Question

# Differentiate ${\mathrm{cot}}^{-1}\left(\frac{1-x}{1+x}\right)$ with respect to x.

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Solution

## $\mathrm{Let},y=co{t}^{-1}\left(\frac{1-x}{1+x}\right)\phantom{\rule{0ex}{0ex}}\mathrm{Put}x=\mathrm{tan}\theta \phantom{\rule{0ex}{0ex}}\therefore y=co{t}^{-1}\left(\frac{1-\mathrm{tan}\theta }{1+\mathrm{tan}\theta }\right)\phantom{\rule{0ex}{0ex}}=co{t}^{-1}\left(\frac{\mathrm{tan}\frac{\mathrm{\pi }}{4}-\mathrm{tan}\theta }{1+\mathrm{tan}\frac{\mathrm{\pi }}{4}\mathrm{tan}\theta }\right)\phantom{\rule{0ex}{0ex}}=co{t}^{-1}\left[\mathrm{tan}\left(\frac{\mathrm{\pi }}{4}-\theta \right)\right]\phantom{\rule{0ex}{0ex}}=co{t}^{-1}\left[cot\left(\frac{\mathrm{\pi }}{2}-\frac{\mathrm{\pi }}{4}+\theta \right)\right]\phantom{\rule{0ex}{0ex}}=\frac{\mathrm{\pi }}{4}+\theta \phantom{\rule{0ex}{0ex}}=\frac{\mathrm{\pi }}{4}+{\mathrm{tan}}^{-1}x\left[\mathrm{Since},x=\mathrm{tan}\theta \right]\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}$ Differentiate it with respect to x, $\frac{dy}{dx}=0+\frac{1}{1+{x}^{2}}\phantom{\rule{0ex}{0ex}}⇒\frac{dy}{dx}=\frac{1}{1+{x}^{2}}$

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