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

 

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best_answer

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} 

Posted by

Gautam harsolia

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