Question

If $y^x-x^y=1$ then $\frac{dy}{dx}$ at $x=1$ is

• A. $2(1-\ln 2)$
• B. $2(1+\ln 2)$
• C. $\ln \left(4e^2\right)$
• D. $\ln \left(\frac{e}{4}\right)$

Answer 1

The correct option is A $2(1-\ln 2)$

at $x=1$ $\Rightarrow$ $y=2$ ,
Let $y^x=u$ &

$x^y=v$ $\Rightarrow u-v=1$

Differentiating $\Rightarrow$ $u^{\prime }-v^{\prime }=0$

$y^x=u$	$x^y=v$
$x\ln y=\ln u$	$y\ln x=\ln v$
$x\frac{1}{y}.\frac{dy}{dx}+\ln y.1=\frac{1}{u}.\frac{du}{dx}$	$y.\frac{1}{x}+\ln x.\frac{dy}{dx}=\frac{1}{v}.\frac{dv}{dx}$
$\frac{du}{dx}=y^x(\frac{x}{y}.\frac{dy}{dx}+\ln y)$	$\frac{dv}{dx}=x^y(\frac{y}{x}+\ln x.\frac{dy}{dx})$

$\frac{du}{dx}-\frac{dv}{dx}=0$
$x.y^{x-1}\frac{dy}{dx}+y^x\ln y-y.x^{y-1}-x^y\ln x.\frac{dy}{dx}=0$
Put $x=1$ & $y=2$
$\frac{dy}{dx}+2\ln 2-2=0$
$\frac{dy}{dx}=2(1-\ln 2)$