Derivative: chain rule applied to $\cos(\pi x)$ 
What is the derivative of the function $f(x)= \cos(\pi x)$? 

I found the derivative to be $f^{\prime}(x)= -\pi\sin(\pi x)$. Am I correct? Can you show me how to find the answer step by step? 
This is a homework question:

What is $x$ equal to if $-\pi\sin(\pi x)=0$?

 A: To find the derivative of this function you need to use the chain rule. 
Let u = $\pi x$, and we know $\frac{d}{du}(\cos(u)) = -\sin(u)$.
Then $$\frac{d}{dx}\cos(\pi x)=\frac{d\cos(u)}{du} \frac{du}{dx}$$
simplifies to your answer that you found.
A: As you wrote in the title, it is just an application of the chain rule: $$\Big(f(g(x))\Big)' = f'(g(x))\cdot g'(x).$$
In you case you can write your function, $\cos(\pi x)$, as the composition of $f(y) = \cos(y)$ and $g(x) = x\pi$. if you know apply the formula you get that your answer is correct since $f'(y) = -\cos(y)$ hence $f'(g(x)) = -\cos(\pi x)$, now just multiply this by $\pi$, i.e. the derivative of $g(x)$.
EDIT: you want to find all $x$ such that $-\pi\sin(\pi x) = 0$ is satisfied. Clearly this is equivalent to find all $x$ such that $\sin(\pi x) = 0$ (just multiplying both sides of the previous eqn by $-\pi^{-1}$). Now you just need to know ow the sine function behaves: it is know that (http://en.wikipedia.org/wiki/Trigonometric_functions) $\sin(k\pi) = 0$ for all integers $k$ and that these are the only zeros of the sine function. Comparing this with your equation you get that $\sin(\pi x) = 0$if and only if $\pi x = k\pi$, which gives $x = k$.
A: The Chain Rule: If $g$ is differentiable at $x$ and $f$ is differentiable at $g(x)$, then the composite function $F=f\circ g$ defined by $F(x)=f(g(x))$ is differentiable at $x$ and $F'$ is given by the product $$F'(x)=f'(g(x))\cdot g'(x).$$
If $f(x)=cos(\pi x)$, then $f'(x)-sin(\pi x)\cdot \pi=-\pi sin(\pi x)$.
If $-\pi sin(\pi x)=0$, then $x=n$ where $x\in \mathbb{Z}$.
A: Regarding the follow up question: Note that $$-\pi\sin(\pi x) = 0 \implies \sin(\pi x)=0$$
Now recall that $\sin\theta=0$ whenever $\theta=n\pi$, where $n\in\Bbb{Z}$ (i.e. $n$ is an integer, in case you're not familiar with that notation).  Therefore $$\sin(\pi x) = 0 \implies \pi x = n \pi \implies x=n.$$
Thus, if you're not restricting yourself to any finite interval, then it follows that $f^{\prime}(x)=-\pi\sin(\pi x)=0$ at every integer.
I hope this makes sense!
