Question

If $y = (\sin^{-1}x)^2 + (\cos^{-1}x)^2$, then $(1 - x^2)\,y'' - x\,y' = $

• A. 1
• B. 4
• C. -4
• D. -1

Hint:

Calculus Tip: When dealing with inverse trigonometric functions and square root denominators, always cross-multiply the denominator to the other side before taking the second derivative. This transforms a messy quotient rule into a much simpler product rule.

Answer 1

The Correct Option is B

Concept: Calculus - Successive Differentiation of Inverse Trigonometric Functions. Let $y_1 = \frac{dy}{dx}$ and $y_2 = \frac{d^2y}{dx^2}$.

Step 1: Find the first derivative ($y_1$). Differentiate $y=(\sin^{-1}x)^{2}+(\cos^{-1}x)^{2}$ with respect to $x$ using the chain rule. $y_1 = 2(\sin^{-1}x)\cdot\frac{1}{\sqrt{1-x^{2}}} + 2(\cos^{-1}x)\cdot\frac{-1}{\sqrt{1-x^{2}}}$.

Step 2: Simplify and rearrange the first derivative equation. Factor out the common denominator: $y_1 = \frac{2(\sin^{-1}x - \cos^{-1}x)}{\sqrt{1-x^{2}}}$. To make further differentiation easier, multiply both sides by the denominator to avoid the quotient rule: $\sqrt{1-x^{2}}\cdot y_1 = 2(\sin^{-1}x - \cos^{-1}x)$.

Step 3: Find the second derivative ($y_2$) by differentiating again. Differentiate both sides with respect to $x$ using the product rule on the left side: $\sqrt{1-x^{2}}\cdot y_2 + y_1\cdot\frac{1}{2\sqrt{1-x^{2}}}(-2x) = 2\left(\frac{1}{\sqrt{1-x^{2}}} - \frac{-1}{\sqrt{1-x^{2}}}\right)$.

Step 4: Simplify the differentiated equation. Simplify the terms on both sides. The left side becomes $\sqrt{1-x^{2}}\cdot y_2 - \frac{x\cdot y_1}{\sqrt{1-x^{2}}}$. The right side simplifies to $2\left(\frac{2}{\sqrt{1-x^{2}}}\right) = \frac{4}{\sqrt{1-x^{2}}}$. So we have: $\sqrt{1-x^{2}}\cdot y_2 - \frac{x\cdot y_1}{\sqrt{1-x^{2}}} = \frac{4}{\sqrt{1-x^{2}}}$.

Step 5: Multiply through to reach the final form. To eliminate the fractional denominators, multiply every single term in the entire equation by $\sqrt{1-x^{2}}$. $\sqrt{1-x^{2}}\cdot(\sqrt{1-x^{2}}\cdot y_2) - \sqrt{1-x^{2}}\cdot\left(\frac{x\cdot y_1}{\sqrt{1-x^{2}}}\right) = \sqrt{1-x^{2}}\cdot\left(\frac{4}{\sqrt{1-x^{2}}}\right)$. This cleanly reduces to $(1-x^{2})y_2 - xy_1 = 4$. This expression perfectly matches the target expression asked for in the question, revealing the answer to be the constant 4. $$ \therefore \text{The value of } (1-x^{2})y_{2}-xy_{1} \text{ is } 4. $$