If x =a (cos θ + θ sin θ), y=a (sin θ – θ cos θ) prove that

(sin θ + θ cosθ) and .

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 :