If (x+y)= x y+3, then d y d x = - Vidyadip

MCQ

If $\log (x+y)=\log x y+3$, then $\frac{d y}{d x}=$

A
$\left(\frac{y}{x}\right)^2$
B
$-\left(\frac{x}{y}\right)^2$
✓
$-\left(\frac{y}{x}\right)^2$
D
$\left(\frac{x}{y}\right)^2$

✓

Answer

Correct option: C.

$-\left(\frac{y}{x}\right)^2$

Given, $\log (x+y)=\log x y+3$
Differentiate with respect to $x,$ we get
$\frac{1}{x+y}\left[1+\frac{d y}{d x}\right]=\frac{1}{x y}\left[x \cdot \frac{d y}{d x}+y\right]$
$\Rightarrow \frac{1}{x+y}+\frac{1}{x+y} \cdot \frac{d y}{d x}=\frac{1}{y} \cdot \frac{d y}{d x}+\frac{1}{x}$
$\Rightarrow \frac{d y}{d x}\left(\frac{1}{x+y}-\frac{1}{y}\right)=\frac{1}{x}-\frac{1}{x+y}$
$\Rightarrow \frac{d y}{d x}\left[\frac{y-x-y}{(x+y) y}\right]=\frac{y}{x(x+y)}$
$\Rightarrow \frac{d y}{d x}\left[\frac{-x}{(x+y) y}\right]=\frac{y}{x(x+y)} $
$\Rightarrow \frac{d y}{d x}=-\left(\frac{y}{x}\right)^2$

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