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If $e^{x+y}=xy,$ what is $\large\frac{dy}{dx}$?

$\begin{array}{1 1}(A)\;\large\frac{y(1-x)}{x(y-1)}\\(B)\;\large\frac{y(1+x)}{x(y-1)}\\(C)\;\large\frac{y(1-x)}{x(1+y)}\\(D)\;\large\frac{y(1+x)}{x(1-y)}\end{array}$

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Given $e^{x+y} = xy$
Taking log on both sides, we get: $ (x+y) \log_e e = \log_e x + \log_e y$
$\Rightarrow x+y = \log_e x + \log_e y$
Differentiating, we get: $dx + dy = \large\frac{1}{x}$$dx + \large\frac{1}{y}$$dy$
Rearranging, we get: $\large\frac{dy}{dx}$$(1 - \large\frac{1}{y}$$) = \large\frac{1}{x}$$-1$
$\Rightarrow \large\frac{dy}{dx} $$ = \large\frac{y(1-x)}{x(y-1)}$
answered Jan 29, 2014 by balaji
 

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