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Question
If y = tan−1 x, show that \[\left( 1 + x^2 \right) \frac{d^2 y}{d x^2} + 2x\frac{dy}{dx} = 0\] ?
Solution
Here,
\[y = \tan^{- 1} x\]
\[\text { Differentiating w . r . t . x, we get }\]
\[\frac{d y}{d x} = \frac{1}{1 + x^2}\]
\[\text { Differentiating again w . r . t . x, we get }\]
\[\frac{d^2 y}{d x^2} = \frac{- 2x}{\left( 1 + x^2 \right)^2}\]
\[ \Rightarrow \frac{d^2 y}{d x^2} = \frac{- 2x}{1 + x^2} \times \frac{1}{1 + x^2}\]
\[ \Rightarrow \frac{d^2 y}{d x^2} = \frac{- 2x\frac{dy}{dx}}{1 + x^2}\]
\[ \Rightarrow \left( 1 + x^2 \right)\frac{d^2 y}{d x^2} = - 2x\frac{dy}{dx}\]
\[ \Rightarrow \left( 1 + x^2 \right)\frac{d^2 y}{d x^2} + 2x\frac{dy}{dx} = 0\]
Hence proved.
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