Question

# Differentiate ${\mathrm{tan}}^{-1}\left(\frac{1-x}{1+x}\right)$ with respect to $\sqrt{1-{x}^{2}},\mathrm{if}-1

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Solution

## $\mathrm{Let},u={\mathrm{tan}}^{-1}\left(\frac{1-x}{1+x}\right)\phantom{\rule{0ex}{0ex}}\mathrm{Put}x=\mathrm{tan}\theta \phantom{\rule{0ex}{0ex}}⇒\theta ={\mathrm{tan}}^{-1}x\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}⇒u={\mathrm{tan}}^{-1}\left(\frac{1-\mathrm{tan}\theta }{1+\mathrm{tan}\theta }\right)\phantom{\rule{0ex}{0ex}}⇒u={\mathrm{tan}}^{-1}\left[\mathrm{tan}\left(\frac{\mathrm{\pi }}{4}-\theta \right)\right]...\left(\mathrm{i}\right)\phantom{\rule{0ex}{0ex}}\mathrm{Here},\phantom{\rule{0ex}{0ex}}-1-\theta >\frac{\mathrm{\pi }}{4}\phantom{\rule{0ex}{0ex}}⇒-\frac{\mathrm{\pi }}{4}<-\theta <\frac{\mathrm{\pi }}{4}\phantom{\rule{0ex}{0ex}}⇒0<\frac{\mathrm{\pi }}{4}-\theta <\frac{\mathrm{\pi }}{2}\phantom{\rule{0ex}{0ex}}\mathrm{So},\mathrm{from}\mathrm{equation}\left(\mathrm{i}\right),\phantom{\rule{0ex}{0ex}}u=\frac{\mathrm{\pi }}{4}-\theta \left[\mathrm{Since},{\mathrm{tan}}^{-1}\left(\mathrm{tan}\theta \right)=\theta ,\mathrm{if}\theta \in \left(-\frac{\mathrm{\pi }}{2},\frac{\mathrm{\pi }}{2}\right)\right]\phantom{\rule{0ex}{0ex}}⇒u=\frac{\mathrm{\pi }}{4}-{\mathrm{tan}}^{-1}x$ Differentiating it with respect to x, $\frac{du}{dx}=0-\left(\frac{1}{1+{x}^{2}}\right)\phantom{\rule{0ex}{0ex}}⇒\frac{du}{dx}=-\frac{1}{1+{x}^{2}}...\left(\mathrm{ii}\right)\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}\mathrm{And}\mathrm{let},v=\sqrt{1-{x}^{2}}$ Differentiating it with respect to x, $\frac{dv}{dx}=\frac{1}{2\sqrt{1-{x}^{2}}}×\frac{d}{dx}\left(1-{x}^{2}\right)\phantom{\rule{0ex}{0ex}}⇒\frac{dv}{dx}=\frac{1}{2\sqrt{1-{x}^{2}}}\left(-2x\right)\phantom{\rule{0ex}{0ex}}⇒\frac{dv}{dx}=\frac{-x}{\sqrt{1-{x}^{2}}}...\left(\mathrm{iii}\right)\phantom{\rule{0ex}{0ex}}\mathrm{Dividing}\mathrm{equation}\left(\mathrm{ii}\right)\mathrm{by}\left(\mathrm{iii}\right),\phantom{\rule{0ex}{0ex}}\frac{\frac{du}{dx}}{\frac{dv}{dx}}=-\frac{1}{1+{x}^{2}}×\frac{\sqrt{1-{x}^{2}}}{-x}\phantom{\rule{0ex}{0ex}}\therefore \frac{du}{dv}=\frac{\sqrt{1-{x}^{2}}}{x\left(1+{x}^{2}\right)}$

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