Answer:

Explanation:
So we have:

And we want to find dy/dx.
So, let's take the derivative of both sides with respect to x:
![(d)/(dx)[y\sin(y)]=(d)/(dx)[x\cos(x)]](https://img.qammunity.org/2021/formulas/mathematics/high-school/8rn91ctala75xc8rhhgx8bexj430x8i70l.png)
Let's do each side individually.
Left Side:
We have:
![(d)/(dx)[y\sin(y)]](https://img.qammunity.org/2021/formulas/mathematics/high-school/3xiiee7z0obmwlfcnbfmffbfv4h00krjxk.png)
We can use the product rule:

So, our derivative is:
![=(d)/(dx)[y]\sin(y)+y(d)/(dx)[\sin(y)]](https://img.qammunity.org/2021/formulas/mathematics/high-school/2351adtsniit8gb5onz5hcngiwz72q2y56.png)
We must implicitly differentiate for y. This gives us:
![=(dy)/(dx)\sin(y)+y(d)/(dx)[\sin(y)]](https://img.qammunity.org/2021/formulas/mathematics/high-school/rgjwp6u4sxzygkzko7ilaetjug4zwoz3sf.png)
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:
![(d)/(dx)[x\cos(x)]](https://img.qammunity.org/2021/formulas/mathematics/high-school/m1u5g770snf8ej3bvi3bs839sf6d4xj4zu.png)
This will be significantly easier since it's just x like normal.
Again, let's use the product rule:
![=(d)/(dx)[x]\cos(x)+x(d)/(dx)[\cos(x)]](https://img.qammunity.org/2021/formulas/mathematics/high-school/tncszrxm9y698dqubkdi40v8458c75skr5.png)
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!