Question

# Differentiate the following with respect to x: (i) ${\mathrm{cos}}^{-1}\left(\mathrm{sin}x\right)$ (ii) ${\mathrm{sin}}^{-1}\left(\frac{{2}^{x+1}}{1+{4}^{x}}\right)$

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Solution

## $\left(\mathrm{i}\right)\mathrm{Let},f\left(x\right)={\mathrm{cos}}^{-1}\left(\mathrm{sin}x\right)\phantom{\rule{0ex}{0ex}}⇒f\left(x\right)={\mathrm{cos}}^{-1}\left[\mathrm{cos}\left(\frac{\mathrm{\pi }}{2}-x\right)\right]\phantom{\rule{0ex}{0ex}}⇒f\left(x\right)=\frac{\mathrm{\pi }}{2}-x\phantom{\rule{0ex}{0ex}}\mathrm{Thus},f\text{'}\left(x\right)=\frac{d}{dx}\left(\frac{\mathrm{\pi }}{2}-x\right)=-1$ $\left(\mathrm{ii}\right)\mathrm{Let},y={\mathrm{sin}}^{-1}\left(\frac{{2}^{x+1}}{1+{4}^{x}}\right)\phantom{\rule{0ex}{0ex}}\mathrm{Put}{2}^{x}=\mathrm{tan}\theta \phantom{\rule{0ex}{0ex}}\therefore y={\mathrm{sin}}^{-1}\left(\frac{{2}^{x}×2}{1+{\left({2}^{x}\right)}^{2}}\right)\phantom{\rule{0ex}{0ex}}⇒y={\mathrm{sin}}^{-1}\left(\frac{2\mathrm{tan}\theta }{1+{\mathrm{tan}}^{2}\theta }\right)\phantom{\rule{0ex}{0ex}}⇒y={\mathrm{sin}}^{-1}\left\{\mathrm{sin}\left(2\theta \right)\right\}\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}⇒y=2\theta \left[\mathrm{Since},{\mathrm{sin}}^{-1}\left(\mathrm{sin}\theta \right)=\theta \right]\phantom{\rule{0ex}{0ex}}⇒y=2{\mathrm{tan}}^{-1}\left({2}^{x}\right)\left[\mathrm{Since},{2}^{x}=\mathrm{tan}\theta \right]$ Differentiate with respect to x using chain rule, $\frac{dy}{dx}=\frac{2}{1+{\left({2}^{x}\right)}^{2}}\frac{d}{dx}\left({2}^{x}\right)\phantom{\rule{0ex}{0ex}}⇒\frac{dy}{dx}=\frac{2×{2}^{x}\mathrm{log}2}{1+{4}^{x}}\phantom{\rule{0ex}{0ex}}\therefore \frac{dy}{dx}=\frac{{2}^{x+1}}{1+{4}^{x}}\mathrm{log}2$

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