Derivative of Logarithm with trig

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Homework Help Overview

The discussion revolves around finding the derivative of a function involving logarithmic and trigonometric components, specifically the expression y = θ(sin(ln θ)) + cos(ln θ). Participants are examining the correctness of their derivative calculations and comparing them to a provided answer in a textbook.

Discussion Character

  • Exploratory, Assumption checking, Problem interpretation

Approaches and Questions Raised

  • Participants present their derivative calculations and question the accuracy of the textbook answer. There is a suggestion to verify the original problem statement for correctness. One participant proposes an alternative formulation of the function that aligns with the textbook's answer.

Discussion Status

The discussion is active, with participants confirming their calculations and exploring the possibility of a miscommunication regarding the problem statement. There is no explicit consensus, but a potential resolution is suggested through a revised function that matches the expected derivative.

Contextual Notes

Participants are working under the assumption that the problem statement may have been copied incorrectly, which could affect the derivative results. The original function's formulation is under scrutiny.

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.
 

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