# Deriving cos(x)^ln(x)?

## Homework Statement

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

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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## Homework Statement

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

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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• 1 person
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.

Can't tell what I did wrong in attempt 2 either Not sure how they got the inner cos outside of ln in the answer.

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

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Can't tell what I did wrong in attempt 2 either 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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• 1 person
I see, I guess the answer key is incorrect. Thanks for the help

I see, I guess the answer key is incorrect. Thanks for the help