If xy.yx=16, then dydx at (2,2) is

If $x^{y} . y^{x} = 16$ , then $\frac{d y}{d x}$ at $(2, 2)$ is

A
$- 1$
B
0
C
1
D
$\frac{1}{2}$

Solution:

$\log_{e} x^{y} + \log_{e} y^{x} = \log_{e} 16$

$y \log x + x \log y = 4 \log_{e} 2$

$\frac{y}{x} + (\log_{e} x) y^{1} + \frac{x}{y} . y^{1} + (\log y) = 0$

$y^{1} = \frac{d y}{d x} = \frac{- (\log_{e} y + \frac{y}{x})}{(\log_{e} x + \frac{x}{y})}$

${\frac{d y}{d x}}_{at (2, 2)} = - 1$

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