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Find $\large\frac{dy}{dx}$ of the function $xy = e^{\large(x-y)}$

$\begin{array}{1 1} \large\frac{y(y-1)}{x(1+y)} \\ \large\frac{y(x+1)}{x(1-y)} \\ \large\frac{x(x-1)}{y(1+y)} \\ \large\frac{y(x-1)}{x(1+y)} \end{array} $

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Toolbox:
  • $\log m^{\large n}=n\log m$
  • $\log e=1$
  • $\log xy=\log x+\log y$
Step 1:
$ xy = e^{\large(x-y)} $
Taking $\log$ on both sides
$ \log xy = \log e^{\large(x-y)} $
$\log xy=\log x+\log y$
$\log x+\log y=(x-y)\log e$
$\log e=1$
Step 2:
Differentiating with respect to $x$
$\large\frac{1}{x}+\large\frac{1}{y}\frac{dy}{dx}$$=(x-y)$
$\large\frac{1}{x}+\large\frac{1}{y}\frac{dy}{dx}$$=1-\large\frac{dy}{dx}$
$\large\frac{1}{x}+\large\frac{1}{y}\frac{dy}{dx}$$=1-\large\frac{1}{x}$
$\large\frac{1}{y}\frac{dy}{dx}+\frac{dy}{dx}$$=1-\large\frac{1}{x}$
$(\large\frac{1}{y}$$+1)\large\frac{dy}{dx}=\large\frac{x-1}{x}$
$\large\big(\frac{1+y}{y}\big)\large\frac{dy}{dx}=\frac{x-1}{x}$
$\large\frac{dy}{dx}=\large\frac{y(x-1)}{x(1+y)}$
answered May 9, 2013 by sreemathi.v
 

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