Question

If $f(x) = \frac{\sin^{-1} x}{\sqrt{1 - x^2}}$, then the value of $(1 - x^2) f'(x) - x f(x)$ is:

• A. $0$
• B. $1$
• C. $2$
• D. $3$
• E. $4$

Hint:

When a derivative question involves $\sqrt{1-x^2}$ and $\sin^{-1}x$, look for ways to eliminate the square root early. Squaring or cross-multiplying usually simplifies the subsequent differentiation steps significantly.

Answer 1

The Correct Option is B

Concept: Instead of differentiating $f(x)$ directly using the quotient rule, it is often simpler to cross-multiply to remove the denominator and then use implicit differentiation and the product rule.

Step 1: Rearrange the function.
Given $f(x) = \frac{\sin^{-1} x}{\sqrt{1 - x^2}}$, we can write: \[ \sqrt{1 - x^2} f(x) = \sin^{-1} x \]

Step 2: Differentiate both sides with respect to $x$.
Using the product rule $(uv)' = u'v + uv'$ on the left side: \[ \frac{d}{dx}(\sqrt{1 - x^2}) f(x) + \sqrt{1 - x^2} f'(x) = \frac{d}{dx}(\sin^{-1} x) \] \[ \frac{-2x}{2\sqrt{1 - x^2}} f(x) + \sqrt{1 - x^2} f'(x) = \frac{1}{\sqrt{1 - x^2}} \]

Step 3: Multiply the entire equation by $\sqrt{1 - x^2}$.
\[ -x f(x) + (1 - x^2) f'(x) = 1 \]

Step 4: Conclusion.
The expression $(1 - x^2) f'(x) - x f(x)$ is equal to $1$.