Question

Given that y = sin(x+y),find the derivative when (x,y)=(π,0)​

Answer 1

Answer:

Shown in the explanation

Step-by-step explanation:

Recall that an implicit function is a relation given by the form:

${\displaystyle R(x_{1},\ldots, x_{n})=0}$

Where $R$ is a function of two or more variables. In this case, that function is:

$y = sin(x+y)$

and is implicit because we can define it as:

$y-sin(x+y)=0$ having two variables.

So, let's take the derivative:

$\frac{d}{dx}\left(y\right)=\frac{d}{dx}\left(\sin \left(x+y\right)\right)$

Applying chain rule:

$\frac{d}{dx}\left(\sin \left(x+y\right)\right)=\cos \left(x+y\right)\left(1+\frac{d}{dx}\left(y\right)\right)$

But:

$\frac{d}{dx}\left(y\right)=y'$

Therefore:

$y'=\cos \left(x+y\right)\left(1+y'\right)$

Isolating $y'$:

$\frac{d}{dx}\left(y\right)=y'=\frac{\cos \left(x+y\right)}{1-\cos \left(x+y\right)}$

When $(x,y)=(\pi,0)$:

$\frac{d}{dx}\left(y\right)|_{(\pi,0)}=\frac{\cos \left(\pi+0\right)}{1-\cos \left(\pi+0\right)} \\ \\ \frac{d}{dx}\left(y\right)|_{(\pi,0)}=\frac{\cos \left(\pi\right)}{1-\cos \left(\pi\right)} \\ \\ \frac{d}{dx}\left(y\right)|_{(\pi,0)}=\frac{-1}{1-(-1)} \\ \\ \boxed{\frac{d}{dx}\left(y\right)|_{(\pi,0)}=-\frac{1}{2}}$