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Q.
If $y^x-x^y=1$, then the value of $\frac{d y}{d x}$ at $x=1$ is equal to
Continuity and Differentiability
Solution:
$ P( x =1, y =2)$
$e ^{ x \ln y }- e ^{ y \ln x }=1$
$\therefore$ On differentiating with respect to $x$, we get
$\Rightarrow y^x\left(\frac{x}{y} \cdot y^{\prime}+\ln y\right)-x^y\left(\frac{y}{x}+y^{\prime} \cdot \ln x\right)=\left.0 \Rightarrow \frac{d y}{d x}\right|_{(x=1, y=2)}=2(1-\ln 2)$