Choose the correct alternative in the following:
If $3 \sin (x y)+4 \cos (x y)=5$, then $\frac{d y}{d x}=$
A. $-\frac{\mathrm{y}}{\mathrm{x}}$
B. $\frac{3 \sin (x y)+4 \cos (x y)}{3 \cos (x y)-4 \sin (x y)}$
C. $\frac{3 \cos (\mathrm{xy})+4 \sin (\mathrm{xy})}{4 \cos (\mathrm{xy})-3 \sin (\mathrm{xy})}$
D. none of these
$3 \sin (x y)+4 \cos (x y)=5$
Differentiating w.r.t $x$ we get,
$\Rightarrow 3\left[\cos (\mathrm{xy}) \cdot\left(1 \cdot \mathrm{y}+\mathrm{x} \cdot \frac{\mathrm{dy}}{\mathrm{dx}}\right)\right]+4\left[-\sin (\mathrm{xy}) \cdot\left(1 \cdot \mathrm{y}+\mathrm{x} \cdot \frac{\mathrm{dy}}{\mathrm{dx}}\right)\right]=0$
(Using Chain Rule)
$\Rightarrow\left[3 y \cos (x y)+3 x \cos (x y) \cdot \frac{d y}{d x}\right]+\left[-4 y \sin (x y)-4 x \sin (x y) \cdot \frac{d y}{d x}\right]=0$
$\Rightarrow \frac{d y}{d x}[3 x \cos (x y)-4 x \sin (x y)]=4 y \sin (x y)-3 y \cos (x y)$
$\Rightarrow \frac{d y}{d x}=-\frac{y[-4 \sin (x y)+3 \cos (x y)]}{x[3 \cos (x y)-4 \sin (x y)]}=-\frac{y}{x}$
$\Rightarrow \frac{d y}{d x}=-\frac{y}{x}=(A)$