Question

If $log(x^2+y^2{)}^{(}0.5)=ta{n}^{-}1\left(\frac{x}{y}\right)$ then show that $\frac{dy}{dx}=\frac{(y-x)}{(y+x)}$

Answer 1

$log\left(\sqrt{x^2+y^2}\right)$
Let$x^2+y^2=t$ $2x+dx+2ydy=dt$
$2x+2y\frac{dy}{dx}=\frac{dt}{dx}$
$\frac{d}{dx}log\sqrt{t}\left(\frac{d}{dt}log\sqrt{t}\right)\left(\frac{dt}{dx}\right)=(\frac{1}{\sqrt{t}}.\frac{1}{2\sqrt{t}})(2x+2y{y}^{\prime })=\frac{1}{2t}\frac{\text{/}2(x+y{y}^{\prime })}{1}=\frac{1}{x^2{y}^2}(x+y{y}^{\prime })$ ......(2)
$\frac{d}{dx}ta{n}^{-1}\left(\frac{x}{y}\right)$
Let $\frac{x}{y}=tx=yt$
or $dx=dy+ydt$
$1-\frac{dy}{dx}+y\frac{dt}{dx}\frac{dt}{dx}=\frac{1-\frac{dy}{dx}t}{y}\frac{d}{dx}ta{n}^{-1}=\frac{d}{dt}\left(ta{n}^{-1}t\right)\frac{dt}{dx}=\frac{1}{1+t^2}\left(\frac{1-y^{\prime }t}{y}\right)=\frac{1}{1+t^2}\left(\frac{1-y^{\prime }t}{y}\right)=\frac{1}{1+\frac{x^2}{y^2}}\left\{\frac{1-y^{\prime }\left(\frac{x}{y}\right)}{y}\right\}=\frac{y^2}{y^2+x^2}\left\{\frac{y-y^{\prime }\left(x\right)}{y^2}\right\}$ ---(1)
Equality $\left(1\right)\&\left(2\right)$
$\frac{y-y^{\prime }x}{x^2+y^2}=\frac{x+y{y}^{\prime }}{x^2+y^2}y-x=y^{\prime }x+y^{\prime }y=y^{\prime }(x+y)\therefore y^{\prime }=\frac{y-x}{x+y}$