# Homework Help: Differentiate with respect to x

1. Dec 22, 2011

### lubo

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

y = 3*cube root 3√(x2-1)

2. Relevant equations

Udu/dx + Vdv/dx (wrong as per below)

Chain Rule now used

3. The attempt at a solution

I have:

Now y = 3*(x2-1)^1/3

let z=x2-1 y= 3(z)^1/3

dz/dx = 2x dy/dz = z^-2/3 = (x2-1)^-2/3

all = dy/dx = dz/dx * dy/dz = 2x*(x2-1)^-2/3 = 2x/(x2-1)^2/3

Instead I have an answer in my book:

dy/dx = 3*1/3(x2-1)^-2/3

which in the end = (x2-1)^-2/3

I believe my answer is correct and it is a typing mistake, but I am not 100% sure.

Thank you for any help in advance.

Last edited: Dec 22, 2011
2. Dec 22, 2011

### Pengwuino

You didn't use/need the product rule here. You just need the chain rule

${{dy}\over{dx}} = {{dy}\over{du}}{{du}\over{dx}}$

where $u = x^2 -1$ and $y(u) = 3u^{1/3}$.

Neither your solution or the books solution is correct, though (are you sure you copied the problem down correctly?). I'm not sure how your last line came to be.

3. Dec 22, 2011

### epenguin

Seeing the answer that came after I started this post, I already thought the quoted book answer doesn't seem right either. Easiest thing is to restart from scratch.

4. Dec 22, 2011

### lubo

Pengwuino and Epenguin - I have anotated the chain Rule. Thanks for that. I am sure the question is copied correctly, but I have made it a little more clearer. It is hard getting used to looking at all the stuff needed for x squared as an example.

Also I am getting used to how this site works.....

Please have another look and check it again. If it is still no good, could you please post what you think the answer is as I have two and this is the best I have.

5. Dec 22, 2011

### Staff: Mentor

This is correct, but you should put parentheses around your exponent.
2x*(x2-1)^(-2/3) = 2x/(x2-1)^(2/3)
This answer is incorrect - it's missing the factor 2x that comes from the chain rule.

6. Dec 22, 2011

### lubo

Thank you. It feels good to have that cleared up.

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