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Question
If \[\log \sqrt{x^2 + y^2} = \tan^{- 1} \left( \frac{y}{x} \right)\] Prove that \[\frac{dy}{dx} = \frac{x + y}{x - y}\] ?
Solution
\[\text{ We have }, \log\sqrt{x^2 + y^2} = \tan^{- 1} \left( \frac{x}{y} \right)\]
\[ \Rightarrow \log \left( x^2 + y^2 \right)^\frac{1}{2} = \tan^{- 1} \left( \frac{y}{x} \right)\]
\[ \Rightarrow \frac{1}{2}\log\left( x^2 + y^2 \right) = \tan^{- 1} \left( \frac{y}{x} \right)\]
Differentiate with respect to x, we get,
\[\Rightarrow \frac{1}{2}\frac{d}{dx}\log\left( x^2 + y^2 \right) = \frac{d}{dx} \tan^{- 1} \left( \frac{y}{x} \right)\]
\[ \Rightarrow \frac{1}{2}\left( \frac{1}{x^2 + y^2} \right)\frac{d}{dx}\left( x^2 + y^2 \right) = \frac{1}{1 + \left( \frac{y}{x} \right)^2}\frac{d}{dx}\left( \frac{y}{x} \right)\]
\[ \Rightarrow \frac{1}{2}\left( \frac{1}{x^2 + y^2} \right)\left[ 2x + 2y\frac{d y}{d x} \right] = \frac{x^2}{\left( x^2 + y^2 \right)}\left[ \frac{x\frac{d y}{d x} - y\frac{d}{dx}\left( x \right)}{x^2} \right]\]
\[ \Rightarrow \left( \frac{1}{x^2 + y^2} \right)\left( x + y\frac{d y}{d x} \right) = \frac{x^2}{\left( x^2 + y^2 \right)}\left[ \frac{x\frac{d y}{d x} - y\frac{d}{dx}\left( x \right)}{x^2} \right]\]
\[ \Rightarrow \left( \frac{1}{x^2 + y^2} \right)\left( x + y\frac{d y}{d x} \right) = \frac{x^2}{\left( x^2 + y^2 \right)}\left[ \frac{x\frac{d y}{d x} - y\left( 1 \right)}{x^2} \right]\]
\[ \Rightarrow x + y\frac{d y}{d x} = x\frac{d y}{d x} - y\]
\[ \Rightarrow y\frac{d y}{d x} - x\frac{d y}{d x} = - y - x\]
\[ \Rightarrow \frac{d y}{d x}\left( y - x \right) = - \left( y + x \right)\]
\[ \Rightarrow \frac{d y}{d x} = \frac{- \left( y + x \right)}{y - x}\]
\[ \Rightarrow \frac{d y}{d x} = \frac{x + y}{x - y}\]
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