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Q.
If $y^{x}=e^{y-x}$, then $\frac{d y}{d x}$ is equal to
Continuity and Differentiability
Solution:
Here, $y^{x}=e^{y-x}$
Taking log on both sides, we get
$\log y ^{ x }=\log e ^{ y - x }$
$\left(\because \log a ^{ b }= b \log a\right.$ and $\left.\log e =1\right)$
$\Rightarrow x \log y=(y-x) \log e $
$\Rightarrow x \log y=y-x \ldots(i)$
On differentiating w.r.t. $x$, we get
$\frac{ d }{ dx }( x \log y )=\frac{ d }{ dx }( y - x ) \,\,\,\,\,$ (using product rule)
$\Rightarrow x\left(\frac{1}{y}\right) \frac{d y}{d x}+\log y(1)=\frac{d y}{d x}-1 $
$\Rightarrow \frac{d y}{d x}\left(\frac{x}{y}-1\right)=-1-\log y$
$\Rightarrow \frac{d y}{d x}\left[\frac{y}{(1+\log y) y}-1\right]=-(1+\log y)$
$\left[\because\right.$ from eq. $\left.(i), x=\frac{y}{(1+\log y)}\right]$
$\Rightarrow \frac{ dy }{ dx }\left[\frac{1-1-\log y }{1+\log y }\right]=-(1+\log y )$
$ \Rightarrow \frac{ dy }{ d x }=-\frac{(1+\log y )^{2}}{-\log y }$
$\Rightarrow \frac{ dy }{ dx }=\frac{(1+\log y )^{2}}{\log y }$