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 $-\pi\sin(\pi x)=0$?


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 $x = k$.
