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