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

prove that

\(\frac{d^2x}{d\theta^2}=\)\(a(cos\theta-\theta sin\theta),\frac{d^2y}{d\theta^2}=\)\(a(sin\theta+ θcos θ)and\frac{d^2y}{dx^2}=\frac{sec^2\theta}{a θ}\)

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 Basic Idea: Second order derivative is nothing but derivative of derivative i.e. \(\frac{d^2y}{dx^2}=\frac{d}{dx}(\frac{dy}{dx})\)

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:

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

= a(-sin θ + θ cos θ + sin θ)

[ differentiated using product rule for θsinθ ]

= a θ cos θ ... eqn 4

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

Similarly,

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

Using equation 4 and 5 :

∴ again differentiating w.r.t x :-

Putting a value in the above equation-

We have :

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