Solving for dy/dx in Derivative of cos(x)^ln(x) | Finding the Derivative

[Esoremada]

Homework Statement

http://puu.sh/4M7BE.png

Homework Equations

ln(a^x) = 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)]

 

[Dick]

Homework Statement

http://puu.sh/4M7BE.png

Homework Equations

ln(a^x) = 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)).

 

[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.

 

[Esoremada]

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

 

[Dick]

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

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

 

[Esoremada]

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

 

[Dick]

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).

 

[Esoremada]

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.