Question

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

• A. $\frac{x+y}{x-y}$
• B. $\frac{x-y}{x+y}$
• C. $\frac{1}{x+y}$
• D. None of these

Answer 1

The correct option is A $\frac{x+y}{x-y}$

Differentiating both sides of the given relation with respect to x, we get,

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

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

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

$\frac{1}{x^2+y^2}[2x+2y\frac{dy}{dx}]=\frac{2}{x^2+y^2}[x\frac{dy}{dx}-y]$

$2[x+y\frac{dy}{dx}]=2[x\frac{dy}{dx}-y]$

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

$\frac{dy}{dx}(y-x)=-(x+y)$

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