If \(y = \sin^{-1}x\) and \((1 - x^2)\displaystyle\frac{d^2y}{dx^2} - x\frac{dy}{dx} = K\), then value of K is:
Answer & explanation
Correct answer: option 3
The correct answer is Option (3) → 0
Differentiating \(y = \sin^{-1}x\):
\( \displaystyle \frac{dy}{dx} = \frac{1}{\sqrt{1 - x^2}} \)
\( \displaystyle \sqrt{1 - x^2}\,\frac{dy}{dx} = 1 \)
Differentiating again, by the product rule:
\( \displaystyle \sqrt{1 - x^2}\,\frac{d^2y}{dx^2} - \frac{x}{\sqrt{1 - x^2}}\,\frac{dy}{dx} = 0 \)
Multiplying by \(\sqrt{1 - x^2}\):
\( \displaystyle (1 - x^2)\frac{d^2y}{dx^2} - x\frac{dy}{dx} = 0 \)
So K = 0.
NCERT (2026-27 edition):
Second Order Derivative
