Question

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

Answer 1

$$\begin{gathered} \mathrm{We}\mathrm{have},\log \sqrt{x^2+y^2}={\tan }^{-1}\left(\frac{x}{y}\right) \\ \Rightarrow \log {\left(x^2+y^2\right)}^{\frac{1}{2}}={\tan }^{-1}\left(\frac{y}{x}\right) \\ \Rightarrow \frac{1}{2}\log \left(x^2+y^2\right)={\tan }^{-1}\left(\frac{y}{x}\right) \end{gathered}$$
Differentiate with respect to x, we get,
$$\begin{gathered} \Rightarrow \frac{1}{2}\frac{d}{dx}\log \left(x^2+y^2\right)=\frac{d}{dx}{\tan }^{-1}\left(\frac{y}{x}\right) \\ \Rightarrow \frac{1}{2}\left(\frac{1}{x^2+y^2}\right)\frac{d}{dx}\left(x^2+y^2\right)=\frac{1}{1+{\left({\displaystyle \frac{y}{x}}\right)}^2}\frac{d}{dx}\left(\frac{y}{x}\right) \\ \Rightarrow \frac{1}{2}\left(\frac{1}{x^2+y^2}\right)\left[2x+2y\frac{\mathit{d}\mathit{y}}{\mathit{d}\mathit{x}}\right]=\frac{x^2}{\left(x^2+y^2\right)}\left[\frac{x{\displaystyle \frac{\mathit{d}\mathit{y}}{\mathit{d}\mathit{x}}}-y{\displaystyle \frac{d}{dx}}\left(x\right)}{x^2}\right] \\ \Rightarrow \left(\frac{1}{x^2+y^2}\right)\left(x+y\frac{\mathit{d}\mathit{y}}{\mathit{d}\mathit{x}}\right)=\frac{x^2}{\left(x^2+y^2\right)}\left[\frac{x{\displaystyle \frac{\mathit{d}\mathit{y}}{\mathit{d}\mathit{x}}}-y{\displaystyle \frac{d}{dx}}\left(x\right)}{x^2}\right] \\ \Rightarrow \left(\frac{1}{x^2+y^2}\right)\left(x+y\frac{\mathit{d}\mathit{y}}{\mathit{d}\mathit{x}}\right)=\frac{x^2}{\left(x^2+y^2\right)}\left[\frac{x{\displaystyle \frac{\mathit{d}\mathit{y}}{\mathit{d}\mathit{x}}}-{\displaystyle y\left(1\right)}}{x^2}\right] \\ \Rightarrow x+y\frac{\mathit{d}\mathit{y}}{\mathit{d}\mathit{x}}=x\frac{\mathit{d}\mathit{y}}{\mathit{d}\mathit{x}}-y \\ \Rightarrow y\frac{\mathit{d}\mathit{y}}{\mathit{d}\mathit{x}}-x\frac{\mathit{d}\mathit{y}}{\mathit{d}\mathit{x}}=-y-x \\ \Rightarrow \frac{\mathit{d}\mathit{y}}{\mathit{d}\mathit{x}}\left(y-x\right)=-\left(y+x\right) \\ \Rightarrow \frac{\mathit{d}\mathit{y}}{\mathit{d}\mathit{x}}=\frac{-\left(y+x\right)}{y-x} \\ \Rightarrow \frac{\mathit{d}\mathit{y}}{\mathit{d}\mathit{x}}=\frac{x+y}{x-y} \end{gathered}$$