If x=yx, then dydx=

If $x = y^{x}$ , then $\frac{d y}{d x} =$

A
$\frac{y (1 - x \log y)}{x}$
B
$\frac{y (1 + x \log y)}{x^{2}}$
C
$\frac{y (1 - x \log y)}{x^{2}}$
D
$\frac{- y (1 - x \log y)}{x}$

Solution:

$x = y^{x}$

$\log x = x \log y$

$\frac{1}{x} = \frac{x}{y} (y^{1}) + \log y$

$\frac{1 - x \log y}{x} = (\frac{x}{y}) y^{1}$

$\frac{d y}{d x} = \frac{y (1 - x \log y)}{x^{2}}$

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