Derivative of A Trigonometric Function

Bashyboy
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Homework Statement


Find the derivative of f(x) = 1/4 sin^2 2x


Homework Equations





The Attempt at a Solution



f'(x) =1/4(sin 2x)^2

f'(x) = 1/4*2(sin 2x)(cos 2x)

f'(x) = 1/4(sin 4x)

That is my answer to the problem, but the book has it as f'(x) = 1/2(sin 4x)
What did I do wrong?
 
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you forgot the factor of 2 when you differentiate (sin(2x))
 
Oh, yes, thank you.
 
Bashyboy said:

Homework Statement


Find the derivative of f(x) = 1/4 sin^2 2x

The Attempt at a Solution


f'(x) =1/4(sin 2x)^2
(Above) This is f(x), not f'(x).
Bashyboy said:
f'(x) = 1/4*2(sin 2x)(cos 2x)

f'(x) = 1/4(sin 4x)

That is my answer to the problem, but the book has it as f'(x) = 1/2(sin 4x)
What did I do wrong?
 
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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