Question

Find the derivative of the function y = ln((x-1)/(x+1))^(1/3) Please show steps.

Answer 1

It's a little hard to read with all the brackets. Can you tell me if this is formatted correctly?\[\huge y=\ln\left[\left(\frac{x-1}{x+1}\right)^{1/3}\right]\]

Answer 2

Yes, that is correct.

Answer 3

Oh boy that looks like a fun one c: hah

Answer 4

Our rules of logarithms will allow us to pull the exponent out as a factor outside of the log.\[\large y=\frac{1}{3}\ln\left(\frac{x-1}{x+1}\right)\]

Answer 5

Remember the derivative of \(\large y=\ln x\)?
That will help us get started on this one.

Answer 6

Yeah, its is 1/ln(x)

Answer 7

close close c:
\(\huge \frac{1}{x}\)

Answer 8

Oh right..I just started calc two yesterday. So I am learning.

Answer 9

So we'll start by taking the derivative of the natural log,
It will give us 1 over THE ARGUMENT (the stuff inside the log).\[\large y=\frac{1}{3}\frac{1}{\left(\frac{x-1}{x+1}\right)}\left(\color{cornflowerblue}{\frac{x-1}{x+1}}\right)'\]
Ignore the blue part for right now, let's just focus on one step at a time.
Do you understand why I put that huge fraction on the bottom?
You put the ENTIRE contents of the log in the bottom when you take that derivative.

Answer 10

Ah, makes sense. So nothing happens to the 1/3 yet? and will we be using quotient rule?

Answer 11

The 1/3 is a constant. You can completely ignore constant factors when taking a derivative.\[\large (3x^2)'=3(x^2)'=3(2x)\]See how I pulled out the 3 and ignored it to take the derivative?

Answer 12

Oh right, ok.

Answer 13

Yes we'll be using the quotient rule on the BLUE PART. c:
Understand where the blue part is coming from?
We have to multiply by the derivative of the INSIDE after we take the derivative of the log.
The prime is to show us that we still need to differentiate it.

Answer 14

Yeah, you used the chain rule.

Answer 15

Remember what the quotient rule looks like? :O

Answer 16

f'(x)g(x)-f(x)g'(x)/g(x)^2

Answer 17

Yah looks good c: so when you throw the blue thing into the quotient rule, whatchu get? :D

Answer 18

1(x+1) - (x-1)(1)/(x+1)^2

Answer 19

the first part in parenthesis

Answer 20

Yep looks good! c: And that will simplify down just a tad further.

Answer 21

Is that it?

Answer 22

The top of the fraction will simplify down. After you distribute that negative to each term in the second set of brackets, the X's will cancel out, leaving us with 2 on top.

So that leaves us with this so far,\[\huge y'=\frac{1}{3}\frac{1}{\left(\frac{x-1}{x+1}\right)}\left(\color{cornflowerblue}{\frac{2}{(x+1)^2}}\right)\]

Answer 23

We can do a bit more simplification from here though.

Answer 24

Remember what you do when you DIVIDE by a fraction?
You can instead write it as MULTIPLYING by it's reciprocal.\[\huge \frac{1}{\left(\frac{x-1}{x+1}\right)} \qquad \rightarrow \qquad \frac{x+1}{x-1}\]

Answer 25

\[\huge y'=\frac{1}{3}\left(\frac{x+1}{x-1}\right)\left(\frac{2}{(x+1)^2}\right)\]

Answer 26

We can then cancel out one of the (x+1)'s from the top and bottom, giving us,\[\large y'=\frac{2}{3}\left(\frac{1}{(x-1)(x+1)}\right)\]

Answer 27

We then have conjugates being multiplied on the bottom, which we can write as the difference of squares.\[\large y'=\frac{2}{3}\left(\frac{1}{x^2-1}\right)\]

Answer 28

Wow this is a really mean problem :O lol

Imma throw it into wolfram a sec, to make sure I didn't screw up anywhere.

Answer 29

Yep looks like I did it correctly.
It's a ton of steps :( If you're really confused on any part you can let me know.

Answer 30

No, all the steps made sense. Thank you very much!