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

MCQ

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

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

✓

Answer

Correct option: A.

$\frac {y}{x}$

a
$x^{m} \cdot y^{n}=(x+y)^{m+n}$

$\Rightarrow m \ln x+n \ln y=(m+n) \ln (x+y)$

Differentiating both sides.

$\therefore \frac{m}{x}+\frac{n}{y} \frac{d y}{d x}=\frac{m+n}{x+y}\left(1+\frac{d y}{d x}\right)$

$\Rightarrow\left(\frac{m}{x}-\frac{m+n}{x+y}\right)=\left(\frac{m+n}{x+y}-\frac{n}{y}\right) \frac{d y}{d x}$

$\Rightarrow \frac{m y-n x}{x(x+y)}=\left(\frac{m y-n x}{y(x+y)}\right) \frac{d y}{d x}$

$\Rightarrow \frac{d y}{d x}=\frac{y}{x}$

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