Derivative of Logarithm with trig

TommG
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Need to find derivative

y = θ(sin(ln θ)) + cos(ln θ)


My work

θ(cos(ln θ))(1/θ) + sin(ln θ) + (-sin(ln θ)(1/θ))

(θcos(ln θ))/θ] + sin(ln θ) + ( (- sin(ln θ))/θ)

cos(ln θ) + [θsin(ln θ) - sin(ln θ)]/ θ

answer in book is 2cos(lnθ)
 
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TommG said:
Need to find derivative

y = θ(sin(ln θ)) + cos(ln θ)


My work

θ(cos(ln θ))(1/θ) + sin(ln θ) + (-sin(ln θ)(1/θ))

(θcos(ln θ))/θ] + sin(ln θ) + ( (- sin(ln θ))/θ)

cos(ln θ) + [θsin(ln θ) - sin(ln θ)]/ θ

answer in book is 2cos(lnθ)

Your answer is correct, and the book's answer is wrong. Are you sure you copied the problem correctly?
 
Ray Vickson said:
Your answer is correct, and the book's answer is wrong. Are you sure you copied the problem correctly?

Yeah it is correct
 
I propose that it was supposed to be
\begin{equation*}
y(\theta) = \theta(\sin(\ln \theta) + \cos(\ln \theta)).
\end{equation*}
 
TommG said:
Yeah it is correct

As Quesadilla points out, the function ##f(\theta) = \theta ( \sin(\ln \theta) + \cos(\ln \theta))## is probably what you want; its derivative agrees with the book's answer.
 
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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