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Question

# Differentiate the following with respect to x: $\left(\mathrm{i}\right){\mathrm{cos}}^{-1}\left(\mathrm{sin}x\right)$ (ii) ${\mathrm{cot}}^{-1}\left(\frac{1-x}{1+x}\right)$

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Solution

## $\left(\mathrm{i}\right)\mathrm{Let},f\left(x\right)={\mathrm{cos}}^{-1}\left(\mathrm{sin}x\right)\phantom{\rule{0ex}{0ex}}⇒f\left(x\right)={\mathrm{cos}}^{-1}\left[\mathrm{cos}\left(\frac{\mathrm{\pi }}{2}-x\right)\right]\phantom{\rule{0ex}{0ex}}⇒f\left(x\right)=\frac{\mathrm{\pi }}{2}-x\phantom{\rule{0ex}{0ex}}\mathrm{Thus},f\text{'}\left(x\right)=\frac{d}{dx}\left(\frac{\mathrm{\pi }}{2}-x\right)=-1$ (ii) ${\mathrm{cot}}^{-1}\left(\frac{1-x}{1+x}\right)=\frac{d}{dx}\left[{\mathrm{tan}}^{-1}\left(\frac{1+x}{1-x}\right)\right]\phantom{\rule{0ex}{0ex}}=\frac{1}{1+{\left(\frac{1+x}{1-x}\right)}^{2}}×\frac{1-x+1+x}{{\left(1-x\right)}^{2}}\phantom{\rule{0ex}{0ex}}=\frac{{\left(1-x\right)}^{2}}{{\left(1-x\right)}^{2}+{\left(1+x\right)}^{2}}×\frac{2}{{\left(1-x\right)}^{2}}\phantom{\rule{0ex}{0ex}}=\frac{{\left(1-x\right)}^{2}}{1-2x+{x}^{2}+1+2x+{x}^{2}}\phantom{\rule{0ex}{0ex}}=\frac{{\left(1-x\right)}^{2}}{2\left(1+{x}^{2}\right)}×\frac{2}{{\left(1-x\right)}^{2}}\phantom{\rule{0ex}{0ex}}=\frac{1}{1+{x}^{2}}$

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