Question

# If $\mathrm{log}\sqrt{{x}^{2}+{y}^{2}}={\mathrm{tan}}^{-1}\left(\frac{y}{x}\right)$, prove that $\frac{dy}{dx}=\frac{x+y}{x-y}$

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Solution

## $\mathrm{We}\mathrm{have},\mathrm{log}\sqrt{{x}^{2}+{y}^{2}}={\mathrm{tan}}^{-1}\left(\frac{x}{y}\right)\phantom{\rule{0ex}{0ex}}⇒\mathrm{log}{\left({x}^{2}+{y}^{2}\right)}^{\frac{1}{2}}={\mathrm{tan}}^{-1}\left(\frac{y}{x}\right)\phantom{\rule{0ex}{0ex}}⇒\frac{1}{2}\mathrm{log}\left({x}^{2}+{y}^{2}\right)={\mathrm{tan}}^{-1}\left(\frac{y}{x}\right)$ Differentiate with respect to x, we get, $⇒\frac{1}{2}\frac{d}{dx}\mathrm{log}\left({x}^{2}+{y}^{2}\right)=\frac{d}{dx}{\mathrm{tan}}^{-1}\left(\frac{y}{x}\right)\phantom{\rule{0ex}{0ex}}⇒\frac{1}{2}\left(\frac{1}{{x}^{2}+{y}^{2}}\right)\frac{d}{dx}\left({x}^{2}+{y}^{2}\right)=\frac{1}{1+{\left(\frac{y}{x}\right)}^{2}}\frac{d}{dx}\left(\frac{y}{x}\right)\phantom{\rule{0ex}{0ex}}⇒\frac{1}{2}\left(\frac{1}{{x}^{2}+{y}^{2}}\right)\left[2x+2y\frac{\mathit{d}\mathit{y}}{\mathit{d}\mathit{x}}\right]=\frac{{x}^{2}}{\left({x}^{2}+{y}^{2}\right)}\left[\frac{x\frac{\mathit{d}\mathit{y}}{\mathit{d}\mathit{x}}-y\frac{d}{dx}\left(x\right)}{{x}^{2}}\right]\phantom{\rule{0ex}{0ex}}⇒\left(\frac{1}{{x}^{2}+{y}^{2}}\right)\left(x+y\frac{\mathit{d}\mathit{y}}{\mathit{d}\mathit{x}}\right)=\frac{{x}^{2}}{\left({x}^{2}+{y}^{2}\right)}\left[\frac{x\frac{\mathit{d}\mathit{y}}{\mathit{d}\mathit{x}}-y\frac{d}{dx}\left(x\right)}{{x}^{2}}\right]\phantom{\rule{0ex}{0ex}}⇒\left(\frac{1}{{x}^{2}+{y}^{2}}\right)\left(x+y\frac{\mathit{d}\mathit{y}}{\mathit{d}\mathit{x}}\right)=\frac{{x}^{2}}{\left({x}^{2}+{y}^{2}\right)}\left[\frac{x\frac{\mathit{d}\mathit{y}}{\mathit{d}\mathit{x}}-y\left(1\right)}{{x}^{2}}\right]\phantom{\rule{0ex}{0ex}}⇒x+y\frac{\mathit{d}\mathit{y}}{\mathit{d}\mathit{x}}=x\frac{\mathit{d}\mathit{y}}{\mathit{d}\mathit{x}}-y\phantom{\rule{0ex}{0ex}}⇒y\frac{\mathit{d}\mathit{y}}{\mathit{d}\mathit{x}}-x\frac{\mathit{d}\mathit{y}}{\mathit{d}\mathit{x}}=-y-x\phantom{\rule{0ex}{0ex}}⇒\frac{\mathit{d}\mathit{y}}{\mathit{d}\mathit{x}}\left(y-x\right)=-\left(y+x\right)\phantom{\rule{0ex}{0ex}}⇒\frac{\mathit{d}\mathit{y}}{\mathit{d}\mathit{x}}=\frac{-\left(y+x\right)}{y-x}\phantom{\rule{0ex}{0ex}}⇒\frac{\mathit{d}\mathit{y}}{\mathit{d}\mathit{x}}=\frac{x+y}{x-y}$

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