This is an application of the chain rule (or it could be something else so if you're in a different section of your course, let me know in a comment). Remember that the chain rule roughly says that you need to take the derivatives of the "outside" and "inside" functions separately and then multiply them together.

The outside function here is \(f(x)=\sin(x)\). Remember that its derivative is \(\cos(x)\)

The inside function is \(g(x)=\pi x. The derivative of this is clearly \pi.

Putting these together we get f′=\cos(\pi x)\times \pi=\pi \cos(\pi x) where the \pi x in the coscos function comes from the fact that we also need to remember to put the same argument of the outside function back in when multiplying.

Let me know if that doesn't make sense.

The outside function here is \(f(x)=\sin(x)\). Remember that its derivative is \(\cos(x)\)

The inside function is \(g(x)=\pi x. The derivative of this is clearly \pi.

Putting these together we get f′=\cos(\pi x)\times \pi=\pi \cos(\pi x) where the \pi x in the coscos function comes from the fact that we also need to remember to put the same argument of the outside function back in when multiplying.

Let me know if that doesn't make sense.