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

$2(1-\log 2)$

We have,

$y^x-x^y=1$            .....(i)

$\Rightarrow e^{x \log y}-e^{y \log x}=1$

Differentiating with respect to x, we get

$y^x\left\{\frac{x}{y} \frac{d y}{d x}+\log y\right\}-x^y\left\{\frac{d y}{d x} \log x+\frac{y}{x}\right\}=0$

Putting x = 1, y = 2, we get

$2\left(\frac{1}{2} \frac{d y}{d x}+\log 2\right)-(0+2)=0$

$\Rightarrow \frac{d y}{d x}=2-2 \log 2$