# Q15  Find dy/dx of the functions given in Exercises 12 to 15.  $xy = e ^{x-y}$

Given function is
$f(x)\Rightarrow xy = e ^{x-y}$
Now, take  take log on both the sides
$\log x+\ log y = (x-y)(1) \ \ \ \ \ \ \ \ \ \ \ \ (\because \log e = 1)\\ \log x+\ log y = (x-y)$
Now, differentiate w.r.t  x
$\frac{1}{x}+\frac{1}{y}\frac{dy}{dx}=1-\frac{dy}{dx}$
By taking similar terms on same side
We get,
$(\frac{1}{y}+1)\frac{dy}{dx}=1-\frac{1}{x}\\ \frac{y+1}{y}.\frac{dy}{dx}= \frac{x-1}{x}\\ \frac{dy}{dx}= \frac{y}{x}.\frac{x-1}{y+1}$
Therefore, the answer is  $\frac{y}{x}.\frac{x-1}{y+1}$

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