Question

If log(x2+y2)=2tan-1(y/x), then show that dy/dx=x+y/x-y.

Answer 1

here is your answer.....

Answer 2

$\bold{\frac{d y}{d x}=\frac{x+y}{x-y} \text { if } \log \left(x^{2}+y^{2}\right)=2 \tan ^{-1} \frac{y}{x}}$

Given:  

$\log \left(x^{2}+y^{2}\right)=2 \tan ^{-1} \frac{y}{x}$

To Prove:

$\frac{d y}{d x}=\frac{x+y}{x-y}$

Solution:  

$\log \left(x^{2}+y^{2}\right)=2 \tan ^{-1} \frac{y}{x}$

We need to find the derivative of y with respect to x.  

Differentiating with respect to x on both sides.  

$\log \left(x^{2}+y^{2}\right)=2 \tan ^{-1} \frac{y}{x}$

$\frac{1}{x^{2}+y^{2}}\left(2 x+2 y\left(\frac{d y}{d x}\right)\right)=\frac{2\left(\frac{x\left(\frac{d y}{d x}\right)-y(1)}{x^{2}}\right)}{\left(1+\frac{y^{2}}{x^{2}}\right)}$

Where we know that $\frac{dy}{dx} =y_1$

$\frac{2\left(x+y y_{1}\right)}{x^{2}+y^{2}}=\frac{2\left(x y_{1}-y\right)}{x^{2}+y^{2}}$

Cancelling $x^2+y^2$ on both sides and multiplying 2 we get  

$2 x+2 y y_{1}=2 x y_{1}-2 y$

Cancelling out 2 on both sides  

$\begin{array}{l}{x+y y_{1}=x y_{1}-y} \\ {x+y=x y_{1}-y y_{1}}\end{array}$

Shifting the term which has $y_1$ on one side  

$\begin{array}{l}{x+y=y_{1}(x-y)} \\ {y_{1}=\frac{x+y}{x-y}}\end{array}$

Substituting $\frac{d y}{d x}=y_{1}$

$\bold{\frac{d y}{d x}=\frac{x+y}{x-y}}$

Hence proved.