Differentiate with respect to x

[lubo]

Homework Statement

y = 3*cube root 3√(x²-1)

Homework Equations

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

Chain Rule now used

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.

 

[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.

 

[epenguin]

Homework Statement

y = 3*cube root 3√(x²-1)

Homework Equations

Udu/dx + Vdv/dx not the appropriate formula, but you compensate for this by applying the right one instead of the one you quote

The Attempt at a Solution

I have:

U=x²-1 v= 3(u)^1/3

du/dx = 2x dv/du = u^-2/3

all = 2x*(x²-1)^-2/3 OK but you then haven't carrie through the whole bracket to the next step= 2x/(x²)^2/3

Instead I have an answer in my book:

dy/dx = 3*1/3(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.

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.

 

[lubo]

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.

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.

 

[Mark44]

Homework Statement

y = 3*cube root 3√(x²-1)

Homework Equations

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

Chain Rule now used

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

This is correct, but you should put parentheses around your exponent.
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

This answer is incorrect - it's missing the factor 2x that comes from the chain rule.

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.

 

[lubo]

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