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

Given, $y={{\tan }^{-1}}x$ $\Rightarrow$   $\tan \,\,y=x$ Now differentiating Eq. (i) w,r.t. $'x'$, we get $\frac{dy}{dx}=\frac{1}{1+{{x}^{2}}}$             Again differentiating w.r.t. $'x'$, we get                         $\frac{{{d}^{2}}y}{d{{x}^{2}}}=-\,{{(1+{{x}^{2}})}^{-\,2}}(2x)$ $=\frac{-\,2x}{{{(1+{{x}^{2}})}^{2}}}=\frac{-\,2\,\,\tan \,\,y}{{{(1+{{\tan }^{2}}\,\,y)}^{2}}}$[from Eq. (ii)] $=\left( \frac{-\,2\,\,\tan \,\,y}{1+{{\tan }^{2}}\,y} \right)\cdot \frac{1}{1+{{\tan }^{2}}\,y}$ $=-\sin 2y\cdot \frac{1}{{{\sec }^{2}}y}$ $\left[ \because \frac{2\,\tan \theta }{1+{{\tan }^{2}}\theta }=\sin 2\theta and\,\,1+{{\tan }^{2}}\theta ={{\sec }^{2}}\theta \right]$                                     $=-\sin 2y\cdot {{\cos }^{2}}y$                                     $=-\,2\sin y\cdot \cos y\cdot {{\cos }^{2}}y$                                     $=-\,2\sin y\cdot {{\cos }^{3}}y$

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