So in this case, the derivative of f(x) = cos pi*x is f'(x) = -pi*sin pi*x.

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How would I do the derivative of

f(x) = cos pi*x

I know f'(x) cos x = - sin x but what about that pi?
 
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Are you familiar with the chain rule ?
 
The chain rule is

\frac{dy}{dx} = \frac{dy}{du} \frac{du}{dx}

Take y = cos u

then you know that dy/du = -sin u.

Now put u = nx, for n a constant.

You know that du/dx = n

Now consider:

y = cos nx

Put u = nx, so that y = cos u. Then the chain rule tells us:

\frac{dy}{dx} = \frac{dy}{du} \frac{du}{dx}
= (-\sin u)(n)
= -n\sin nx
 
There are two things I don't understand about this problem. First, when finding the nth root of a number, there should in theory be n solutions. However, the formula produces n+1 roots. Here is how. The first root is simply ##\left(r\right)^{\left(\frac{1}{n}\right)}##. Then you multiply this first root by n additional expressions given by the formula, as you go through k=0,1,...n-1. So you end up with n+1 roots, which cannot be correct. Let me illustrate what I mean. For this...
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