Question:
Find $\frac{d y}{d x}$ :
$2 x+3 y=\sin x$
Solution:
The given relationship is
Differentiating this relationship with respect to x, we obtain
$\frac{d}{d x}(2 x+3 y)=\frac{d}{d x}(\sin x)$
$\Rightarrow \frac{d}{d x}(2 x)+\frac{d}{d x}(3 y)=\cos x$
$\Rightarrow 2+3 \frac{d y}{d x}=\cos x$
$\Rightarrow 3 \frac{d y}{d x}=\cos x-2$
$\therefore \frac{d y}{d x}=\frac{\cos x-2}{3}$