Question

Find the derivative of ln of absolute value of cos of pi/x. The answer is (pi/x^2)*tan(pi/x).

Answer 1

\[\Large\bf\sf \ln\left|\cos \frac{\pi}{x}\right|\]
Mmm ok absolute can be a bit of a doozy.
Do you remember this derivative?
\[\Large\bf\sf \frac{d}{dx}|x|\]It'll help when we apply one of our chains.

Answer 2

No. Can you show the work?

Answer 3

Sure, if we need to show the work for that.. we would do it like this:\[\Large\bf\sf |x|\quad:=\quad \sqrt{x^2}\]We'll take the derivative from this form.\[\Large\bf\sf \frac{d}{dx}\sqrt{x^2}\quad=\quad \frac{2x}{2\sqrt{x^2}}\quad=\quad \frac{x}{\sqrt{x^2}}\quad=\quad \frac{x}{|x|}\]

\[\Large\bf\sf \implies \frac{d}{dx}|x|\quad=\quad \frac{x}{|x|}\]

Answer 4

So there is a certain part in our problem when we chain rule, we'll have to take care of:\[\Large\bf\sf \frac{d}{dx}|\cos \frac{\pi}{x}|\]And we'll use this derivative rule.

Answer 5

So it's cos pi/x divided by absolute value of cos pi/x?

Answer 6

For that particular part, yes good.
We would have to chain again though right? Because of the pi/x inside.

Answer 7

\[\Large\bf\sf \color{royalblue}{\left(\ln\left|\cos \frac{\pi}{x}\right|\right)'}\quad=\quad \frac{1}{\left|\cos \dfrac{\pi}{x}\right|}\cdot \color{royalblue}{\left(\left|\cos\dfrac{\pi}{x}\right|\right)'}\]

Answer 8

First step make sense?
Log gives us one over the contents of the log right?
Then we chain.

Answer 9

Okay, how do we find the derivative of the inside?

Answer 10

The inside is where we apply our fancy rule from before:\[\Large\bf\sf \frac{d}{dx}\left|\cos \frac{\pi}{x}\right|\quad=\quad \left(\left|\cos \frac{\pi}{x}\right|\right)'\]

Answer 11

\[\Large\bf\sf =\frac{1}{\left|\cos \dfrac{\pi}{x}\right|}\cdot \frac{\cos\dfrac{\pi}{x}}{\left|\cos\dfrac{\pi}{x}\right|}\color{royalblue}{\left(\frac{\pi}{x}\right)'}\]

Answer 12

Understand the second step? :o
Took the derivative of absolute cosine, then chained again.

Answer 13

Yes. Let me work it out. Please don't leave.

Answer 14

k

Answer 15

Oh crap I skipped a step, sorry about that.
Our next inner chain is the stuff INSIDE of the absolute bars, not the stuff inside of cosine.
I jumped ahead too far.

Answer 16

\[\Large\bf\sf =\frac{1}{\left|\cos \dfrac{\pi}{x}\right|}\cdot \frac{\cos\dfrac{\pi}{x}}{\left|\cos\dfrac{\pi}{x}\right|}\color{royalblue}{\left(\cos\frac{\pi}{x}\right)'}\]

Answer 17

I got cos pi/x divided by absolute value of cos pi/x squared times -sin pi/x and times -pi/x^2, how do I simplify this?

Answer 18

\[\Large\bf\sf =\quad \frac{1}{\cos^2\dfrac{\pi}{x}}\cdot \left(\sin\frac{\pi}{x}\cos\frac{\pi}{x}\right)\cdot\frac{\pi}{x^2}\]Ok good.
Cancel the negatives.
Then you can DROP the absolute bars sinse the inside is a square.
(Squares are always positive, so the absolute bars aren't doing anything for us anymore.

Answer 19

See any nice cancellations or simplifications from there? :o

Answer 20

Got it. Thanks. But is there another easier way to do this?

Answer 21

If the absolute thing is confusing, you can just remember the definition of absolute and rewrite your problem as:\[\Large\bf\sf \frac{d}{dx}\ln \sqrt{\cos^2\frac{\pi}{x}}\]Because:\[\Large\bf\sf |x|\quad:=\quad \sqrt{x^2}\]And use normal rules of differentiation to do it from there.
It's still a ton of chains, but it doesn't deal with the absolute anymore.

This one is actually worth memorizing I think, it shows up later in Calc 2 and stuff.\[\Large\bf\sf \int\limits \tan x\;dx \quad=\quad -\ln|\cos x|\]\[\Large\bf\sf \implies \frac{d}{dx}\ln|\cos x|\quad=\quad -\tan x\]

Answer 22

Thanks for the extra info!