Shown in the explanation
Recall that an implicit function is a relation given by the form:
[tex]{\displaystyle R(x_{1},\ldots, x_{n})=0}[/tex]
Where [tex]R[/tex] is a function of two or more variables. In this case, that function is:
[tex]y = sin(x+y)[/tex]
and is implicit because we can define it as:
[tex]y-sin(x+y)=0[/tex] having two variables.
So, let's take the derivative:
[tex]\frac{d}{dx}\left(y\right)=\frac{d}{dx}\left(\sin \left(x+y\right)\right) \\ \\[/tex]
Applying chain rule:
[tex]\frac{d}{dx}\left(\sin \left(x+y\right)\right)=\cos \left(x+y\right)\left(1+\frac{d}{dx}\left(y\right)\right)[/tex]
But:
[tex]\frac{d}{dx}\left(y\right)=y'[/tex]
Therefore:
[tex]y'=\cos \left(x+y\right)\left(1+y'\right)[/tex]
Isolating [tex]y'[/tex]:
[tex]\frac{d}{dx}\left(y\right)=y'=\frac{\cos \left(x+y\right)}{1-\cos \left(x+y\right)}[/tex]
When [tex](x,y)=(\pi,0)[/tex]:
[tex]\frac{d}{dx}\left(y\right)|_{(\pi,0)}=\frac{\cos \left(\pi+0\right)}{1-\cos \left(\pi+0\right)} \\ \\ \frac{d}{dx}\left(y\right)|_{(\pi,0)}=\frac{\cos \left(\pi\right)}{1-\cos \left(\pi\right)} \\ \\ \frac{d}{dx}\left(y\right)|_{(\pi,0)}=\frac{-1}{1-(-1)} \\ \\ \boxed{\frac{d}{dx}\left(y\right)|_{(\pi,0)}=-\frac{1}{2}}[/tex]
