Solving g(t): Seeking Help with Chain Rule

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Homework Statement



g(t) = 4ln(5ln(4t))

Homework Equations



What is the derivative of g(t)?

The Attempt at a Solution



I have tried to use the chain rule in many different ways, and still can't come up with the correct answer. Anybody care to walk me through this? Is the derivative of ln(4t) still 1/t?
 
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(ln(4t))'=(1/(4t))*4=1/t. Yes. It's still 1/t. For your problem g'=4*(1/(5*ln(4t)) times the derivative of 5*ln(4t). What do you make it?
 
4*(1/(5*ln(4t)) = 4/(5*ln(4t))
Deriv 5*ln(4t) = 5/t

5/t * 4/(5*ln(4t)) = 20/(5t*ln(4t)) = 4/(t*ln(4t)) =

4*(ln(4t))^-1*t^-1?
 
I agree with that.
 
So does the answer key. Thanks for your help.
 
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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