Answer:
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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