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Question
If y = cos−1 x, find \[\frac{d^2 y}{d x^2}\] in terms of y alone ?
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Solution
Here,
\[y = \cos^{- 1} x\]
\[\text { Differentiating w . r . t . x, we get }\]
\[\frac{d y}{d x} = \frac{- 1}{\sqrt{1 - x^2}}\]
\[\text { Differentiating again w . r . t . x, we get }\]
\[\frac{d^2 y}{d x^2} = \frac{- 2x}{2 \sqrt{1 - x^2}^{3/2}} = \frac{- x}{\left( 1 - x^2 \right)^{3/2}}\]
\[\text { Now, } \]
\[y = \cos^{- 1} x\]
\[ \Rightarrow x = \cos y\]
\[ \Rightarrow \frac{d^2 y}{d x^2} = \frac{- \cos y}{\left( 1 - \cos^2 y \right)^{3/2}} = - \frac{\cos y}{\left( \sin^2 y \right)^{3/2}} = - \cot y \ {cosec}^2 y\]
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