Deriving cos(x)^ln(x)?

1. Oct 9, 2013

1. The problem statement, all variables and given/known data

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

2. Relevant equations

ln(ax) = x*ln(a)

3. 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)]

Last edited by a moderator: May 6, 2017
2. Oct 9, 2013

Dick

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)).

Last edited by a moderator: May 6, 2017
3. Oct 9, 2013

lurflurf

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. Oct 9, 2013

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]

Last edited by a moderator: May 6, 2017
5. Oct 9, 2013

Dick

Your second attempt is correct. You are trying to compare it to a wrong answer.

Last edited by a moderator: May 6, 2017
6. Oct 9, 2013

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

7. Oct 9, 2013

Dick

No, no, no. Compare your answer with the key answer c). Not with d).

8. Oct 10, 2013