Using implicit differentiation, find $y^\prime$ if $\sin(x + y) = y^2\cos(x)$

Question

$$\text{Find $y^\prime$ if} \quad\sin(x + y) = y^2\cos(x)$$

So I know I am supposed to find the derivative of each side.

The RHS would come out to be $2y(dy/dx)-\sin(x) $ but I can not figure out how to differentiate the LHS.

Firstly, can someone confirm or deny that I have done the RHS correctly and then explain how I would differentiate the LHS?

Thank you!

Answer 1

$ \sin(x + y) = \underbrace{y^2\cos(x)}_{\text{product of 2 functions}}$

so the derivative of the product of 2 functions is:

$(f(x)g(x))'=f'(x)g(x)+f(x)g'(x)$

$ \cos(x + y)(1+y') = 2yy'\cos(x)-y^2\sin(x)$

Answer 2

I recommend the formal approach.

Say $F(x,y) = 0$. Take the $d$ of that and get $dF(x,y) = 0$. But $$dF(x,y) = \frac{\partial F}{\partial x} dx + \frac{\partial F}{\partial y} dy$$ and so $$\frac{dy}{dx} = - \frac{\frac{\partial F}{\partial x}}{\frac{\partial F}{\partial y}}$$

This is also part of the implicit function theorem.

So now $F(x,y) = \sin(x+y) - y^2 \cos x$, so $$\frac{\partial F}{\partial x} = \cos(x+y) + y^2 \sin x\\ \frac{\partial F}{\partial y}= \cos(x+y)- 2 y \cos x\\ \frac{dy}{dx} = - \frac{\cos(x+y) + y^2 \sin x}{ \cos(x+y)- 2 y \cos x}$$