MCQ
જો ${x^m}{y^n} = 2{(x + y)^{m + n}},$ તો ${{dy} \over {dx}} = . . . . .$
- A$x + y$
- B$x/y$
- ✓$y/x$
- D$x - y$
$\Rightarrow m\log x + n\log y = \log 2 + (m + n)\log (x + y)$
Differentiating both sides w.r.t. $x,$
$\frac{m}{x} + \frac{n}{y}\,\frac{{dy}}{{dx}} = \frac{{m + n}}{{x + y}}\left[ {1 + \frac{{dy}}{{dx}}} \right]$
==> $\frac{{dy}}{{dx}} = \frac{y}{x}$.
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