Question
If $2 x y+3 x+y-4=0$ then find $\frac{d y}{d x}$.
✓
Answer
$ 2 x y+3 x+y-4=0$
$\therefore 2 x y+y=4-3 x$
$\therefore y(2 x+1)=4-3 x$
$\therefore y=\frac{4-3 x}{2 x+1} $
Here, take $u=4-3 x$ and $v=2 x+1$.
$ \therefore \frac{d u}{d x}=-3 \text { and } \frac{d v}{d x}=2 $
Now, $y=\frac{u}{v}$
$ \therefore \frac{d y}{d x} =\frac{v \frac{d u}{d x}-u \frac{d v}{d x}}{v^{2}}$
$ =\frac{(2 x+1)(-3)-(4-3 x)(2)}{(2 x+1)^{2}}$
$ =\frac{-6 x-3-8+6 x}{(2 x+1)^{2}}$
$ =\frac{-11}{(2 x+1)^{2}} $
$\therefore 2 x y+y=4-3 x$
$\therefore y(2 x+1)=4-3 x$
$\therefore y=\frac{4-3 x}{2 x+1} $
Here, take $u=4-3 x$ and $v=2 x+1$.
$ \therefore \frac{d u}{d x}=-3 \text { and } \frac{d v}{d x}=2 $
Now, $y=\frac{u}{v}$
$ \therefore \frac{d y}{d x} =\frac{v \frac{d u}{d x}-u \frac{d v}{d x}}{v^{2}}$
$ =\frac{(2 x+1)(-3)-(4-3 x)(2)}{(2 x+1)^{2}}$
$ =\frac{-6 x-3-8+6 x}{(2 x+1)^{2}}$
$ =\frac{-11}{(2 x+1)^{2}} $
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