Question

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

Hint:

For proofs involving second derivatives of inverse trigonometric or implicit functions, it is often much cleaner to differentiate once, rearrange the equation to eliminate fractions or roots, and then differentiate again implicitly. This avoids complicated applications of the quotient or chain rule.

Answer 1

Step 1: Understanding the Concept:
This problem requires finding the first and second derivatives of the given function and then substituting them into the provided differential equation to prove the identity. An alternative, often simpler method, involves implicit differentiation to avoid complex expressions.
Step 2: Key Formula or Approach:
1. Differentiate \( y = \sin^{-1} x \) to find \( \frac{dy}{dx} \). 2. Rearrange the equation to remove the square root. 3. Differentiate the rearranged equation again with respect to \( x \) using the product rule to find a relationship involving \( \frac{d^2y}{dx^2} \). 4. Simplify the result to match the target equation.
Step 3: Detailed Explanation:
Given the function: \[ y = \sin^{-1} x \] First, we find the first derivative: \[ \frac{dy}{dx} = \frac{1}{\sqrt{1 - x^2}} \] To avoid dealing with the square root in the next differentiation, we can rearrange this equation: \[ \sqrt{1 - x^2} \frac{dy}{dx} = 1 \] Squaring both sides gives: \[ (1 - x^2) \left(\frac{dy}{dx}\right)^2 = 1 \] Now, we differentiate this equation with respect to \( x \) using the product rule on the left side: \[ \frac{d}{dx}\left[(1 - x^2) \left(\frac{dy}{dx}\right)^2\right] = \frac{d}{dx}(1) \] \[ \frac{d}{dx}(1 - x^2) \cdot \left(\frac{dy}{dx}\right)^2 + (1 - x^2) \cdot \frac{d}{dx}\left(\frac{dy}{dx}\right)^2 = 0 \] \[ (-2x) \left(\frac{dy}{dx}\right)^2 + (1 - x^2) \cdot 2\left(\frac{dy}{dx}\right)\frac{d^2y}{dx^2} = 0 \] We can divide the entire equation by \( 2 \frac{dy}{dx} \) (since \( \frac{dy}{dx} \neq 0 \) for \( |x|<1 \)): \[ -x \frac{dy}{dx} + (1 - x^2) \frac{d^2y}{dx^2} = 0 \] Rearranging the terms to match the required form: \[ (1 - x^2) \frac{d^2y}{dx^2} - x \frac{dy}{dx} = 0 \] Step 4: Final Answer:
We have successfully derived the given differential equation starting from \( y = \sin^{-1} x \). Thus, the statement is proven.