CUET · MATHS · PYQ PAPER 2023
If \(y =\sin ^{-1} x\), then
- A \(\frac{d^2 y}{d x^2}=\frac{x}{1+x^2} \frac{d y}{d x}\)
- B \(\frac{d^2 y}{d x^2}=\frac{x}{1-x^2} \frac{d y}{d x}\)
- C \(\frac{d^2 y}{d x^2}=\frac{-x}{1+x^2} \frac{d y}{d x}\)
- D \(\frac{d y}{d x}=\frac{x}{1+x^2} \frac{d^2 y}{d x^2}\)
Answer & Solution
Correct Answer
(B) \(\frac{d^2 y}{d x^2}=\frac{x}{1-x^2} \frac{d y}{d x}\)
Step-by-step Solution
Detailed explanation
\(\frac{dy}{dx} = (1-x^2)^{-1/2}\) \(\frac{d^2y}{dx^2} = -\frac{1}{2}(1-x^2)^{-3/2}(-2x)\) \(\frac{d^2y}{dx^2} = x(1-x^2)^{-3/2}\) \(\frac{d^2y}{dx^2} = \frac{x}{1-x^2} \cdot (1-x^2)^{-1/2}\) \(\frac{d^2y}{dx^2} = \frac{x}{1-x^2} \frac{dy}{dx}\)
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