If \(y = \cos^{-1}x\), \(-1 \lt x \lt 1\), then \(\displaystyle\frac{d^{2}y}{dx^{2}}\), written in terms of y alone, is equal to
Answer & explanation
Correct answer: option 1
The correct answer is Option (1) → \(-\mathrm{cosec}^{2}y\cot y\)
Since \(y = \cos^{-1}x\), we have \(x = \cos y\) with \(0 \lt y \lt \pi\), so \(\sin y \gt 0\) and \(\sqrt{1 - x^{2}} = \sin y\).
\( \displaystyle \frac{dy}{dx} = -\frac{1}{\sqrt{1 - x^{2}}} = -\frac{1}{\sin y} = -\mathrm{cosec}\,y \)
For the second derivative, differentiate with respect to x using the chain rule (y depends on x):
\( \displaystyle \frac{d^{2}y}{dx^{2}} = \frac{d}{dy}\left(-\mathrm{cosec}\,y\right) \cdot \frac{dy}{dx} \)
\( \displaystyle = \mathrm{cosec}\,y\cot y \cdot \left(-\mathrm{cosec}\,y\right) \)
\( = -\mathrm{cosec}^{2}y\cot y \)