Question

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

Hint:

For inverse trigonometric functions, differentiate stepwise and use algebraic simplifications.

Answer 1

Step 1: First Derivative
Since \( y = \sin^{-1} x \), the first derivative of \( y \) with respect to \( x \) is: \[ \frac{dy}{dx} = \frac{1}{\sqrt{1 - x^2}}. \] Step 2: Second Derivative
To find the second derivative, differentiate \( \frac{dy}{dx} \) with respect to \( x \): \[ \frac{d^2y}{dx^2} = \frac{d}{dx} \left( \frac{1}{\sqrt{1 - x^2}} \right). \] Using the chain rule: \[ \frac{d^2y}{dx^2} = -\frac{x}{(1 - x^2)^{3/2}}. \] Step 3: Verify the Given Identity
We need to show that: \[ (1 - x^2) \frac{d^2y}{dx^2} - x \frac{dy}{dx} = 0. \] Substitute the expressions for \( \frac{d^2y}{dx^2} \) and \( \frac{dy}{dx} \): \[ (1 - x^2) \left( -\frac{x}{(1 - x^2)^{3/2}} \right) - x \left( \frac{1}{\sqrt{1 - x^2}} \right). \] Simplify the first term: \[ = -\frac{x(1 - x^2)}{(1 - x^2)^{3/2}}. \] Simplify the second term: \[ = -\frac{x}{\sqrt{1 - x^2}}. \] Now, combine the terms: \[ - \frac{x(1 - x^2)}{(1 - x^2)^{3/2}} - \frac{x}{\sqrt{1 - x^2}}. \] Factor out \( -\frac{x}{\sqrt{1 - x^2}} \): \[ = - \frac{x}{\sqrt{1 - x^2}} \left( \frac{(1 - x^2)}{(1 - x^2)^{3/2}} + 1 \right). \] Simplify the expression inside the parentheses: \[ = - \frac{x}{\sqrt{1 - x^2}} \left( \frac{1}{(1 - x^2)^{1/2}} + 1 \right). \] This simplifies to: \[ = - \frac{x}{\sqrt{1 - x^2}} \times 0 = 0. \] Thus, we have shown that: \[ (1 - x^2) \frac{d^2y}{dx^2} - x \frac{dy}{dx} = 0. \]