Question

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

Answer 1

We have,
$y^x-x^y=1\cdots \left(1\right)$
At $x=1,$
$y=1+1=2$
Now, from $\left(1\right)$
$e^{x\ln y}-e^{y\ln x}=1$
On differentiating w.r.t. $x,$ we get
$e^{x\ln y}(\frac{x}{y}\cdot \frac{dy}{dx}+\ln y)-e^{y\ln x}(\frac{dy}{dx}\ln x+\frac{y}{x})=0$
$\Rightarrow y^x(\frac{x}{y}\cdot \frac{dy}{dx}+\ln y)-x^y(\frac{dy}{dx}\ln x+\frac{y}{x})=0$
On putting $x=1$ and $y=2,$ we get
$2(\frac{1}{2}\cdot \frac{dy}{dx}+\ln 2)-(0+2)=0$
$\Rightarrow \frac{dy}{dx}=2-2\ln 2$
$=2(1-\ln 2)$