What is the derivative of $y=\cos(x+y)$? I know it'll be solved using chain rule but I don't know how to use chain rule formula.
 A: Differentiating the sides wrt $x$ yields:
$$\frac{dy}{dx}=(1+\frac{dy}{dx})(-\sin(x+y))$$
A: Take the derivative of both sides

$$y' = -\sin(x+y)\cdot (1+y')$$

Now just solve,
$$y' = {-\sin(x+y)\over 1+\sin(x+y)}$$
A: Implicit differentiation is also useful.
Consider $$F=y-\cos(x+y)=0$$ Computing the partial derivatives $$F'_x=\sin(x+y)\qquad, \qquad F'_y=1+\sin(x+y)$$ By the implicit function theorem $$\frac{dy}{dx}=-\frac{F'_x}{F'_y}=-\frac{\sin(x+y)}{1+\sin(x+y)}$$
A: We have $y=\cos(x+y)$
Differentiating using chain rule
$ \dfrac{\mathrm d y}{\mathrm d x} =−\sin⁡(x+y)(x+y) \; \Rightarrow$
$\dfrac{\mathrm d y}{\mathrm d x}=−\sin⁡(x+y)\left(1+\dfrac{\mathrm d y}{\mathrm d x}\right)$
Rewriting, 
$\dfrac{\mathrm dy}{\mathrm dx}=−\sin⁡(x+y)−\sin⁡(x+y)\dfrac{\mathrm dy}{\mathrm dx} \; \Rightarrow$
$\dfrac{\mathrm dy}{\mathrm dx}+\sin⁡(x+y)\dfrac{\mathrm dy}{\mathrm dx}=−\sin⁡(x+y)$
Taking $\dfrac{\mathrm dy}{\mathrm dx}$ common,
$\dfrac{\mathrm dy}{\mathrm dx}\left(1+\sin⁡(x+y)\right)=−\sin⁡(x+y)$
Bringing $(1+\sin⁡(x+y))$ to other side,
$\dfrac{\mathrm dy}{\mathrm dx}=−\dfrac{\sin⁡(x+y)}{1+\sin⁡(x+y)}$ will be the final answer.
