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: Understanding the Concept:
Differentiate \(y=\sin^{-1}x\) once to get \(dy/dx\) in a form that's easy to differentiate again, then differentiate once more and combine to match the target identity.

Step 2: First derivative:
\(\dfrac{dy}{dx}=\dfrac{1}{\sqrt{1-x^2}}\).

Step 3: Rewriting to avoid the square root before differentiating again:
\(\dfrac{dy}{dx}\cdot\sqrt{1-x^2}=1\). Squaring: \(\left(\dfrac{dy}{dx}\right)^2(1-x^2)=1\).

Step 4: Differentiating this squared relation implicitly:
\(2\dfrac{dy}{dx}\dfrac{d^2y}{dx^2}(1-x^2)+\left(\dfrac{dy}{dx}\right)^2(-2x)=0\).

Step 5: Simplifying:
Divide throughout by \(2\dfrac{dy}{dx}\) (nonzero since \(dy/dx=1/\sqrt{1-x^2}\neq0\)): \((1-x^2)\dfrac{d^2y}{dx^2}-x\dfrac{dy}{dx}=0\).

Final Answer:
Hence \(\boxed{(1-x^2)\dfrac{d^2y}{dx^2}-x\dfrac{dy}{dx}=0}\), as required.