1. ## Derivatives

Hi guys,

I've got this exercise where i've gotten stuck on. Diﬀerentiate $\displaystyle ln(cos(1/x ))$ on its domain of deﬁnedness. I'm hoping I can get some guidance/help if i've gone wrong in what i've already done and in completing it.

$\displaystyle d, dx ln(cos(1/ x ))$

I choose to use the chain rule as there is a composition of functions.
We take $\displaystyle 1/x$ to be the inner function and cos to be the outer function.

$\displaystyle d,dx cos = -sin$
$\displaystyle d,dx 1/ x = -1/x^2$

which leaves me with

$\displaystyle d, dx ln( -sin. -1/x^2)$

From here I don't how to conitue or if i even should

Please help!. I apologize for how the maths terms are displayed, I'm quite new to the site and don't know any of the commands to make fractions symbols or anything else for that matter. Can't seem to find a help page that shows the maths commands, hence d by dx been display as it is, as well as 1 over x squared.

2. Originally Posted by mathsToday
Hi guys,

I've got this exercise where i've gotten stuck on. Diﬀerentiate $\displaystyle ln(cos(1/x ))$ on its domain of deﬁnedness. I'm hoping I can get some guidance/help if i've gone wrong in what i've already done and in completing it.

$\displaystyle d, dx ln(cos(1/ x ))$

I choose to use the chain rule as there is a composition of functions.
We take $\displaystyle 1/x$ to be the inner function and cos to be the outer function.

$\displaystyle d,dx cos = -sin$
$\displaystyle d,dx 1/ x = -1/x^2$

which leaves me with

$\displaystyle d, dx ln( -sin. -1/x^2)$

From here I don't how to conitue or if i even should

Please help!. I apologize for how the maths terms are displayed, I'm quite new to the site and don't know any of the commands to make fractions symbols or anything else for that matter. Can't seem to find a help page that shows the maths commands, hence d by dx been display as it is, as well as 1 over x squared.

$\displaystyle \frac{d}{dx}ln \left ( cos \left [ \frac{1}{x} \right ] \right )$

$\displaystyle = \frac{1}{cos \left [ \frac{1}{x} \right ] } \cdot -sin \left [ \frac{1}{x} \right ] \cdot -\frac{1}{x^2}$

-Dan

3. Originally Posted by topsquark
$\displaystyle \frac{d}{dx}ln \left ( cos \left [ \frac{1}{x} \right ] \right )$

$\displaystyle = \frac{1}{cos \left [ \frac{1}{x} \right ] } \cdot -sin \left [ \frac{1}{x} \right ] \cdot -\frac{1}{x^2}$

-Dan
Um.. did you get that by using the chain rule alone?

4. Hello,

You have to use the chain rule twice

5. Originally Posted by Moo
Hello,

You have to use the chain rule twice
oh okay, i'll try figure this out.

6. Originally Posted by Moo
Hello,

You have to use the chain rule twice
$\displaystyle \frac{d}{dx}ln \left ( cos \left [ \frac{1}{x} \right ] \right )$

Since there is two cases where there us an outer and inner function we use the chain rule twice.