Question

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

• A. $2(1-\log 2)$
• B. $2(1+\log 2)$
• C. $2-\log 2$
• D. $2+\log 2$

Answer 1

Answer: (C)

$2-\log 2$

We have
$y^x-x^y=\mathrm{1....}\left(i\right)$
$\Rightarrow e^{x\log y}-e^{y\log x}=1$
On differentiating w.r.t. $x$ we get
$y^x(\frac{x}{y}.\frac{dy}{dx}+\log y)-x^y(\frac{dy}{dx}\log x+\frac{y}{x})=0$
On putting $x=1$ and $y=2$ we get
$2(\frac{1}{2}.\frac{dy}{dx}+\log 2)-(0+2)=0$
$\frac{dy}{dx}=2-2\log 2$