Differentiation

(i) x^-3

Formula:-

= n

Differentiating w.r.t x,

= -3

= -3

(ii) =

Formula:-

= n

Differentiating w.r.t x,

=

=

(i) =

Formula:-

= n

Differentiating w.r.t x,

= -1

= -

(ii) =

Formula:-

= n

Differentiating w.r.t x,

=

=

(iii) =

Formula:-

= n

Differentiating w.r.t x,

=

=

(i) 3x^-5

Formula:-

= n

Differentiating with respect to x,

= 3(-5)

= -15

(ii) 1/5x =

Formula:-

= n

Differentiating with respect to x,

=

= -

(iii) =

Formula:-

= n

Differentiating with respect to x,

=

=

(i) 6x⁵ + 4x³ – 3x² + 2x – 7

Formula:-

= n

Differentiating with respect to x,

6x⁵ + 4x³ – 3x² + 2x – 730x^5-1 + 12x^3-1 – 6x^2-1 + 2x^1-1 + 0

= 30x⁴ + 12x² – 6x¹ + 2x

(ii)

Formula:-

= n

Differentiating with respect to x,

=

=

(iii) ax³ + bx² + cx + d, where a, b, c, d are constants

Formula:-

= n

Differentiating with respect to x,

ax³ + bx² + cx + d) = 3ax^3-1 + 2bx^2-1 + c + d×0

= 3ax² + 2bx + c

(i)

=

Formulae:

= n

- cosec²x

a^x = log_n(a)×a^x

Differentiating with respect to x,

= 4.3x^3-1 + 3.log_n(2).2^x + 6×- + 5 ×- cosec²x

= 12x² + 3.log_n(2).2^x -3 -5 cosec²x

(ii)

=

= n

a^x = log_n(a)×a^x

Differentiating with respect to x,

=

=

Formulae: -

- cosec²x

- sinx

secx tanx

- cosecx cotx

sec²x

cosx

0,k is constant

(i)

=

Differentiating with respect to x,

=

=

(ii) -5 tan x + 4 tan x cos x – 3 cot x sec x + 2sec x – 13

= -5 tan x + 4 sinx – 3 cosecx+ 2sec x – 13

Differentiating with respect to x,

-5 tan x + 4 sinx – 3 cosecx+ 2sec x – 13

= -5 sec²x + 4cosx -3(- cosecx cotx) + 2 secx tanx – 0

= -5 sec²x + 4cosx + 3 cosecx cotx + 2 secx tanx

Formula:

Chain rule -

Where u and v are the functions of x.

(i) (2x + 3) (3x – 5)

Applying, Chain rule

Here, u = 2x + 3

V = 3x -5

(2x + 3) (3x – 5)

= (2x + 3)(3x^1-1+0) + (3x – 5)(2x^1-1+0)

= 6x + 9 + 6x -10

= 12x -1

(ii) x(1 + x)³

Applying, Chain rule

Here, u = x

V = (1 + x)³

x(1 + x)³

= x×3×(1 + x)² + (1 + x)³(1)

= (1 + x)²(3x+x+1)

= (1 + x)²(4x+1)

(iii) = (x^1/2 + x^-1)(x – x^-1/2 )

Applying, Chain rule

Here, u = (x^1/2 + x^-1)

V = (x – x^-1/2 )

(x^1/2 + x^-1)(x – x^-1/2 )

= (x^1/2 + x^-1)(x – x^-1/2 ) + (x – x^-1/2 )(x^1/2 + x^-1)

= (x^1/2 + x^-1)(1+ x^-3/2) + (x – x^-1/2 )(x^-1/2 – x^-2)

= x^1/2 + x^-1 + x^-1 + x^-5/2 + x^1/2 – x^-1 - x^-1 + x^-5/2

= x^1/2 + x^-5/2

(iv)

Differentiation of composite function can be done by

Here, f(g) = g² , g(x) =

= 2g×(1 + )

= 2( (1 + )

= 2(x + - + )

= 2(x + )

(v)

Differentiation of composite function can be done by

Here, f(g) = g³ , g(x) = x² -

= 3g²×(2x - )

= 3 (2x - )

= 3(2x³ - - + )

= 3(2x³ - + )

(vi) (2x² + 5x – 1) (x – 3)

Applying, Chain rule

Here, u = (2x² + 5x – 1)

V = (x – 3)

(2x² + 5x – 1) (x – 3)

(2x² + 5x – 1)(2x² + 5x – 1)

= (2x² + 5x – 1)×1 + (x – 3)(4x + 5)

= 2x² + 5x – 1 + 4x² -7x -15

= 6x² -2x -16

Formula:

(i)

Applying, quotient rule

=

=

=

(ii)

Applying, quotient rule

=

=

=

=

=

(iii)

Applying, quotient rule

=

=

=

=

(iv)

Applying, quotient rule

=

=

=

=

=

=

(v)

Applying, quotient rule

=

=

(vi)

Applying, quotient rule

=

=

=

Formulae:

= n

- cosec²x

- cosecx cotx

sec²x

cosx

(i) If y = 6x⁵ – 4x⁴ – 2x² + 5x – 9, find at x = -1.

Differentiating with respect to x,

6x⁵ - 4x⁴ – 2x² + 5x – 9

30x⁴ -16x³ – 4x + 5

(_{x= -1} = 30(-1)⁴ -16(-1)³ – 4(-1) + 5

= 30+16+4+5

= 55

(ii) If y = (sin x + tan x), find at .

Differentiating with respect to x,

sinx + tanx) = cos x + sec² x

Substituting

_{x = π/3} = cos + sec²

= + 4

=

(iii) If , find at .

Differentiating with respect to x,

2cosec x-3cot x) = 2(- cosecx cotx) – 3(- cosec²x)

Substituting

_{x = π/4} = 2(- cosec cot) – 3(- cosec²)

= - 2× + 3×2

= 6 - 2×

To show:

Differentiating with respect to x

= =

Now,

LHS =

LHS = )

LHS = +

LHS = 2

∴ LHS = RHS

To prove:.

Differentiating y with respect to x

=

Now,

LHS =

LHS =

LHS = ()()

LHS =

∴ LHS = RHS

Formula:

Using double angle formula:

2cos²x – 1

= 1 – 2 sin² x

∴1 + cos 2 x = 2cos²x

1 - cos 2 x = 2sin²x

∴

=

= cotx

Differentiating y with respect to x

=

- cosec² x

Formula:

Using Half angle formula,

cos x =

∴y = cos x

Differentiating y with respect to x

Let f(x) = ax + b

We need to find the derivative of f(x) i.e. f’(x)

We know that,

…(i)

f(x) = ax + b

f(x + h) = a(x + h) + b

= ax + ah + b

Putting values in (i), we get

f’(x) = a

Hence, f’(x) = a

We need to find the derivative of f(x) i.e. f’(x)

We know that,

…(i)

Putting values in (i), we get

Taking ‘h’ common from both the numerator and denominator, we get

Putting h = 0, we get

Hence,

Let f(x) = 3x² + 2x – 5

We need to find the derivative of f(x) i.e. f’(x)

We know that,

…(i)

f(x) = 3x² + 2x – 5

f(x + h) = 3(x + h)² + 2(x + h) – 5

= 3(x² + h² + 2xh) + 2x + 2h – 5

[∵(a + b)² = a² + b² + 2ab]

= 3x² + 3h² + 6xh + 2x + 2h – 5

Putting values in (i), we get

Putting h = 0, we get

f’(x) = 3(0) + 6x + 2

= 6x + 2

Hence, f’(x) = 6x + 2

Let f(x) = x³ – 2x² + x + 3

We need to find the derivative of f(x) i.e. f’(x)

We know that,

…(i)

f(x) = x³ – 2x² + x + 3

f(x + h) = (x + h)³ – 2(x + h)² + (x + h) + 3

Putting values in (i), we get

Using the identities:

(a + b)³ = a³ + b³ + 3ab² + 3a²b

(a + b)² = a² + b² + 2ab

Putting h = 0, we get

f’(x) = (0)² + 2x(0) + 3x² – 2(0) – 4x + 1

= 3x² – 4x + 1

Hence, f’(x) = 3x² – 4x + 1

Let f(x) = x⁸

We need to find the derivative of f(x) i.e. f’(x)

We know that,

…(i)

f(x) = x⁸

f(x + h) = (x + h)⁸

Putting values in (i), we get

[Add and subtract x in denominator]

= 8x^8-1

= 8x⁷

Hence, f’(x) = 8x⁷

Let

We need to find the derivative of f(x) i.e. f’(x)

We know that,

…(i)

Putting values in (i), we get

[Add and subtract x in denominator]

= (-3)x^-3-1

= -3x^-4

Hence,

Let

We need to find the derivative of f(x) i.e. f’(x)

We know that,

…(i)

Putting values in (i), we get

[Add and subtract x in denominator]

= (-5)x^-5-1

= -5x^-6

Hence,

Let

We need to find the derivative of f(x) i.e. f’(x)

We know that,

…(i)

Putting values in (i), we get

Now rationalizing the numerator by multiplying and divide by the conjugate of

Using the formula:

(a + b)(a – b) = (a² – b²)

Putting h = 0, we get

Hence,

Let

We need to find the derivative of f(x) i.e. f’(x)

We know that,

…(i)

Putting values in (i), we get

Now rationalizing the numerator by multiplying and divide by the conjugate of

Using the formula:

(a + b)(a – b) = (a² – b²)

Putting h = 0, we get

Hence,

Let

We need to find the derivative of f(x) i.e. f’(x)

We know that,

…(i)

Putting values in (i), we get

Now rationalizing the numerator by multiplying and divide by the conjugate of

Using the formula:

(a + b)(a – b) = (a² – b²)

Putting h = 0, we get

Hence,

Let

We need to find the derivative of f(x) i.e. f’(x)

We know that,

…(i)

Putting values in (i), we get

Now rationalizing the numerator by multiplying and divide by the conjugate of

Using the formula:

(a + b)(a – b) = (a² – b²)

Putting h = 0, we get

Hence,

Let

We need to find the derivative of f(x) i.e. f’(x)

We know that,

…(i)

Putting values in (i), we get

Now rationalizing the numerator by multiplying and divide by the conjugate of

Using the formula:

(a + b)(a – b) = (a² – b²)

Putting h = 0, we get

Hence,

Let

We need to find the derivative of f(x) i.e. f’(x)

We know that,

…(i)

Putting values in (i), we get

Now rationalizing the numerator by multiplying and divide by the conjugate of

Using the formula:

(a + b)(a – b) = (a² – b²)

Putting h = 0, we get

Hence,

Let

We need to find the derivative of f(x) i.e. f’(x)

We know that,

…(i)

Putting values in (i), we get

Putting h = 0, we get

Hence,

Let

We need to find the derivative of f(x) i.e. f’(x)

We know that,

…(i)

Putting values in (i), we get

Putting h = 0, we get

Hence,

Let

We need to find the derivative of f(x) i.e. f’(x)

We know that,

…(i)

Putting values in (i), we get

Putting h = 0, we get

Hence,

Let

We need to find the derivative of f(x) i.e. f’(x)

We know that,

…(i)

Putting values in (i), we get

Now rationalizing the numerator by multiplying and divide by the conjugate of

Using the formula:

(a + b)(a – b) = (a² – b²)

Using the formula:

Putting h = 0, we get

Hence,

Let

We need to find the derivative of f(x) i.e. f’(x)

We know that,

…(i)

Putting values in (i), we get

Now rationalizing the numerator by multiplying and divide by the conjugate of

Using the formula:

(a + b)(a – b) = (a² – b²)

Using the formula:

Putting h = 0, we get

Hence,

Let f(x) = tan²x

We need to find the derivative of f(x) i.e. f’(x)

We know that,

…(i)

f(x) = tan²x

f(x + h) = tan²(x + h)

Putting values in (i), we get

Using:

[∵ sin A cos B – sin B cos A = sin(A – B)

& sin A cos B + sin B cos A = sin(A + B)]

Putting h = 0, we get

[∵ sin2x = 2sinxcosx]

= 2tanx sec²x

Hence, f’(x) = 2tanx sec²x

Let f(x) = sin (2x + 3)

We need to find the derivative of f(x) i.e. f’(x)

We know that,

…(i)

f(x) = sin (2x + 3)

f(x + h) = sin [2(x + h) + 3]

Putting values in (i), we get

Using the formula:

Putting h = 0, we get

= 2cos(2x + 0 + 3)

= 2cos(2x + 3)

Hence, f’(x) = 2cos (2x + 3)

Let f(x) = tan (3x + 1)

We need to find the derivative of f(x) i.e. f’(x)

We know that,

…(i)

f(x) = tan (3x + 1)

f(x + h) = tan [3(x + h) + 1]

Putting values in (i), we get

Using the formula:

Putting h = 0, we get

= 3sec²(3x+ 1)

Hence, f’(x) = 3sec²(3x+ 1)

To find: Differentiation of x² sin x

Formula used: (i) (uv)′ = u′v + uv′ (Leibnitz or product rule)

(ii)

(iii)

Let us take u = x² and v = sinx

Putting the above obtained values in the formula:-

(uv)′ = u′v + uv′

(x² sin x)’ = 2x × sinx + x² × cosx

= 2xsinx + x²cosx

Ans) 2xsinx + x²cosx

To find: Differentiation of e^x cos x

Formula used: (i) (uv)′ = u′v + uv′ (Leibnitz or product rule)

(ii)

(iii)

Let us take u = e^x and v = cosx

Putting the above obtained values in the formula:-

(uv)′ = u′v + uv′

(e^x cosx)’ = e^x× cosx + e^x× -sinx

= e^xcosx - e^xsinx

= e^x (cosx - sinx)

Ans) e^x (cosx - sinx)

To find: Differentiation of e^x cot x

Formula used: (i) (uv)′ = u′v + uv′ (Leibnitz or product rule)

(ii)

(iii)

Let us take u = e^x and v = cotx

Putting the above obtained values in the formula:-

(uv)′ = u′v + uv′

(e^x cotx)’ = e^x× cotx + e^x× -cosec²x

= e^xcotx - e^xcosec²x

= e^x (cotx - cosec²x)

Ans) e^x (cotx - cosec²x)

To find: Differentiation of xⁿ cot x

Formula used: (i) (uv)′ = u′v + uv′ (Leibnitz or product rule)

(ii)

(iii)

Let us take u = xⁿ and v = cotx

Putting the above obtained values in the formula :-

(uv)′ = u′v + uv′

(xⁿ cotx)’ = nx^n-1× cotx + xⁿ× -cosec²x

= nx^n-1cotx - xⁿcosec²x

= xⁿ (nx^-1cotx - cosec²x)

Ans) xⁿ (nx^-1cotx - cosec²x)

To find: Differentiation of x³ sec x

Formula used: (i) (uv)′ = u′v + uv′ (Leibnitz or product rule)

(ii)

(iii)

Let us take u = x³ and v = sec x

Putting the above obtained values in the formula :-

(uv)′ = u′v + uv′

(x³ sec x)’ = 3x²× secx + x³× secx tanx

= 3x²secx + x³secx tanx

= x²secx(3 + x tanx)

Ans) x²secx(3 + x tanx)

To find: Differentiation of (x² + 3x + 1) sin x

Formula used: (i) (uv)′ = u′v + uv′ (Leibnitz or product rule)

(ii)

(iii)

Let us take u = x² + 3x + 1 and v = sin x

Putting the above obtained values in the formula :-

(uv)′ = u′v + uv′

[(x² + 3x + 1) sin x]’ = (2x + 3) × sinx + (x² + 3x + 1) × cosx

= sinx (2x + 3) + cosx (x² + 3x + 1)

Ans) (2x + 3) sinx + (x² + 3x + 1) cosx

To find: Differentiation of x⁴ tan x

Formula used: (i) (uv)′ = u′v + uv′ (Leibnitz or product rule)

(ii)

(iii)

Let us take u = x⁴ and v = tan x

Putting the above obtained values in the formula:-

(uv)′ = u′v + uv′

(x⁴ tan x)’ = 4x³× tanx + x⁴× sec²x

= 4x³tanx + x⁴sec²x

= x³ (4tanx + xsec²x)

Ans) x³ (4tanx + xsec²x)

To find: Differentiation of (3x – 5) (4x² – 3 + e^x)

Formula used: (i) (uv)′ = u′v + uv′ (Leibnitz or product rule)

(ii)

(iii)

Let us take u = (3x – 5) and v = (4x² – 3 + e^x)

Putting the above obtained values in the formula :-

(uv)′ = u′v + uv′

[(3x – 5)(4x² – 3 + e^x)]’ = 3×(4x² – 3 + e^x) + (3x – 5)×(8x + e^x)

= 12x² – 9 + 3e^x+ 24x² + 3xe^x – 40x - 5e^x

= 36x² + x(3e^x – 40) – 9 - 2e^x

Ans) 36x² + x(3e^x – 40) – 9 - 2e^x

To find: Differentiation of (x² – 4x + 5) (x³ – 2)

Formula used: (i) (uv)′ = u′v + uv′ (Leibnitz or product rule)

(ii)

Let us take u = (x² – 4x + 5) and v = (x³ – 2)

Putting the above obtained values in the formula:-

(uv)′ = u′v + uv′

[(x² – 4x + 5) (x³ – 2)]’ = (2x – 4)×(x³ – 2) + (x² – 4x + 5)×(3x²)

= 2x⁴ – 4x - 4x³ + 8 + 3x⁴ – 12x³ + 15x²

= 5x⁴ - 16x³ + 15x² – 4x + 8

Ans) 5x⁴ - 16x³ + 15x² – 4x + 8

To find: Differentiation of (x² + 2x – 3) (x² + 7x + 5)

Formula used: (i) (uv)′ = u′v + uv′ (Leibnitz or product rule)

(ii)

Let us take u = (x² + 2x – 3) and v = (x² + 7x + 5)

Putting the above obtained values in the formula :-

(uv)′ = u′v + uv′

[(x² + 2x – 3) (x² + 7x + 5)]’

= (2x + 2) × (x² + 7x + 5) + (x² + 2x – 3) × (2x + 7)

= 2x³ + 14x² + 10x + 2x² + 14x + 10 + 2x³ + 7x² + 4x² + 14x – 6x – 21

= 4x³ + 27x² + 32x – 11

Ans) 4x³ + 27x² + 32x – 11

To find: Differentiation of (tan x + sec x) (cot x + cosec x)

Formula used: (i) (uv)′ = u′v + uv′ (Leibnitz or product rule)

(ii)

(iii)

(iv)

(v)

Let us take u = (tan x + sec x) and v = (cot x + cosec x)

Putting the above obtained values in the formula:-

(uv)′ = u′v + uv′

[(tan x + sec x) (cot x + cosec x)]’

= [secx (secx + tanx)] × [(cot x + cosec x)] + [(tan x + sec x)] × [-cosecx (cosecx + cotx)]

= (secx +tanx) [secx(cotx + cosecx) - cosecx(cosecx + cotx)]

= (secx + tanx) (secx – cosecx) (cotx + cosecx)

Ans) (secx + tanx) (secx – cosecx) (cotx + cosecx)

To find: Differentiation of (x³ cos x – 2^x tan x)

Formula used: (i) (uv)′ = u′v + uv′ (Leibnitz or product rule)

(ii)

(iii)

(iv)

(v)

Here we have two function (x³ cos x) and (2^x tan x)

We have two differentiate them separately

Let us assume g(x) = (x³ cos x)

And h(x) = (2^x tan x)

Therefore, f(x) = g(x) – h(x)

⇒ f’(x) = g’(x) – h’(x) … (i)

Applying product rule on g(x)

Let us take u = x³ and v = cos x

Putting the above obtained values in the formula:-

(uv)′ = u′v + uv′

[x³ cosx]’ = 3x² × cosx + x³ × -sinx

= 3x²cosx - x³sinx

= x² (3cosx – x sinx)

g’(x) = x² (3cosx – x sinx)

Applying product rule on h(x)

Let us take u = 2^x and v = tan x

Putting the above obtained values in the formula:-

(uv)′ = u′v + uv′

[2^x tan x]’ = 2^x log2× tanx + 2^x × sec²x

= 2^x (log2tanx + sec²x)

h’(x) = 2^x (log2tanx + sec²x)

Putting the above obtained values in eqn. (i)

f’(x) = x² (3cosx – x sinx) - 2^x (log2tanx + sec²x)

Ans) x² (3cosx – x sinx) - 2^x (log2tanx + sec²x)

Let us take u = 2^x and v = x

Putting the above obtained values in the formula:-

Let us take u = logx and v = x

Putting the above obtained values in the formula:-

Let us take u = e^x and v = (1+x)

Putting the above obtained values in the formula:-

Let us take u = e^x and v = (1+x²)

Putting the above obtained values in the formula:-

Let us take u = (2x² – 4) and v = (3x² + 7)

Putting the above obtained values in the formula:-

Let us take u = (x² + 3x – 1) and v = (x + 2)

Putting the above obtained values in the formula:-

Let us take u = (x² – 1) and v = (x² + 7x + 1)

Putting the above obtained values in the formula:-

Let us take u = (5x² + 6x + 7) and v = (2x² + 3x + 4)

Putting the above obtained values in the formula:-

Let us take u = (x) and v = (a² + x²)

Putting the above obtained values in the formula:-

Let us take u = (x⁴) and v = (sinx)

Putting the above obtained values in the formula:-

Let us take u = and v =

Putting the above obtained values in the formula:-

Let us take u = (cosx) and v = (logx)

Putting the above obtained values in the formula:-

Let us take u = (2cotx) and v =

Putting the above obtained values in the formula:-

Let us take u = (sinx) and v = (1 + cosx)

Putting the above obtained values in the formula:-

Let us take u = (1 + sinx) and v = (1 - sinx)

Putting the above obtained values in the formula:-

Let us take u = (1 - cosx) and v = (1 + cosx)

Putting the above obtained values in the formula:-

Let us take u = (sinx - cosx) and v = (sinx + cosx)

Putting the above obtained values in the formula:-

Let us take u = (secx - tanx) and v = (secx + tanx)

Putting the above obtained values in the formula:-

Let us take u = () and v = ()

Putting the above obtained values in the formula:-

(v) (uv)′ = u′v + uv′ (Leibnitz or product rule)

Let us take u = () and v = ()

Applying Product rule

(gh)′ = g′h + gh′

Taking g = e^x and h = sinx

= e^xsinx + e^xcosx

u’ = e^xsinx + e^xcosx

Putting the above obtained values in the formula:-

(v) (uv)′ = u′v + uv′ (Leibnitz or product rule)

Let us take u = () and v = ()

Applying Product rule

(gh)′ = g′h + gh′

Taking g = and h =

= +

u’ = -

u’ =

Putting the above obtained values in the formula:-

(iv) (uv)′ = u′v + uv′ (Leibnitz or product rule)

Let us take u = e^x(x-1) and v = (x+1)

Applying Product rule

(gh)′ = g′h + gh′

Taking g = and h = x – 1

[e^x(x-1)]’ = e^x(x-1) + e^x (1)

= e^x(x-1) + e^x

u’ = e^xx

Putting the above obtained values in the formula:-

(iv) (uv)′ = u′v + uv′ (Leibnitz or product rule)

Let us take u = (x tanx) and v = (secx + tanx)

Applying Product rule for finding u’

(gh)′ = g′h + gh′

Taking g = xand h = tanx

[]’ = (1) (tanx) + x (sec²x)

= tanx + xsec²x

u’ = tanx + xsec²x

Putting the above obtained values in the formula:-

Let us take u = () and v = ()

Putting the above obtained values in the formula:-

(iv) (uv)′ = u′v + uv′ (Leibnitz or product rule)

Let us take u = () and v = ()

Applying Product rule for finding the term xcosx in u’

(gh)′ = g′h + gh′

Taking g = xand h = cosx

[]’ = (1) (cosx) + x (-sinx)

[]’ = cosx – x sinx

Applying the above obtained value for finding u’

u’ = cosx – (cosx – x sinx)

u’ = x sinx

Applying Product rule for finding the term xsinx in v’

(gh)′ = g′h + gh′

Taking g = xand h = sinx

[]’ = (1) (sinx) + x (cosx)

[]’ = sinx + x cosx

Applying the above obtained value for finding v’

v’ = sinx + x cosx - sinx

v’ = x cosx

Putting the above obtained values in the formula:-

Let us take u = cosx and v = sinx

u’ = (cosx)’ = -sinx

v’ = (sinx)’ = cosx

Putting the above obtained values in the formula:-

Ans).

(ii)

Let us take u = 1 and v = cosx

u’ = (1)’ = 0

v’ = (cosx)’ = -sinx

Putting the above obtained values in the formula:-

Ans).

To Find: Differentiation

NOTE : When 2 functions are in the product then we used product rule i.e

Formula used: (sin nu) = cos (nu)(nu)

Let us take y = sin 4x.

So, by using the above formula, we have

(sin4x) = cos (4x) (4x) = 4cos4x.

Differentiation of y = sin 4x is 4cos4x

To Find: Differentiation

NOTE : When 2 functions are in the product then we used product rule i.e

Formula used: (cos nu) = - sin (nu)(nu).

Let us take y = cos5x.

So, by using the above formula, we have

(cos5x) = - sin(5x) (5x) = - 5sin5x.

Differentiation of y = cos 5x is - 5sin5x

To Find: Differentiation

NOTE : When 2 functions are in the product then we used product rule i.e

Formula used: (tan nu) = sec² (nu).(nu).

Let us take y = tan3x

So, by using the above formula, we have

tan3x = sec²(3x) (3x) = 3sec²(3x)

Differentiation of y = tan3x is 3sec²(3x)

To Find: Differentiation

NOTE : When 2 functions are in the product then we used product rule i.e

Formula used: (cos nu) = - sin nu(nu) and = nx^{n - 1}

Let us take y = cos x³

So, by using the above formula, we have

cos x³ = - sin(x³)(x³) = - 3x² sin(x³)

Differentiation of y = cos x³ is - 3x² sin(x³)

To Find: Differentiation

NOTE : When 2 functions are in the product then we used product rule i.e

Formula used: (cot^a nu) = acot^{a - 1}(nu) (cot nu) (nu) and = nx^{n - 1}

Let us take y = cot²x

So, by using the above formula, we have

cot²x = 2cot(x) = - 2cotx (cosec²x).

Differentiation of y = cot²x is - 2cotx (cosec²x)

To Find: Differentiation

NOTE : When 2 functions are in the product then we used product rule i.e

Formula used: (tan^anu) = atan^{a - 1}nu and = nx^{n - 1}

Let us take y = tan³x

So, by using the above formula, we have

tan³x = 3tan²(x) = 3tan²x (sec²x).

Differentiation of y = tan³x is 3tan²x (sec²x)

To Find: Differentiation

NOTE : When 2 functions are in the product then we used product rule i.e

Formula used: (tan ) = sec²() (nu).

Let us take y = tan

So, by using the above formula, we have

tan = sec²() ()(x). = .

Differentiation of y = tan is .

To Find: Differentiation

NOTE : When 2 functions are in the product then we used product rule i.e

Formula used: () = () and = nx^{n - 1}

Let us take y =

So, by using the above formula, we have

= ()

Differentiation of y = is

To Find: Differentiation

NOTE : When 2 functions are in the product then we used product rule i.e

Formula used: () = and = nx^{n - 1}

Let us take y =

So, by using the above formula, we have

= = - cosec²x.

Differentiation of y = is - cosec²x

To Find: Differentiation

NOTE : When 2 functions are in the product then we used product rule i.e

Formula used: () = ()(nu) and = nx^{n - 1}

Let us take y =

So, by using the above formula, we have

= ()(x) = cosx

Differentiation of y = is cosx

To Find: Differentiation

NOTE : When 2 functions are in the product then we used product rule i.e

Formula used: (yⁿ) = ny^{n - 1}

Let us take y = (5 + 7x)⁶

So, by using the above formula, we have

(5 + 7x)⁶ = 6(5 + 7x)⁵(5 + 7x) = 6(5 + 7x)⁵7 = 42(5 + 7x)⁵

Differentiation of y = (5 + 7x)⁶ is 42(5 + 7x)⁵

To Find: Differentiation

NOTE : When 2 functions are in the product then we used product rule i.e

Formula used: (yⁿ) = ny^{n - 1}

Let us take y= (3 - 4x)⁵

So, by using the above formula, we have

(3 - 4x)⁵ = 4(3 - 4x)⁵(3 - 4x)= 4(3 - 4x)⁵( - 4) = - 16(3 - 4x)⁵

Differentiation of y = (3 - 4x)⁵is - 16(3 - 4x)⁵

To Find: Differentiation

NOTE : When 2 functions are in the product then we used product rule i.e

Formula used: (yⁿ) = ny^{n - 1}

Let us take y= (3x² - x + 1)⁴

So, by using the above formula, we have

(3x² - x + 1)⁴ = 4(3x² - x + 1)³(3x² - x + 1) = 4(3x² - x + 1)³(36x - 1) = 4(3x² - x + 1)³(6x - 1)

Differentiation of y = (3x² - x + 1)⁴ is 4(3x² - x + 1)³(6x - 1)

To Find: Differentiation

NOTE : When 2 functions are in the product then we used product rule i.e

Formula used: (yⁿ) = ny^{n - 1}

Let us take y= (ax² + bx + c)

So, by using the above formula, we have

(ax² + bx + c) = 2ax + b

Differentiation of y = (ax² + bx + c) is 2ax + b

To Find: Differentiation

NOTE : When 2 functions are in the product then we used product rule i.e

Formula used: (yⁿ) = ny^{n - 1}

Let us take y= = (x² - x + 3) ^{- 3}

So, by using the above formula, we have

(x² - x + 3) ^{- 3} = - 3(x² - x + 3) ^{- 4}(2x - 1) = - 3 (2x - 1)

Differentiation of y = (x² - x + 3) ^{- 3} is

To Find: Differentiation

NOTE : When 2 functions are in the product then we used product rule i.e

Formula used: sin² (ax + b) = 2 sin (ax + b)sin(ax + b)(ax + b)

Let us take y = sin² (2x + 3)

So, by using above formula, we have

sin² (2x + 3) = 2 sin (2x + 3)sin(2x + 3)(2x + 3) = 4sin(2x + 3)cos(2x + 3).

Differentiation of y = sin² (2x + 3)is 4sin(2x + 3)cos(2x + 3)

To Find: Differentiation

NOTE : When 2 functions are in the product then we used product rule i.e

Formula used: (cos^a nu) = acos^{a - 1}nu (cos nu) (nu)

Let us take y = cos²(x³)

So, by using the above formula, we have

cos²(x³) = 2 cosx³ ( - sin (x³))3x² = - 6x² cos(x³)sin x³

Differentiation of y = cos²(x³) is - 6x² cos(x³)sin x³

To Find: Differentiation

NOTE : When 2 functions are in the product then we used product rule i.e

Formula used: () = ()(u^a)

Let us take y =

So, by using the above formula, we have

= ()(x³) = ()3x² =

Differentiation of y = is

To Find: Differentiation

NOTE : When 2 functions are in the product then we used product rule i.e

Formula used: () = (u)

Let us take y =

So, by using the above formula, we have

= () = =

Differentiation of y = is

To Find: Differentiation

NOTE : When 2 functions are in the product then we used product rule i.e

Formula used: () = (). ()

Let us take y =

So, by using the above formula, we have

= cot

Differentiation of y = is

To Find: Differentiation

NOTE : When 2 functions are in the product then we used product rule i.e

Let us take y = cos 3x sin 5x

So, by using the above formula, we have

sin5 ( - 3sin 3) + cos 3(5cos 5) = 5cos (3 cos (5) - 3 sin () 3sin (3)

Differentiation of y = cos 3x sin 5x is 5cos (3 cos (5) - 3 sin () 3sin (3)

To Find: Differentiation

NOTE : When 2 functions are in the product then we used product rule i.e

Let us take y = sin x sin 2x

So, by using the above formula, we have

sin (2cos 2) + sin (sin) = 2sin)cos 2 + sin (sin)

Differentiation of y = sin x sin 2x is 2sin)cos 2 + sin (sin)

Let y = cos(sin ) , z = sin and w =

Formula :

According to the chain rule of differentiation

Let y = e^2x sin 3x , z = e^2x and w = sin 3x

Formula :

According to product rule of differentiation

Let y = e^3x cos 2x , z = e^3x and w = cos 2x

Formula :

According to the product rule of differentiation

Let y = e^-5x cot 4x , z = e^-5x and w = cot 4x

Formula :

According to the product rule of differentiation

Let y = cos (x³ . e^x) , z = x³ . e^x , m = e^x and w = x³

Formula :

According to the product rule of differentiation

According to the chain rule of differentiation

Let y = e^(xsinx+cosx) , z = x sin x+ cos x, m = x and w = sin x

Formula :

According to the product rule of differentiation

= x cosx

According to the chain rule of differentiation

Let y = , u = , v =

Formula :

According to the quotient rule of differentiation

If y =

( a² – b² = (a - b)(a + b) )

Let y = , u = , v =

Formula :

According to the quotient rule of differentiation

If y =

( a² – b² = (a - b)(a + b)

Let y = , u =, v = , z=

Formula :

According to the quotient rule of differentiation

If z =

According to chain rule of differentiation

=

=

=

Let y = , u =, v = , z=

Formula :

According to the quotient rule of differentiation

If z =

According to the chain rule of differentiation

=

=

=

Let y = , u =1+sin x, v = 1 – sin x , z=

Formula :

According to the quotient rule of differentiation

If z =

According to the chain rule of differentiation

=

=

=

Let y = , u = , v = , z=

Formula :

According to the quotient rule of differentiation

If z =

According to chain rule of differentiation

=

=

=

Let y = , u = , v =

Formula:

According to the quotient rule of differentiation

If y =

=

Let y = sin( ) , z =

Formula :

�

According to the chain rule of differentiation

Let y = e^x log (sin 2x) , z = e^x and w = log (sin 2x)

Formula :

According to the product rule of differentiation

Let y = cos ( ) , u =, v = , z=

Formula :

According to the quotient rule of differentiation

If z =

According to the chain rule of differentiation

=

=

Let y = sin ( ) , u =, v = , z=

Formula :

According to the quotient rule of differentiation

If z =

According to the chain rule of differentiation

=

Let y = , u = , v =

Formula:

According to the quotient rule of differentiation

If y =

=

Let , , u =, v =

Formula:

According to the quotient rule of differentiation

If y =

( )

HENCE PROVED.

Let , , u =, v =

Formula:

According to the quotient rule of differentiation

If y =

[ cos a cos b - sin a sin b = cos (a + b)]

=

HENCE PROVED.

Let y = , u =, v = , z=

Formula :

According to quotient rule of differentiation

If z =

According to the chain rule of differentiation

=

=

=

(Muliplying and dividing by 1-x )

=

=

Therefore

HENCE PROVED

y =

u =1-sin x, v = 1 + sin x , z=

Formula :

According to quotient rule of differentiation

If z =

According to the chain rule of differentiation

=

=

=

( Multiplying and dividing by )

=

=

=

=

=

=

=

=

=

=

HENCE PROVED