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

 

 

 

 
 
 
 
 

Answers (1)

Given \log(x^2 + y^2) = 2\tan^{-1}\left (\frac{y}{x} \right ). Show that \frac{dy}{dx} = \frac{x + y}{x -y}

Differentiating both sides w.r.t x we get

\frac{1}{x^2+y^2}\left(2x +2y\frac{dy}{dx}\right) = 2\frac{1}{1+\frac{y^2}{x^2}}\times \left [ \frac{\frac{dy}{dx}x - y}{x^2}\right ]

x +y\frac{dy}{dx} = {\frac{dy}{dx}x - y}

x +y = {\frac{dy}{dx}x - y\frac{dy}{dx}}

\frac{dy}{dx} = \frac{x + y}{x-y}

Hence proved.

Posted by

Ravindra Pindel

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