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 :\(\frac{dy}{dx}\)\(=\frac{\frac{dy}{d\theta}}{\frac{dx}{d\theta}}\)
Given,
y = sin3θ ……equation 1
x = cos θ ……equation 2
To prove: \(y\frac{d^2y}{dx^2}+(\frac{dy}{dx})^2\)\(=3sin^2\theta(5cos^2\theta-1)\)
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 \(\frac{d^2y}{dx^2}\)
As, \(\frac{d^2y}{dx^2}=\frac{d}{dx}(\frac{dy}{dx})\)
So, lets first find dy/dx using parametric form and differentiate it again.
Applying chain rule to differentiate sin3θ :
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: