If x^y . y^x = 1, then dy dx = - Vidyadip

MCQ

If ${x^y}.{y^x} = 1$, then ${{dy} \over {dx}} =$

A
${{y\,(y + x\log y)} \over {x(y\log x + x)}}$
B
${{y\,(x + y\log x)} \over {x(y + x\log y)}}$
✓
$ - {{y(y + x\log y)} \over {x(x + y\log x)}}$
D
None of these

✓

Answer

Correct option: C.

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

c
(c) $y\log x + x\log y = 0$;

$\therefore$ ${{dy} \over {dx}} = - \frac{{{\partial f} \over {\partial x}} }{{{\partial f} \over {\partial y}} }$

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