Question

Find  $$\frac{dy}{dx}$$ in the following case $$\sin xy+\cos (x+y)=1$$ ?

 

Answer 1

$$\text{ We have, }\sin xy+\cos (x+y)=1$$

Differentiating with respect to x, we get,

$$\frac{d}{dx}(\sin xy)+\frac{d}{dx}\cos (x+y)=\frac{d}{dx}\left(1\right)$$
$$\Rightarrow \cos xy\frac{d}{dx}\left(xy\right)-\sin (x+y)\frac{d}{dx}(x+y)=0$$
$$\Rightarrow \cos xy[x\frac{dy}{dx}+y\frac{d}{dx}\left(x\right)]-\sin (x+y)[1+\frac{dy}{dx}]=0$$
$$\Rightarrow \cos xy[x\frac{dy}{dx}+y\left(1\right)]-\sin (x+y)-\sin (x+y)\frac{dy}{dx}=0$$
$$\Rightarrow x\cos xy\frac{dy}{dx}+y\cos xy-\sin (x+y)-\sin (x+y)\frac{dy}{dx}=0$$
$$\Rightarrow [x\cos xy-\sin (x+y)]\frac{dy}{dx}=[\sin (x+y)-y\cos xy]$$
$$\Rightarrow \frac{dy}{dx}=\left[\frac{\sin (x+y)-y\cos xy}{x\cos xy-\sin (x+y)}\right]$$