Question

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

Hint:

For proving differential identities, always calculate $\dfrac{dy}{dx}$ and $\dfrac{d^{2}y}{dx^{2}}$, then simplify the given expression.

Answer 1

We are given: \[ y = \sin^{-1}x \]

Step 1: First derivative. \[ \frac{dy}{dx} = \frac{1}{\sqrt{1-x^{2}}} \]

Step 2: Second derivative. \[ \frac{d^{2}y}{dx^{2}} = \frac{d}{dx}\left(\frac{1}{\sqrt{1-x^{2}}}\right) = \frac{1}{2}(1-x^{2})^{-\tfrac{3}{2}} \cdot (2x) = \frac{x}{(1-x^{2})^{\tfrac{3}{2}}} \]

Step 3: Substitute in given expression. \[ (1-x^{2}) \cdot \frac{d^{2}y}{dx^{2}} - x \cdot \frac{dy}{dx} \] \[ = (1-x^{2}) \cdot \frac{x}{(1-x^{2})^{3/2}} - x \cdot \frac{1}{\sqrt{1-x^{2}}} \] \[ = \frac{x(1-x^{2})}{(1-x^{2})^{3/2}} - \frac{x}{\sqrt{1-x^{2}}} \] \[ = \frac{x}{\sqrt{1-x^{2}}} - \frac{x}{\sqrt{1-x^{2}}} = 0 \] Hence Proved. \[ \boxed{(1 - x^{2}) \frac{d^{2}y}{dx^{2}} - x \frac{dy}{dx} = 0} \]