Question

If \(y=\sin^{-1}x\), then show that \((1-x^2)\dfrac{d^2y}{dx^2}-x\dfrac{dy}{dx}=0\).

Hint:

Differentiate \(y'=(1-x^2)^{-1/2}\) again (either via the squared relation or the chain rule directly) and simplify.

Answer 1

Step 1: Direct second differentiation instead of the squared-relation trick:
Differentiate \(\dfrac{dy}{dx}=(1-x^2)^{-1/2}\) directly using the chain rule: \(\dfrac{d^2y}{dx^2}=-\dfrac12(1-x^2)^{-3/2}\cdot(-2x)=x(1-x^2)^{-3/2}\).

Step 2: Substituting into the target expression:
\((1-x^2)\dfrac{d^2y}{dx^2}=(1-x^2)\cdot x(1-x^2)^{-3/2}=x(1-x^2)^{-1/2}\).

Step 3: Comparing with x·dy/dx:
\(x\dfrac{dy}{dx}=x\cdot(1-x^2)^{-1/2}\), which is identical to the expression just found for \((1-x^2)\dfrac{d^2y}{dx^2}\).

Final Answer:
So \((1-x^2)\dfrac{d^2y}{dx^2}=x\dfrac{dy}{dx}\), i.e. \(\boxed{(1-x^2)\dfrac{d^2y}{dx^2}-x\dfrac{dy}{dx}=0}\), matching the implicit-differentiation method.