Answer:

Step-by-step explanation:
So we have:

And we want to find dy/dx.
So, let's take the derivative of both sides with respect to x:
![\frac{d}{dx}[y\sin(y)]=\frac{d}{dx}[x\cos(x)]](https://tex.z-dn.net/?f=%5Cfrac%7Bd%7D%7Bdx%7D%5By%5Csin%28y%29%5D%3D%5Cfrac%7Bd%7D%7Bdx%7D%5Bx%5Ccos%28x%29%5D)
Let's do each side individually.
Left Side:
We have:
![\frac{d}{dx}[y\sin(y)]](https://tex.z-dn.net/?f=%5Cfrac%7Bd%7D%7Bdx%7D%5By%5Csin%28y%29%5D)
We can use the product rule:

So, our derivative is:
![=\frac{d}{dx}[y]\sin(y)+y\frac{d}{dx}[\sin(y)]](https://tex.z-dn.net/?f=%3D%5Cfrac%7Bd%7D%7Bdx%7D%5By%5D%5Csin%28y%29%2By%5Cfrac%7Bd%7D%7Bdx%7D%5B%5Csin%28y%29%5D)
We must implicitly differentiate for y. This gives us:
![=\frac{dy}{dx}\sin(y)+y\frac{d}{dx}[\sin(y)]](https://tex.z-dn.net/?f=%3D%5Cfrac%7Bdy%7D%7Bdx%7D%5Csin%28y%29%2By%5Cfrac%7Bd%7D%7Bdx%7D%5B%5Csin%28y%29%5D)
For the sin(y), we need to use the chain rule:

Our u(x) is sin(x) and our v(x) is y. So, u'(x) is cos(x) and v'(x) is dy/dx.
So, our derivative is:

Simplify:

And we are done for the right.
Right Side:
We have:
![\frac{d}{dx}[x\cos(x)]](https://tex.z-dn.net/?f=%5Cfrac%7Bd%7D%7Bdx%7D%5Bx%5Ccos%28x%29%5D)
This will be significantly easier since it's just x like normal.
Again, let's use the product rule:
![=\frac{d}{dx}[x]\cos(x)+x\frac{d}{dx}[\cos(x)]](https://tex.z-dn.net/?f=%3D%5Cfrac%7Bd%7D%7Bdx%7D%5Bx%5D%5Ccos%28x%29%2Bx%5Cfrac%7Bd%7D%7Bdx%7D%5B%5Ccos%28x%29%5D)
Differentiate:

So, our entire equation is:

To find our derivative, we need to solve for dy/dx. So, let's factor out a dy/dx from the left. This yields:

Finally, divide everything by the expression inside the parentheses to obtain our derivative:

And we're done!