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Deriving cos(x)^ln(x)?

  • Thread starter Esoremada
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  • #1
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Homework Statement



http://puu.sh/4M7BE.png [Broken]

Homework Equations



ln(ax) = x*ln(a)

The Attempt at a Solution



ln(y) = cos(x)*ln(ln(x))

dy/dx * 1/y = -sinx*ln(ln(x)) + cosx/(x*lnx)

No clue how to solve this, there's no ln(ln(x)) in the possible answers


Attempt 2

y = cos(x)^ln(x)
ln(y) = ln(x)*ln[cos(x)]
1/y * dy/dx = 1/x * ln(cos(x)) + ln(x) * 1/cos(x) * (-sin(x))
dy/dx = y[ ln(cos(x))/x - ln(x) * sin(x) / cos(x)]
dy/dx = [cos(x)ln(x)][ ln(cos(x))/x - (ln(x)sin(x))/cos(x)]
 
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Answers and Replies

  • #2
Dick
Science Advisor
Homework Helper
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Homework Statement



http://puu.sh/4M7BE.png [Broken]

Homework Equations



ln(ax) = x*ln(a)

The Attempt at a Solution



ln(y) = cos(x)*ln(ln(x))

dy/dx * 1/y = -sinx*ln(ln(x)) + cosx/(x*lnx)

No clue how to solve this, there's no ln(ln(x)) in the possible answers
You got off to a bad start. If y=(cos(x))^(ln(x)) then ln(y)=ln(x)*ln(cos(x)). NOT cos(x)*ln(ln(x)).
 
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  • #3
lurflurf
Homework Helper
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The power rule is

$$(u^v)^\prime=v \, u^{v-1} \, u^\prime +u^{v} \log(u) \, v^\prime $$

You can derive it by writing

$$u^v=\exp(v \log(u))$$

Then differentiate both sides.
 
  • #4
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Can't tell what I did wrong in attempt 2 either :eek:

Not sure how they got the inner cos outside of ln in the answer.

http://puu.sh/4MaXv.png [Broken]
 
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  • #5
Dick
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Can't tell what I did wrong in attempt 2 either :eek:

Not sure how they got the inner cos outside of ln in the answer.

http://puu.sh/4MaXv.png [Broken]
Your second attempt is correct. You are trying to compare it to a wrong answer.
 
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  • #6
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I see, I guess the answer key is incorrect. Thanks for the help
 
  • #7
Dick
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I see, I guess the answer key is incorrect. Thanks for the help
No, no, no. Compare your answer with the key answer c). Not with d).
 
  • #8
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Ah, just went through the news feed. They noted that it should be C not D a couple of days ago, didn't catch that.
 

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