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  • If y = tan^-1x, find d^2y/dx^2 in terms of y alone.

If y = tan^{-1}x  , find   \frac{d^2y}{dx^2}   in terms of y alone.

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Solution

Given: y = tan^{-1}x

\\\Rightarrow \tan y=x

On differentiating Eq. (i) with respect to x, we get 

\\\frac{d y}{d x}=\frac{1}{1+x^{2}}
Now, again differentiating the above with respect to x, we get

\\\frac{d^{2} y}{d x^{2}}=\frac{\left(1+x^{2}\right) \times \frac{d}{d x}(1)-1 \times \frac{d}{d x}\left(1+x^{2}\right)}{\left(1+x^{2}\right)^{2}} \\\Rightarrow \frac{\mathrm{d}^{2} \mathrm{y}}{\mathrm{dx}^{2}}=\frac{0-2 \mathrm{x}}{\left(1+\mathrm{x}^{2}\right)^{2}} \\\Rightarrow \frac{\mathrm{d}^{2} \mathrm{y}}{\mathrm{dx}^{2}}=\frac{-2 \tan \mathrm{y}}{\left(1+\tan ^{2} \mathrm{y}\right)^{2}} \\\Rightarrow \frac{\mathrm{d}^{2} \mathrm{y}}{\mathrm{dx}^{2}}=\frac{-2 \tan\mathrm{y}}{\left(\mathrm{sec}^{2} \mathrm{y}\right)^{2}}

\\ \Rightarrow \frac{d^{2} y}{d x^{2}}=\frac{-2 \sin y}{\cos y} \times \cos ^{4} y \\ \Rightarrow \frac{d^{2} y}{d x^{2}}=-2 \sin y \cos ^{3} y

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infoexpert22

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