Question

# If ${x}^{y}={e}^{x-y},\mathrm{then}\frac{dy}{dx}$ is (a) $\frac{1+x}{1+\mathrm{log}x}$ (b) $\frac{1-\mathrm{log}x}{1+\mathrm{log}x}$ (c) not defined (d) $\frac{\mathrm{log}x}{{\left(1+\mathrm{log}x\right)}^{2}}$

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Solution

## (d) $\frac{\mathrm{log}x}{{\left(1+\mathrm{log}x\right)}^{2}}$ $\mathrm{We}\mathrm{have},{x}^{y}={e}^{x-y}\phantom{\rule{0ex}{0ex}}\mathrm{Taking}\mathrm{log}\mathrm{on}\mathrm{both}\mathrm{sides}\mathrm{we}\mathrm{get},\phantom{\rule{0ex}{0ex}}⇒y\mathrm{log}x=\left(x-y\right){\mathrm{log}}_{e}e\phantom{\rule{0ex}{0ex}}⇒y\mathrm{log}x=x-y\phantom{\rule{0ex}{0ex}}⇒y\mathrm{log}x+y=x\phantom{\rule{0ex}{0ex}}⇒y\left(1+\mathrm{log}x\right)=x\phantom{\rule{0ex}{0ex}}⇒y=\frac{x}{1+\mathrm{log}x}\phantom{\rule{0ex}{0ex}}$ $⇒\frac{dy}{dx}=\frac{\left(1+\mathrm{log}x\right)×1-x×\left(0+\frac{1}{x}\right)}{{\left(1+\mathrm{log}x\right)}^{2}}\phantom{\rule{0ex}{0ex}}⇒\frac{dy}{dx}=\frac{1+\mathrm{log}x-1}{{\left(1+\mathrm{log}x\right)}^{2}}\phantom{\rule{0ex}{0ex}}⇒\frac{dy}{dx}=\frac{\mathrm{log}x}{{\left(1+\mathrm{log}x\right)}^{2}}$

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