Higher Order Derivatives

Basic idea:

√Second order derivative is nothing but derivative of derivative i.e.

√The idea of chain rule of differentiation: If f is any real-valued function which is the composition of two functions u and v, i.e. f = v(u(x)). For the sake of simplicity just assume t = u(x)

Then f = v(t). By chain rule, we can write the derivative of f w.r.t to x as:

√Product rule of differentiation-

√Apart from these remember the derivatives of some important functions like exponential, logarithmic, trigonometric etc..

Let’s solve now:

Given, y = x³ + tan x

We have to find

As

So lets first find and differentiate it again.

∴

[∵ tan x) = sec² x & ]

∴

Differentiating again with respect to x :

[ differentiated sec²x using chain rule, let t = sec x and z = t² ∴ ]

√Basic Idea: Second order derivative is nothing but derivative of derivative i.e.

√The idea of chain rule of differentiation: If f is any real-valued function which is the composition of two functions u and v, i.e. f = v(u(x)). For the sake of simplicity just assume t = u(x)

Then f = v(t). By chain rule, we can write the derivative of f w.r.t to x as:

√Product rule of differentiation-

√Apart from these remember the derivatives of some important functions like exponential, logarithmic, trigonometric etc..

Let’s solve now:

Given, y = sin ( log x )

We have to find

As

So lets first find dy/dx and differentiate it again.

∴

differentiating using the chain rule,

let, t = log x and y = sin t

∵ [using chain rule]

[∵ log x) = & ]

Differentiating again with respect to x :

[ using product rule of differentiation]

√Basic Idea: Second order derivative is nothing but derivative of derivative i.e.

√The idea of chain rule of differentiation: If f is any real-valued function which is the composition of two functions u and v, i.e. f = v(u(x)). For the sake of simplicity just assume t = u(x)

Then f = v(t). By chain rule, we can write the derivative of f w.r.t to x as:

√Product rule of differentiation-

Apart from these remember the derivatives of some important functions like exponential, logarithmic, trigonometric etc..

Let’s solve now:

Given, y = log (sin x)

We have to find

As

So lets first find dy/dx and differentiate it again.

∴

differentiating using cthe hain rule,

let, t = sin x and y = log t

∵ [using chain rule]

[∵ = & ]

Differentiating again with respect to x:

[ ∵ ]

√Basic Idea: Second order derivative is nothing but derivative of derivative i.e.

√The idea of chain rule of differentiation: If f is any real-valued function which is the composition of two functions u and v, i.e. f = v(u(x)). For the sake of simplicity just assume t = u(x)

Then f = v(t). By chain rule, we can write the derivative of f w.r.t to x as:

√Product rule of differentiation-

Apart from these remember the derivatives of some important functions like exponential, logarithmic, trigonometric etc..

Let’s solve now:

Given, y = e^x sin 5x

We have to find

As,

So lets first find dy/dx and differentiate it again.

∴

Let u = e^x and v = sin 5x

As, y = uv

∴ Using product rule of differentiation:

∴

[ ∵ ]

Again differentiating w.r.t x:

Again using the product rule :

[∵

√Basic Idea: Second order derivative is nothing but derivative of derivative i.e.

√The idea of chain rule of differentiation: If f is any real-valued function which is the composition of two functions u and v, i.e. f = v(u(x)). For the sake of simplicity just assume t = u(x)

Then f = v(t). By chain rule, we can write the derivative of f w.r.t to x as:

√Product rule of differentiation-

Apart from these remember the derivatives of some important functions like exponential, logarithmic, trigonometric etc..

Let’s solve now:

Given, y = e^6x cos 3x

We have to find

As,

So lets first find dy/dx and differentiate it again.

∴

Let u = e^6x and v = cos 3x

As, y = uv

∴ Using product rule of differentiation:

∴

[ ∵ ]

Again differentiating w.r.t x:

Again using the product rule :

√Basic Idea: Second order derivative is nothing but derivative of derivative i.e.

√The idea of chain rule of differentiation: If f is any real-valued function which is the composition of two functions u and v, i.e. f = v(u(x)). For the sake of simplicity just assume t = u(x)

Then f = v(t). By chain rule, we can write the derivative of f w.r.t to x as:

√Product rule of differentiation-

Apart from these remember the derivatives of some important functions like exponential, logarithmic, trigonometric etc..

Let’s solve now:

Given, y = x³ log x

We have to find

As

So lets first find dy/dx and differentiate it again.

∴

Let u = x³ and v = log x

As, y = uv

∴ Using product rule of differentiation:

∴

[ ∵ ]

Again differentiating w.r.t x:

Again using the product rule :

[ ∵ ]

Basic idea:

√Second order derivative is nothing but derivative of derivative i.e.

√The idea of chain rule of differentiation: If f is any real-valued function which is the composition of two functions u and v, i.e. f = v(u(x)). For the sake of simplicity just assume t = u(x)

Then f = v(t). By chain rule, we can write the derivative of f w.r.t to x as:

√Product rule of differentiation-

√Apart from these remember the derivatives of some important functions like exponential, logarithmic, trigonometric etc..

Let’s solve now:

Given, y = tan ^–1 x

We have to find

As

So lets first find dy/dx and differentiate it again.

∴ [∵ tan^–1 x) = ]

∴ [∵ tan^–1 x) = ]

Differentiating again with respect to x :

Differentiating using chain rule,

let t = 1 +x² and z = 1/t

∵ [ from chain rule of differentiation]

∴ [∵ ]

∴

√Basic Idea: Second order derivative is nothing but derivative of derivative i.e.

√The idea of chain rule of differentiation: If f is any real-valued function which is the composition of two functions u and v, i.e. f = v(u(x)). For the sake of simplicity just assume t = u(x)

Then f = v(t). By chain rule, we can write the derivative of f w.r.t to x as:

√Product rule of differentiation-

Apart from these remember the derivatives of some important functions like exponential, logarithmic, trigonometric etc..

Let’s solve now:

Given, y = x cos x

We have to find

As

So lets first find dy/dx and differentiate it again.

∴

Let u = x and v = cos x

As, y = uv

∴ Using product rule of differentiation:

∴

[ ∵ ]

Again differentiating w.r.t x:

Again using the product rule :

[ ∵ ]

√Basic Idea: Second order derivative is nothing but derivative of derivative i.e.

√The idea of chain rule of differentiation: If f is any real-valued function which is the composition of two functions u and v, i.e. f = v(u(x)). For the sake of simplicity just assume t = u(x)

Then f = v(t). By chain rule, we can write the derivative of f w.r.t to x as:

√Product rule of differentiation-

Apart from these remember the derivatives of some important functions like exponential, logarithmic, trigonometric etc..

Let’s solve now:

Given, y = log ( log x )

We have to find

As,

So lets first find dy/dx and differentiate it again.

∴

Let y = log t and t = log x

Using chain rule of differentiation:

∴ [∵ log x) = ]

Again differentiating w.r.t x:

As,

Where u = and v =

∴ using product rule of differentiation:

∴ [ use chain rule to find ]

[ ∵ ]

Basic idea:

√Second order derivative is nothing but derivative of derivative i.e.

√The idea of chain rule of differentiation: If f is any real-valued function which is the composition of two functions u and v, i.e. f = v(u(x)). For the sake of simplicity just assume t = u(x)

Then f = v(t). By chain rule, we can write the derivative of f w.r.t to x as:

√Product rule of differentiation-

√Apart from these remember the derivatives of some important functions like exponential, logarithmic, trigonometric etc..

Let’s solve now:

Given,

y=e^–x cos x

TO prove :

sin x.

Clearly from the expression to be proved we can easily observe that we need to just find the second derivative of given function.

Given, y = e–^x cos x

We have to find

As,

So lets first find dy/dx and differentiate it again.

∴

Let u = e^–x and v = cos x

As, y = u*v

∴ using product rule of differentiation:

∴

[ ∵ ]

Again differentiating w.r.t x:

Again using the product rule :

[∵

Basic idea:

√Second order derivative is nothing but derivative of derivative i.e.

√The idea of chain rule of differentiation: If f is any real-valued function which is the composition of two functions u and v, i.e. f = v(u(x)). For the sake of simplicity just assume t = u(x)

Then f = v(t). By chain rule, we can write the derivative of f w.r.t to x as:

√Product rule of differentiation-

√Apart from these remember the derivatives of some important functions like exponential, logarithmic, trigonometric etc..

Let’s solve now:

Given, y = x+ tan x …..equation 1

As we have to prove: cos²

We notice a second-order derivative in the expression to be proved so first take the step to find the second order derivative.

Let’s find

As

So lets first find dy/dx and differentiate it again.

∴ [∵ tan x) = sec² x & ]

∴

Differentiating again with respect to x :

[ differentiated sec²x using chain rule, let t = sec x and z = t² ∴ ]

……….equation 2

As we got an expression for the second order, as we need cos²x term with

Multiply both sides of equation 1 with cos²x:

∴ we have,

[∵ cos x × sec x = 1]

From equation 1:

tan x = y – x

….proved

Basic idea:

√Second order derivative is nothing but derivative of derivative i.e.

√The idea of chain rule of differentiation: If f is any real-valued function which is the composition of two functions u and v, i.e. f = v(u(x)). For the sake of simplicity just assume t = u(x)

Then f = v(t). By chain rule, we can write the derivative of f w.r.t to x as:

√Product rule of differentiation-

√Apart from these remember the derivatives of some important functions like exponential, logarithmic, trigonometric etc..

Let’s solve now:

As we have to prove :

We notice a third order derivative in the expression to be proved so first take the step to find the third order derivative.

Given, y = x³ log x

Let’s find -

As

So lets first find dy/dx and differentiate it again.

∴

differentiating using product rule:

[log x) = ]

Again differentiating using product rule:

[log x) = ]

Again differentiating using product rule:

[log x) = ]

Again differentiating w.r.t x :

Basic idea:

√Second order derivative is nothing but derivative of derivative i.e.

√The idea of chain rule of differentiation: If f is any real-valued function which is the composition of two functions u and v, i.e. f = v(u(x)). For the sake of simplicity just assume t = u(x)

Then f = v(t). By chain rule, we can write the derivative of f w.r.t to x as:

√Product rule of differentiation-

√Apart from these remember the derivatives of some important functions like exponential, logarithmic, trigonometric etc..

Let’s solve now:

As we have to prove: cos x cose³ x

We notice a third order derivative in the expression to be proved so first take the step to find the third order derivative.

Given, y = log (sin x)

Let’s find –

As

So lets first find dy/dx and differentiate it again.

∴

differentiating using the chain rule,

let, t = sin x and y = log t

∵ [using chain rule]

[∵ = & ]

Differentiating again with respect to x :

[ ∵ ]

Differentiating again with respect to x:

using the chain rule and

[ ∵ cot x = cos x/sin x]

Basic idea:

√Second order derivative is nothing but derivative of derivative i.e.

√The idea of chain rule of differentiation: If f is any real-valued function which is the composition of two functions u and v, i.e. f = v(u(x)). For the sake of simplicity just assume t = u(x)

Then f = v(t). By chain rule, we can write the derivative of f w.r.t to x as:

√Product rule of differentiation-

√Apart from these remember the derivatives of some important functions like exponential, logarithmic, trigonometric etc..

Let’s solve now:

Given, y = 2sin x+ 3cos x …..equation 1

As we have to prove : .

We notice a second-order derivative in the expression to be proved so first take the step to find the second order derivative.

Let’s find

As

So lets first find dy/dx and differentiate it again.

∴

[∵ sin x) = cosx & ]

∴

Differentiating again with respect to x :

From equation 1 we have :

y = 2 sin x + 3 cos x

∴

∴

Basic idea:

√Second order derivative is nothing but derivative of derivative i.e.

√The idea of chain rule of differentiation: If f is any real-valued function which is the composition of two functions u and v, i.e. f = v(u(x)). For the sake of simplicity just assume t = u(x)

Then f = v(t). By chain rule, we can write the derivative of f w.r.t to x as:

√Product rule of differentiation-

√Apart from these remember the derivatives of some important functions like exponential, logarithmic, trigonometric etc..

Let’s solve now:

Given, y = …..equation 1

As we have to prove : ..

We notice a second-order derivative in the expression to be proved so first take the step to find the second order derivative.

Let’s find

As

So, lets first find dy/dx and differentiate it again.

As y is the product of two functions u and v

Let u = log x and v = 1/x

Using product rule of differentiation:

[∵ log x) = & ]

Again using the product rule to find :

[∵ log x) = & ]

∴ ….. proved

Idea of parametric form of differentiation:

If y = f (θ) and x = g(θ) i.e. y is a function of θ and x is also some other function of θ.

Then dy/dθ = f’(θ) and dx/dθ = g’(θ)

We can write :

Given,

x = a sec θ ……equation 1

y = b tan θ ……equation 2

to prove : .

We notice a second order derivative in the expression to be proved so first take the step to find the second order derivative.

Let’s find

As,

So, lets first find dy/dx using parametric form and differentiate it again.

…..equation 3

Similarly, ……equation 4

[∵

Differentiating again w.r.t x :

…..equation 5 [ using chain rule]

From equation 3:

∴

Putting the value in equation 5 :

From equation 1:

y = b tan θ

∴ …..proved.

Basic idea:

√Second order derivative is nothing but derivative of derivative i.e.

√The idea of chain rule of differentiation: If f is any real-valued function which is the composition of two functions u and v, i.e. f = v(u(x)). For the sake of simplicity just assume t = u(x)

Then f = v(t). By chain rule, we can write the derivative of f w.r.t to x as:

√Product rule of differentiation-

√Apart from these remember the derivatives of some important functions like exponential, logarithmic, trigonometric etc..

The idea of parametric form of differentiation:

If y = f (θ) and x = g(θ), i.e. y is a function of θ and x is also some other function of θ.

Then dy/dθ = f’(θ) and dx/dθ = g’(θ)

We can write :

Given,

x =a (cos θ + θ sin θ) ……equation 1

y =a (sin θ – θ cos θ) ……equation 2

to prove :

(sin θ + θ cosθ)

.

We notice a second order derivative in the expression to be proved so first take the step to find the second order derivative.

Let’s find

As

[ differentiated using product rule for θsinθ ]

..eqn 4

Again differentiating w.r.t θ using product rule:-

∴

Similarly,

∴ ………….equation 5

Again differentiating w.r.t θ using product rule:-

∴

∵

Using equation 4 and 5 :

As

∴ again differentiating w.r.t x :-

[using chain rule]

∵

Putting a value in the above equation-

We have :

Basic idea:

√Second order derivative is nothing but derivative of derivative i.e.

√The idea of chain rule of differentiation: If f is any real-valued function which is the composition of two functions u and v, i.e. f = v(u(x)). For the sake of simplicity just assume t = u(x)

Then f = v(t). By chain rule, we can write the derivative of f w.r.t to x as:

√Product rule of differentiation-

√Apart from these remember the derivatives of some important functions like exponential, logarithmic, trigonometric etc..

Let’s solve now:

Given,

y=e^x cos x

TO prove :

Clearly from the expression to be proved we can easily observe that we need to just find the second derivative of given function.

Given, y = e^x cos x

We have to find

As

So lets first find dy/dx and differentiate it again.

∴

Let u = e^x and v = cos x

As, y = u*v

∴ Using product rule of differentiation:

∴

[ ∵ ]

Again differentiating w.r.t x:

Again using the product rule :

[∵

[∵ –sin x = cos (x + π/2)]

Idea of parametric form of differentiation:

If y = f (θ) and x = g(θ) i.e. y is a function of θ and x is also some other function of θ.

Then dy/dθ = f’(θ) and dx/dθ = g’(θ)

We can write :

Given,

x = a cos θ ……equation 1

y = b sin θ ……equation 2

to prove : .

We notice a second order derivative in the expression to be proved so first take the step to find the second order derivative.

Let’s find

As

So, lets first find dy/dx using parametric form and differentiate it again.

…..equation 3

Similarly, ……equation 4

[∵

Differentiating again w.r.t x :

…..equation 5

[ using chain rule and ]

From equation 3:

∴

Putting the value in equation 5 :

From equation 1:

y = b sin θ

∴ …..proved.

Idea of parametric form of differentiation:

If y = f (θ) and x = g(θ) i.e. y is a function of θ and x is also some other function of θ.

Then dy/dθ = f’(θ) and dx/dθ = g’(θ)

We can write :

Given,

x = a (1 – cos ³θ) ……equation 1

y = a sin ³ θ, ……equation 2

to prove : .

We notice a second order derivative in the expression to be proved so first take the step to find the second order derivative.

Let’s find

As

So, lets first find dy/dx using parametric form and differentiate it again.

…..equation 3 [using chain rule]

Similarly,

……equation 4

[∵

Differentiating again w.r.t x :

…..equation 5

[ using chain rule and ]

From equation 3:

∴

Putting the value in equation 5 :

Put θ = π/6

∴

Idea of parametric form of differentiation:

If y = f (θ) and x = g(θ) i.e. y is a function of θ and x is also some other function of θ.

Then dy/dθ = f’(θ) and dx/dθ = g’(θ)

We can write :

Given,

x = a (θ + sin θ) ……equation 1

y = a (1+ cos θ) ……equation 2

to prove :

We notice a second-order derivative in the expression to be proved so first take the step to find the second order derivative.

Let’s find

As,

So, lets first find dy/dx using parametric form and differentiate it again.

[∵ from equation 2] …..equation 3

Similarly,

……equation 4

[∵

[∵ from equation 2] …..equation 5

Differentiating again w.r.t x :

Using product rule and chain rule of differentiation together:

[using equation 3 and 5]

[ from equation 1]

∴ ….proved

Idea of parametric form of differentiation:

If y = f (θ) and x = g(θ) i.e. y is a function of θ and x is also some other function of θ.

Then dy/dθ = f’(θ) and dx/dθ = g’(θ)

We can write :

Given,

x = a (θ – sin θ) ……equation 1

y = a (1+ cos θ) ……equation 2

to find :

As,

So, lets first find dy/dx using parametric form and differentiate it again.

…..equation 3

Similarly,

……equation 4

[∵

…..equation 5

Differentiating again w.r.t x :

Using product rule and chain rule of differentiation together:

Apply chain rule to determine

[using equation 3]

[ ∵ ]

[ ∵1– cos θ = 2sin² θ/2]

∴

Idea of parametric form of differentiation:

If y = f (θ) and x = g(θ) i.e. y is a function of θ and x is also some other function of θ.

Then dy/dθ = f’(θ) and dx/dθ = g’(θ)

We can write :

Given,

y = a (θ + sin θ) ……equation 1

x = a (1– cos θ) ……equation 2

to prove : .

We notice a second-order derivative in the expression to be proved so first take the step to find the second order derivative.

Let’s find

As

So, lets first find dy/dx using parametric form and differentiate it again.

…..equation 3

Similarly,

……equation 4

[∵

…..equation 5

Differentiating again w.r.t x :

Using product rule and chain rule of differentiation together:

[using equation 4]

As we have to find

∴ put θ = π/2 in above equation:

=

Idea of parametric form of differentiation:

If y = f (θ) and x = g(θ) i.e. y is a function of θ and x is also some other function of θ.

Then dy/dθ = f’(θ) and dx/dθ = g’(θ)

We can write :

Given,

y = a (θ + sin θ) ……equation 1

x = a (1+ cos θ) ……equation 2

to prove : .

We notice a second-order derivative in the expression to be proved so first take the step to find the second order derivative.

Let’s find

As,

So, lets first find dy/dx using parametric form and differentiate it again.

…..equation 3

Similarly,

……equation 4

[∵

…..equation 5

Differentiating again w.r.t x :

Using product rule and chain rule of differentiation together:

[using equation 4]

As we have to find

∴ put θ = π/2 in above equation:

=

The idea of parametric form of differentiation:

If y = f (θ) and x = g(θ), i.e. y is a function of θ and x is also some other function of θ.

Then dy/dθ = f’(θ) and dx/dθ = g’(θ)

We can write :

Given,

y = sin³θ ……equation 1

x = cos θ ……equation 2

To prove:

We notice a second-order derivative in the expression to be proved so first take the step to find the second order derivative.

Let’s find

As,

So, lets first find dy/dx using parametric form and differentiate it again.

………….equation 3

Applying chain rule to differentiate sin³θ :

…………..equation 4

………..equation 5

Again differentiating w.r.t x:

Applying product rule and chain rule to differentiate:

[using equation 3 to put the value of dθ/dx]

Multiplying y both sides to approach towards the expression we want to prove-

[from equation 1, substituting for y]

Adding equation 5 after squaring it:

Given,

y = sin (sin x) ……equation 1

To prove:

We notice a second-order derivative in the expression to be proved so first take the step to find the second order derivative.

Let’s find

As

So, lets first find dy/dx

Using chain rule, we will differentiate the above expression

Let t = sin x ⟹

…….equation 2

Again differentiating with respect to x applying product rule:

Using chain rule again in the next step-

[using equation 1 : y =sin (sin x)]

And using equation 2, we have:

Note: y₂ represents second order derivative i.e. and y₁ = dy/dx

Given,

y = (sin^–1 x)² ……equation 1

to prove : (1–x²) y₂–xy₁–2=0

We notice a second–order derivative in the expression to be proved so first take the step to find the second order derivative.

Let’s find

As

So, lets first find dy/dx

Using chain rule we will differentiate the above expression

Let t = sin^–1 x => [using formula for derivative of sin^–1x]

And y = t²

…….equation 2

Again differentiating with respect to x applying product rule:

[using ]

Using equation 2 :

∴ (1–x²) y₂–xy₁–2=0 ……proved

Note: y₂ represents second order derivative i.e. and y₁ = dy/dx

Given,

y = (sin^–1 x)² ……equation 1

to prove : (1–x²) y₂–xy₁–2=0

We notice a second order derivative in the expression to be proved so first take the step to find the second order derivative.

Let’s find

As,

So, lets first find dy/dx

Using chain rule we will differentiate the above expression

Let t = sin^–1 x => [using formula for derivative of sin^–1x]

And y = t²

…….equation 2

Again differentiating with respect to x applying product rule:

[using ]

Using equation 2 :

∴ (1–x²) y₂–xy₁–2=0 ……proved

Note: y₂ represents second order derivative i.e. and y₁ = dy/dx

Given,

y = e^tan–1x ……equation 1

to prove : (1+x²)y₂+(2x–1)y₁=0

We notice a second order derivative in the expression to be proved so first take the step to find the second order derivative.

Let’s find

As,

So, lets first find dy/dx

Using chain rule we will differentiate the above expression

Let t = tan^–1 x => []

And y = e^t

…….equation 2

Again differentiating with respect to x applying product rule:

Using chain rule we will differentiate the above expression-

[using & ]

Using equation 2 :

∴ (1+x²)y₂+(2x–1)y₁=0 ……proved

Note: y₂ represents second order derivative i.e. and y₁ = dy/dx

Given,

y = 3 cos (log x) + 4 sin (log x) ……equation 1

to prove: x²y₂+xy₁+ y =0

We notice a second order derivative in the expression to be proved so first take the step to find the second order derivative.

Let’s find

As,

So, lets first find dy/dx

Let, log x = t

∴ y = 3cos t + 4sin t …………….equation 2

………….equation 3

……...equation 4

Again differentiating w.r.t x:

Using product rule of differentiation we have

Using equation 2,3 and 4 we can substitute above equation as:

Multiplying x² both sides:

∴ x²y₂+xy₁+ y =0 ………..proved

Note: y₂ represents second order derivative i.e. and y₁ = dy/dx

Given,

y = e^2x(ax + b) ……equation 1

to prove: y₂–4y₁+4y = 0

We notice a second order derivative in the expression to be proved so first take the step to find the second order derivative.

Let’s find

As,

So, lets first find dy/dx

∵ y = e^2x(ax + b)

Using product rule to find dy/dx:

……..equation 2

Again differentiating w.r.t x using product rule:

…….equation 3

In order to prove the expression try to get the required form:

Subtracting 4*equation 2 from equation 3:

Using equation 1:

∴ y₂–4y₁+4y = 0 ……..proved

Note: y₂ represents second order derivative i.e. and y₁ = dy/dx

Given,

x = sin

y = ……equation 1

to prove: (1–x²)y₂–xy₁–a² y = 0

We notice a second order derivative in the expression to be proved so first take the step to find the second order derivative.

Let’s find

As,

So, lets first find dy/dx

∵ y =

Let t = asin^–1 x => []

And y = e^t

…….equation 2

Again differentiating with respect to x applying product rule:

Using chain rule and equation 2:

[using ]

Using equation 1 and equation 2 :

∴ (1–x²)y₂–xy₁–a²y = 0……proved

Note: y₂ represents second order derivative i.e. and y₁ = dy/dx

Given,

log y = tan^–1 X

∴ y = ……equation 1

to prove : (1+x²)y₂+(2x–1)y₁=0

We notice a second order derivative in the expression to be proved so first take the step to find the second order derivative.

Let’s find

As

So, lets first find dy/dx

Using chain rule, we will differentiate the above expression

Let t = tan^–1 x => []

And y = e^t

…….equation 2

Again differentiating with respect to x applying product rule:

Using chain rule we will differentiate the above expression-

[using & ]

Using equation 2 :

∴ (1+x²)y₂+(2x–1)y₁=0 ……proved

Formula: –

Given: –

Y = tan ^{– 1}x

Differentiating w.r.t x

Using formula(ii)

Again Differentiating w.r.t x

Using formula(iii)

Hence proved.

Formula: –

Given: –

Differentiating w.r.t x

Using formula(ii)

Using formula(i)

Squaring both sides

Differentiating w.r.t x

Using formual(iii)

Hence proved

Formula: –

Given: –

Y = (tan ^{– 1}x)²

Then

Using formula (ii)&(i)

Again differentiating with respect to x on both the sides,we obtain

using formula(i)&(iii)

Hence proved.

Formula: –

Given: –

Y = cotx

Differentiating w.r.t. x

Using formula (ii)

Differentiating w.r.t x

Using formual (iii)

Hence proved.

Formula: –

Given: –

Differentiating w.r.t x

Again Differentiating w.r.t x

Formula: –

Given: –

Differentiating w.r.t x

Differentiating w.r.t x

Adding and subtracting on RHS

Formula: –

Given: –

y = e^x(sinx + cosx)

differentiating w.r.t x

Differentiating w.r.t x

Adding and subtracting y on RHS

Hence proved

Formula: –

Given: – y = cos ^{– 1}x

Then,

y = cos ^{– 1} x

⇒x = cosy

Putting x = cosy in equation(i), we obtain

Formula: –

Given: –

Taking logarithm on both sides we obtain

By squaring both sides, wee obtain

Again differentiating both sides with respect to x,we obtain

Hence proved

Formula: –

Given: –

y = 500e^7x + 600e ^{– 7x}

Hence proved.

Formula: –

Given: –

x = 2cost – cos2t

y = 2sint – sin2t

differentiating w.r.t t

Dividing both

Differentiating w.r.t t

Dividing

Putting

Formula: –

Given: –

x = 4z² + 5,y = 6z² + 72 + 3

Differentiating both w.r.t z

and

differentiating w.r.t z

Dividing

Formula: –

Given: –

Y = log(1 + cosx)

Differentiating w.r.t x

Differentiating w.r.t.x

Differentiating w.r.t x

Formula: –

Given: –

y = sin(logx)

Hence proved.

Formula: –

Given: –

y = 3e^2x + 2e^3x

Hence

Given: –

y = (cot ^{– 1}x)²

differentiating w.r.t x

Differentiating w.r.t x

Hence proved

Formula: –

Given: –

Y = cosec ^{– 1}x

We know that

Let y = cosec ^{– 1}x

Since x>1,|x| = x

Differentiating the above function with respect to x

Thus

Similarly

Hence proved.

Formula: –

Given: –

,y = sint

Differentiating with respect to t ,we have

Now find the value of

Now

We have

Differentiating with w.r.t t

At

Now putting

Formula: –

Given: –

Formula: –

Given: –

Putting

Formula: –

Given: –

,y = sint

Differentiating with respect to t ,we have

Now find the value of

Formula: –

Given: –

x = a (cos 2t + 2t sin 2t)

and y = a (sin 2t – 2t cos 2t)

Formula: –

given: –

x = 3 cot t – 2 cos³ t, y = 3 sin t – 2 sin³ t

differentiating both w.r.t t

And

differentiating both w.r.t t

Now,

= cot t

differentiating both w.r.t x

Formula: –

Given: –

x = asint – bcost,y = accost + bsint

differentiating both w.r.t t

,

Dividing both

Differentiating w.r.t t

Putting the value

Dividing them

Hence proved.

Formula: –

Given: –

y = Asin3x + Bcos3x

differentiating w.r.t x

Again differentiating w.r.t x

Now adding

But given,

12A – 6B = 10

– (12B + 6A) = 0

6A = – 12B

A = – 2B

Puttuing A

And ,

Formula: –

Given: –

Differentiating w.r.t t

Differentiating w.r.t t

Hence proved

Formula: –

Given: –

y = xⁿ(acos(logx) + bsin(logx))

y = axⁿcos(logx) + bxⁿsin(logx)

= xⁿ (na + b)[(n – 1) cos(logx) – sin (logx) ] + (bn – a) xⁿ [(n – 1) sin(logx) + cos(logx)] + (1 – 2n)x^{n – 1}cos(logx)(na + b) + (1 – 2n)x^{n – 1}sin(logx)(bn – a) + a(1 + n²)xⁿcos(logx) + bxⁿ(1 + n²)sin(logx)

Formula: –

Given: –

Now

Given:

x=a cos nt-b sin nt

= -n² (a cos nt-b sin nt )

= - n² x

Given:

y = 2at, x = at²

Given:

=n(n+1)[a xⁿ⁺¹ +bx^-n]

=n(n+1)y

Given:

Let y=2 cos x cos 3x

So y=cos 2x+cos 4x

=(-2)¹ (sin 2x+2¹ sin 4x )

=(-2)² (cos 2x+2² cos 4x )

=(-2)³ (cos 2x+2³ cos 4x )

=(-2)⁴ (cos 2x+2⁴ cos 4x )

For every odd degree; differential = =(-2)ⁿ (cos 2x+2ⁿ cos 4x );n={1,3,5…}

For every even degree; differential =(-2)ⁿ (cos 2x+2ⁿ cos 4x );n={0,2,4…}

So,

=(-2)²⁰ (cos 2x+2²⁰ cos 4x );

Given:

Given:

Given:

f(x) = (cos x + i sinx) (cos 2x + i sin 2x) (cos 3x + i sin 3x) …. (cos nx + i sin nx)

Since

So,

Given:

Given:

Given:

Given:

Since,

So,

Also,

So,

Given:

xy₁

x² y₂

x² y₂ + xy₁

Given:

x = 2at, y = at²

Given:

x = f(t) and y = g(t)

Given:

(1 – x²) y₂

(1 – x²) y₂ – xy₁

Given:

(1 – x²) y₂

= 2+ xy₁

Given:

(cos² x)y₂

y₁

Given:

Divide numerator and denominator by ;

We get:

Given:

2xy₁

Given:

Given:

Given:

Given:

So, at k=r;

Also,

So,

Given:

xy₁

(1)

x² y₂

(2)

x² y₂ + (3 – 2n) xy₁

Given:

Differentiate w.r.t. ‘x’ on both sides;

Since

So,

Given:

Hence, y is a constant.

Given:

y=axⁿ⁺¹ +bx^-n

λy=n(n+1)a x^n-1+2 +n(n+1)bx^-n-2+2

λy=n(n+1)[a x^(n+1)+bx^(-n)]

λy=n(n+1)

λ=n(n+1)

Given:

y=a cos nt-b sin nt

λy= -n² (a cos nt-b sin nt )

λy= - n² y

λ= -n²

Given:

x=t² ;y=t³

Given:

x = 2at, y = at²

=t

Given:

x = f(t) and y = g(t)

Given:

Given:

=e^-x

Given:

Given: