Simple derivative of exponential function

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The derivative of the function y=e^(cos(t)+ln(t)) is calculated using the chain rule, resulting in y'=e^(cos(t)+ln(t))*(-sin(t)+1/t). The book provides a different but equivalent form, y'=e^(cos(t))*(1-tsin(t)). The discussion centers on whether these two expressions represent the same derivative and explores the properties of exponential and logarithmic functions. Clarifications on the identities a^(b+c) and e^(ln(t)) are also sought to understand the equivalence. Both forms of the derivative are indeed valid and represent the same mathematical relationship.
Jrlinton
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1. Homework Statement
Find derivative of
y=e^(cos(t)+lnt)

Homework Equations

The Attempt at a Solution


So just using the chain rule:
y'=e^(cos(t)+lnt)*(-sin(t)+1/t)
The answer in the back of the book is
y'=e^(cos(t))*(1-tsin(t))
 
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Just trying to figure out where I went wrong or if these are just two forms of the same answer?
 
The answers are identical. What is ##a^{b+c}## and what ##e^{\ln t}##?
 
Thank you
 
Question: A clock's minute hand has length 4 and its hour hand has length 3. What is the distance between the tips at the moment when it is increasing most rapidly?(Putnam Exam Question) Answer: Making assumption that both the hands moves at constant angular velocities, the answer is ## \sqrt{7} .## But don't you think this assumption is somewhat doubtful and wrong?

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