If x^y = y^x , then dy dx = - Vidyadip

MCQ

If ${x^y} = {y^x},$ then ${{dy} \over {dx}} = $

A
${{y(x{{\log }_e}y + y)} \over {x(y{{\log }_e}x + x)}}$
✓
${{y(x{{\log }_e}y - y)} \over {x(y{{\log }_e}x - x)}}$
C
${{x(x{{\log }_e}y - y)} \over {y(y{{\log }_e}x - x)}}$
D
${{x(x{{\log }_e}y + y)} \over {y(y{{\log }_e}x + x)}}$

✓

Answer

Correct option: B.

${{y(x{{\log }_e}y - y)} \over {x(y{{\log }_e}x - x)}}$

b
(b) ${x^y} = {y^x} \Rightarrow y{\log _e}x = x{\log _e}y$

Differentiating w.r.t. $x$ of $y,$ we get

${\log _e}x\frac{{dy}}{{dx}} + \frac{y}{x} = {\log _e}y + x\frac{1}{y}\frac{{dy}}{{dx}}$

$\therefore \frac{{dy}}{{dx}} = \frac{{y(x{{\log }_e}y - y)}}{{x(y{{\log }_e}x - x)}}$.

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