have to find dy/dx

As we can observe that y is a product of two functions say u and v where,

u = xⁿ and v = tan x

∴ y = uv

As we know that to find the derivative of product of two function we apply product rule of differentiation.

By product rule, we have –

…equation 1

As, u = xⁿ

∴ …equation 2 {∵ }

As, v = tan x

…equation 3 {∵ }

∴ from equation 1, we can find dy/dx

∴

⇒ {using equation 2 & 3}

Hence,

Let, y = xⁿ log _a x

We have to find dy/dx

As we can observe that y is a product of two functions say u and v where,

u = xⁿ and v = log _a x

∴ y = uv

As we know that to find the derivative of product of two function we apply product rule of differentiation.

By product rule, we have –

…equation 1

As, u = xⁿ

∴ …equation 2 {∵ }

As, v = log_a x

…equation 3 {∵ }

∴ from equation 1, we can find dy/dx

∴

⇒ {using equation 2 & 3}

Hence,

Let, y = (x³ + x² + 1) sin x

We have to find dy/dx

As we can observe that y is a product of two functions say u and v where,

u = x³ + x² + 1 and v = sin x

∴ y = uv

As we know that to find the derivative of product of two function we apply product rule of differentiation.

By product rule, we have –

…equation 1

As, u = x³ + x² + 1

∴

⇒ …equation 2 {∵ }

As, v = sin x

…equation 3 {∵ }

∴ from equation 1, we can find dy/dx

∴

⇒ {using equation 2 & 3}

Hence,

Let, y = cos x sin x

We have to find dy/dx

As we can observe that y is a product of two functions say u and v where,

u = cos x and v = sin x

∴ y = uv

As we know that to find the derivative of product of two function we apply product rule of differentiation.

By product rule, we have –

…equation 1

As, u = cos x

∴ …equation 2 {∵ }

As, v = sin x

…equation 3 {∵ }

∴ from equation 1, we can find dy/dx

∴

⇒ {using equation 2 & 3}

⇒

Hence,

Let, y =

We have to find dy/dx

As we can observe that y is a product of three functions say u, v & w where,

u = x ^{– 1/2}

v = 2^x

w = cot x

∴ y = uvw

As we know that to find the derivative of product of three function we apply product rule of differentiation.

By product rule, we have –

…equation 1

As, u = x ^{– 1/2}

∴ …equation 2 {∵ }

As, v = 2^x

…equation 3 {∵ }

As, w = cot x

∴ …equation 4 {∵ }

∴ from equation 1, we can find dy/dx

∴

using equation 2, 3 & 4,we have –

⇒

Hence,

Let, y = x² sin x log x

We have to find dy/dx

As we can observe that y is a product of three functions say u, v & w where,

u = x²

v = sin x

w = log x

∴ y = uvw

As we know that to find the derivative of product of three function we apply product rule of differentiation.

By product rule, we have –

…equation 1

As, u = x²

∴ …equation 2 {∵ }

As, v = sin x

…equation 3 {∵ }

As, w = log x

∴ …equation 4 {∵ }

∴ from equation 1, we can find dy/dx

∴

using equation 2, 3 & 4, we have –

⇒

Hence,

Let, y = x⁵ e^x + x⁶ log x

Let, A = x⁵ e^x and B = x⁶ log x

∴ y = A + B

⇒

We have to find dA/dx first

As we can observe that A is a product of two functions say u and v where,

u = x⁵ and v = e^x

∴ A = uv

As we know that to find the derivative of product of two function we apply product rule of differentiation.

By product rule, we have –

…equation 1

As, u = x⁵

∴ …equation 2 {∵ }

As, v = e^x

…equation 3 {∵ }

∴ from equation 1, we can find dy/dx

∴

⇒ {using equation 2 & 3}

Hence,

…..equation 4

Now, we will find dB/dx first

As we can observe that A is a product of two functions say m and n where,

m = x⁶ and n = log x

∴ B = mn

As we know that to find the derivative of product of two function we apply product rule of differentiation.

By product rule, we have –

…equation 5

As, m = x⁶

∴ …equation 6 {∵ }

As, v = log x

…equation 7 {∵ }

∴ from equation 5, we can find dy/dx

∴

⇒ {using equation 6 & 7}

Hence,

…..equation 8

As,

∴ from equation 4 and 8, we have –

Let, y = (x sin x + cos x)(x cos x – sin x)

We have to find dy/dx

As we can observe that y is a product of two functions say u and v where,

u = x sin x + cos x and v = x cos x – sin x

∴ y = uv

As we know that to find the derivative of product of two function we apply product rule of differentiation.

By product rule, we have –

…equation 1

As, u = x sin x + cos x

∴

Using algebra of derivatives –

⇒

∵

∴ {using product rule}

⇒ …equation 2

As, v = x cos x – sin x

Using algebra of derivatives –

⇒

∵

∴ {using product rule}

⇒ …equation 3

∴ from equation 1, we can find dy/dx

∴

using equation 2 & 3, we get –

⇒

⇒

As, we know that: cos²x – sin²x = cos 2x & 2sin x cos x = sin 2x

Hence,

Let, y = (x sin x + cos x)( e^x + x² log x)

We have to find dy/dx

As we can observe that y is a product of two functions say u and v where,

u = x sin x + cos x and v = (e^x + x² log x)

∴ y = uv

As we know that to find the derivative of product of two function we apply product rule of differentiation.

By product rule, we have –

…equation 1

As, u = x sin x + cos x

∴

Using algebra of derivatives –

⇒

∵

∴ {using product rule}

⇒ …equation 2

As, v = e^x + x² log x

Using algebra of derivatives –

⇒

∵

∴ {using product rule}

⇒ …equation 3

∴ from equation 1, we can find dy/dx

∴

using equation 2 & 3, we get –

⇒

⇒

Hence,

Let, y = (1 – 2 tan x)(5 + 4 sin x)

We have to find dy/dx

As we can observe that y is a product of two functions say u and v where,

u = (1 – 2tan x) and v = (5 + 4sin x)

∴ y = uv

As we know that to find the derivative of product of two function we apply product rule of differentiation.

By product rule, we have –

…equation 1

As, u = (1 – 2tan x)

∴

⇒

⇒ …..equation 2 {∵ }

As, v = 5 + 4sin x

⇒

⇒ …equation 3 {∵ }

∴ from equation 1, we can find dy/dx

∴

using equation 2 & 3, we get –

⇒

⇒

∵ sin x = tan x cos x , so we get –

⇒

Hence,

Let, y = (x² + 1) cos x

We have to find dy/dx

As we can observe that y is a product of two functions say u and v where,

u = x² + 1 and v = cos x

∴ y = uv

As we know that to find the derivative of product of two function we apply product rule of differentiation.

By product rule, we have –

…equation 1

As, u = x² + 1

∴

⇒ …equation 2 {∵ }

As, v = cos x

…equation 3 {∵ }

∴ from equation 1, we can find dy/dx

∴

⇒ {using equation 2 & 3}

Hence,

Let, y = sin² x = sin x sin x

We have to find dy/dx

As we can observe that y is a product of two functions say u and v where,

u = sin x and v = sin x

∴ y = uv

As we know that to find the derivative of product of two function we apply product rule of differentiation.

By product rule, we have –

…equation 1

As, u = sin x

∴ …equation 2 {∵ }

As, v = sin x

…equation 3 {∵ }

∴ from equation 1, we can find dy/dx

∴

⇒ {using equation 2 & 3}

⇒

Hence,

Let, y =

Using change of base formula for logarithm, we can write y as –

As, y = 1/2

We know that

∴

Let, y = e^x log √x tan x = e^x log x^1/2 tan x = 1/2 e^x log x tan x

We have to find dy/dx

As we can observe that y is a product of three functions say u, v & w where,

u = log x

v = e^x

w = tan x

∴ y = 1/2 uvw

As we know that to find the derivative of product of three function we apply product rule of differentiation.

By product rule, we have –

…equation 1

As, u = log x

∴ …equation 2 {∵ }

As, v = e^x

…equation 3 {∵ }

As, w = tan x

∴ …equation 4 {∵ }

∴ from equation 1, we can find dy/dx

∴

using equation 2, 3 & 4,we have –

⇒

Hence,

Let, y = x³ e^x cos x

We have to find dy/dx

As we can observe that y is a product of three functions say u, v & w where,

u = x³

v = cos x

w = e^x

∴ y = uvw

As we know that to find the derivative of product of three function we apply product rule of differentiation.

By product rule, we have –

…equation 1

As, u = x³

∴ …equation 2 {∵ }

As, v = cos x

…equation 3 {∵ }

As, w = e^x

∴ …equation 4 {∵ }

∴ from equation 1, we can find dy/dx

∴

using equation 2, 3 & 4, we have –

⇒

Hence,

Let, y = {∵ cos π/4 = 1/√2 }

We have to find dy/dx

As we can observe that y is a product of two functions say u and v where,

u = x² and v = cosec x

∴ y = (1/√2) uv

As we know that to find the derivative of product of two function we apply product rule of differentiation.

By product rule, we have –

…equation 1

As, u = x²

∴ …equation 2 {∵ }

As, v = cosec x

⇒ …equation 3 {∵ }

∴ from equation 1, we can find dy/dx

∴

⇒ {using equation 2 & 3}

Hence,

Let, y = x⁴ (5sin x – 3cos x)

We have to find dy/dx

As we can observe that y is a product of two functions say u and v where,

u = x⁴ and v = 5sin x – 3cos x

∴ y = uv

As we know that to find the derivative of product of two function we apply product rule of differentiation.

By product rule, we have –

…equation 1

As, u = x⁴

∴ …equation 2 {∵ }

As, v = 5sin x – 3cos x

Using:

⇒ …equation 3

∴ from equation 1, we can find dy/dx

∴

⇒ {using equation 2 & 3}

Hence,

Let, y = (2x² – 3) sin x

We have to find dy/dx

As we can observe that y is a product of two functions say u and v where,

u = 2x² – 3 and v = sin x

∴ y = uv

As we know that to find the derivative of product of two function we apply product rule of differentiation.

By product rule, we have –

…equation 1

As, u = 2x² – 3

∴ …equation 2 {∵ }

As, v = sin x

…equation 3 {∵ }

∴ from equation 1, we can find dy/dx

∴

⇒ {using equation 2 & 3}

Hence,

Let, y = x⁵ (3 – 6x ^{– 9})

We have to find dy/dx

As we can observe that y is a product of two functions say u and v where,

u = x⁵ and v = 3 – 6x ^{– 9}

∴ y = uv

As we know that to find the derivative of product of two function we apply product rule of differentiation.

By product rule, we have –

…equation 1

As, u = x⁵

∴ …equation 2 {∵ }

As, v = 3 – 6x ^{– 9}

…equation 3 {∵ }

∴ from equation 1, we can find dy/dx

∴

⇒ {using equation 2 & 3}

⇒

Hence,

Let, y = x ^{– 4} (3 – 4x ^{– 5})

We have to find dy/dx

As we can observe that y is a product of two functions say u and v where,

u = x ^{– 4} and v = 3 – 4x ^{– 5}

∴ y = uv

As we know that to find the derivative of product of two function we apply product rule of differentiation.

By product rule, we have –

…equation 1

As, u = x ^{– 4}

∴ …equation 2 {∵ }

As, v = 3 – 4x ^{– 5}

…equation 3 {∵ }

∴ from equation 1, we can find dy/dx

∴

⇒ {using equation 2 & 3}

⇒

Hence,

Let, y = x ^{– 3} (5 + 3x)

We have to find dy/dx

As we can observe that y is a product of two functions say u and v where,

u = x ^{– 3} and v = (5 + 3x)

∴ y = uv

As we know that to find the derivative of product of two function we apply product rule of differentiation.

By product rule, we have –

…equation 1

As, u = x ^{– 3}

∴ …equation 2 {∵ }

As, v = 5 + 3x

…equation 3 {∵ }

∴ from equation 1, we can find dy/dx

∴

⇒ {using equation 2 & 3}

⇒

Hence,

Let, y = (ax + b)/(cx + d)

We have to find dy/dx

As we can observe that y is a product of two functions say u and v where,

u = ax + b and v = cx + d

∴ y = uv

As we know that to find the derivative of product of two function we apply product rule of differentiation.

By product rule, we have –

…equation 1

As, u = ax + b

∴ …equation 2 {∵ }

As, v =

As,

…equation 3

∴ from equation 1, we can find dy/dx

∴

⇒ {using equation 2 & 3}

⇒

Hence,

Let, y = (ax + b)ⁿ(cx + d)^m

We have to find dy/dx

As we can observe that y is a product of two functions say u and v where,

u = (ax + b)ⁿ and v = (cx + d)^m

∴ y = uv

As we know that to find the derivative of product of two function we apply product rule of differentiation.

By product rule, we have –

…equation 1

As, u = (ax + b)ⁿ

As,

∴ …equation 2

As, v = (cx + d)^m

As,

…equation 3

∴ from equation 1, we can find dy/dx

∴

{using equation 2 & 3}

⇒

⇒

Hence,

Let, y = (1 + 2 tan x)(5 + 4 cos x)

⇒ y = 5 + 4 cos x + 10 tan x + 8 tan x cos x

⇒ y = 5 + 4 cos x + 10 tan x + 8 sin x {∵ tan x cos x = sin x}

Differentiating y w.r.t x –

Using algebra of derivatives, we have –

Use formula of derivative of above function to get the result.

⇒

∴ …equation 1

Derivative using product rule –

We have to find dy/dx

As we can observe that y is a product of two functions say u and v where,

u = (1 + 2tan x) and v = (5 + 4cos x)

∴ y = uv

As we know that to find the derivative of product of two function we apply product rule of differentiation.

By product rule, we have –

…equation 2

As, u = (1 + 2tan x)

∴

⇒

⇒ …..equation 3 {∵ }

As, v = 5 + 4cos x

⇒

⇒ …equation 4 {∵ }

∴ from equation 2, we can find dy/dx

∴

using equation 3 & 4, we get –

⇒

⇒

∵ sin x = tan x cos x , so we get –

⇒

⇒

⇒ [∵ 1 – sin² x = cos² x ]

∴

Hence,

….equation 5

Clearly from equation 1 and 5 we observed that both equations gave identical results.

Hence, Results are verified

Let, y = (3x² + 2)² = (3x² + 2)(3x² + 2)

⇒ y = 9x⁴ + 6x² + 6x² + 4

⇒ y = 9x⁴ + 12x² + 4

Differentiating y w.r.t x –

Using algebra of derivatives, we have –

Use formula of derivative of above function to get the result.

⇒ {∵

∴ …equation 1

Derivative using product rule –

We have to find dy/dx

As we can observe that y is a product of two functions say u and v where,

u = (3x² + 2) and v = (3x² + 2)

∴ y = uv

As we know that to find the derivative of product of two function we apply product rule of differentiation.

By product rule, we have –

…equation 2

As, u = (3x² + 2)

∴

⇒

⇒ …..equation 3 {∵ }

As, v = (3x² + 2)

⇒

⇒ …equation 4 {∵ }

∴ from equation 2, we can find dy/dx

∴

using equation 3 & 4, we get –

⇒

⇒

Hence,

….equation 5

Clearly from equation 1 and 5 we observed that both equations gave identical results.

Hence, Results are verified

Let, y = (x + 2)(x + 3)

⇒ y = x² + 5x + 6

Differentiating y w.r.t x –

Using algebra of derivatives, we have –

Use formula of derivative of above function to get the result.

⇒ {∵

∴ …equation 1

Derivative using product rule –

We have to find dy/dx

As we can observe that y is a product of two functions say u and v where,

u = (x + 2) and v = (x + 3)

∴ y = uv

As we know that to find the derivative of product of two function we apply product rule of differentiation.

By product rule, we have –

…equation 2

As, u = (x + 2)

∴

⇒

⇒ …..equation 3 {∵ }

As, v = (x + 3)

⇒

⇒ …equation 4 {∵ }

∴ from equation 2, we can find dy/dx

∴

using equation 3 & 4, we get –

⇒

⇒

Hence,

….equation 5

Clearly from equation 1 and 5 we observed that both equations gave identical results.

Hence, Results are verified

Let, y = (3 sec x – 4 cosec x)( – 2 sin x + 5 cos x)

⇒ y = – 6 sec x sin x + 15 sec x cos x + 8 sin x cosec x – 20cosec x cos x

⇒ y = – 6tan x + 15 + 8 – 20 cot x {∵ tan x cos x = sin x}

⇒ y = – 6 tan x – 20 cot x + 23

Differentiating y w.r.t x –

Using algebra of derivatives, we have –

Use formula of derivative of above function to get the result.

⇒

∴ …equation 1

Derivative using product rule –

We have to find dy/dx

As we can observe that y is a product of two functions say u and v where,

u = (3 sec x – 4 cosec x) and v = ( – 2 sin x + 5 cos x)

∴ y = uv

As we know that to find the derivative of product of two function we apply product rule of differentiation.

By product rule, we have –

…equation 2

As, u = (3 sec x – 4 cosec x)

∴

Use the formula: &

⇒

⇒ …..equation 3

As, v = – 2 sin x + 5 cos x

⇒

⇒ …equation 4 {∵ }

∴ from equation 2, we can find dy/dx

∴

using equation 3 & 4, we get –

⇒

∵ sin x = tan x cos x , so we get –

⇒

⇒

⇒ = 20(1 + cot² x) – 6(1 + tan² x)

∴ [∵ 1 + tan² x = sec² x & 1 + cot² x = cosec² x ]

∴

Hence,

….equation 5

Clearly from equation 1 and 5 we observed that both equations gave identical results.

Hence, Results are verified

Let, y =

We have to find dy/dx

As we can observe that y is a fraction of two functions say u and v where,

u = x² + 1 and v = x + 1

∴ y = u/v

As we know that to find the derivative of fraction of two function we apply quotient rule of differentiation.

By quotient rule, we have –

…equation 1

As, u = x² + 1

∴ …equation 2 {∵ }

As, v = x + 1

…equation 3 {∵ }

∴ from equation 1, we can find dy/dx

∴

⇒ {using equation 2 and 3}

⇒

⇒

Hence,

Let, y =

We have to find dy/dx

As we can observe that y is a fraction of two functions say u and v where,

u = 2x – 1 and v = x² + 1

∴ y = u/v

As we know that to find the derivative of fraction of two function we apply quotient rule of differentiation.

By quotient rule, we have –

…equation 1

As, u = 2x – 1

∴ …equation 2 {∵ }

As, v = x² + 1

…equation 3 {∵ }

∴ from equation 1, we can find dy/dx

∴

⇒ {using equation 2 and 3}

⇒

⇒

Hence,

Let, y =

We have to find dy/dx

As we can observe that y is a fraction of two functions say u and v where,

u = x + e^x and v = 1 + log x

∴ y = u/v

As we know that to find the derivative of fraction of two function we apply quotient rule of differentiation.

By quotient rule, we have –

…equation 1

As, u = x + e^x

∴

∵ , so we get –

⇒ …equation 2

As, v = 1 + log x

⇒ …equation 3 {∵ }

∴ from equation 1, we can find dy/dx

∴

⇒ {using equation 2 and 3}

⇒

⇒

Hence,

Let, y =

We have to find dy/dx

As we can observe that y is a fraction of two functions say u and v where,

u = e^x – tan x and v = cot x – xⁿ

∴ y = u/v

As we know that to find the derivative of fraction of two function we apply quotient rule of differentiation.

By quotient rule, we have –

…equation 1

As, u = e^x – tan x

∴

∵ , so we get –

⇒ …equation 2

As, v = cot x – xⁿ

∵ , so we get –

⇒ …equation 3

∴ from equation 1, we can find dy/dx

∴

using equation 2 and 3, we get –

⇒

⇒

Hence,

Let, y =

We have to find dy/dx

As we can observe that y is a fraction of two functions say u and v where,

u = ax² + bx + c and v = px² + qx + r

∴ y = u/v

As we know that to find the derivative of fraction of two function we apply quotient rule of differentiation.

By quotient rule, we have –

…equation 1

As, u = ax² + bx + c

∴ …equation 2 {∵ }

As, v = px² + qx + r

∵ , so we get –

…equation 3

∴ from equation 1, we can find dy/dx

∴

⇒ {using equation 2 and 3}

⇒

⇒

⇒

Hence,

Let, y =

We have to find dy/dx

As we can observe that y is a fraction of two functions say u and v where,