If x^p y^q = (x + y)^ p + q , then dy dx = - Vidyadip

MCQ

If ${x^p}{y^q} = {(x + y)^{p + q}},$ then ${{dy} \over {dx}} = $

✓
${y \over x}$
B
$ - {y \over x}$
C
${x \over y}$
D
$ - {x \over y}$

✓

Answer

Correct option: A.

${y \over x}$

a
(a) Taking $\log $ both sides,

$p\log x + q\log y = (p + q)\log (x + y)$

==> $\frac{p}{x} + \frac{q}{y}\frac{{dy}}{{dx}} = \frac{{p + q}}{{x + y}}\left( {1 + \frac{{dy}}{{dx}}} \right) $

$\Rightarrow \frac{{dy}}{{dx}} = \frac{y}{x}$.

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