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Calculus help. Fairly easy stuff.

FrogDog

Diamond Member
I have to find the derivative of this:

y = [(x - 1) / (x + 1)] ^ 3

I have to do it using the chain rule. I should be able to do this but I'm not getting the answer that's in my book, which is

6(x - 1) ^ 2 / (x + 1) ^ 4

So what I'm doing is letting u = (x - 1) / (x + 1) so that y = u ^ 3. Then I'm solving for the derivate of both u(x) and y(u) and multiplying them together.

u(x) = (x - 1) / (x + 1)
(I'm getting 0 for u'(x) and this must be what I'm doing wrong. I'm using the quotient rule, not sure why I'm getting the wrong answer here)

y(u) = 3u ^ 2
After I sub in the original function for u I end up with this for y'(u):

3{ [(x - 1) / (x + 1)] ^ 2 }

Then multiplying those two derivatives together which should give me the derivative of f(x).

The answer I got was 0 obviously.

Well, if anyone can make sense of all that jibrish I'd love to see someone tell me what I'm doing wrong. 🙂 If you want me to post what I did to get the derivative of u(x), the one I got 0 for, I will, but it's a lot of writing so I won't post it initially.

Edit: Oh for god's sake, after writing all that I found my problem. Doh!
 
y = [(x - 1) / (x + 1)] ^ 3

let u = (x-1)/(x+1)

therefore, y= u^3

dy/dx = du/dx*dy/du

du/dx = [(x+1) - (x-1)]/ (x+1)^2

du/dx = 2/(x+1)^2

and dy/du = 3u^2 = 3[(x-1)/(x+1)]^2

and finally, dy/dx = 6*[(x-1)^2]/(x+1)^4
 


<< y = [(x - 1) / (x + 1)] ^ 3

let u = (x-1)/(x+1)

therefore, y= u^3

dy/dx = du/dx*dy/du

du/dx = [(x+1) - (x-1)]/ (x+1)^2

du/dx = 2/(x+1)^2

and dy/du = 3u^2 = 3[(x-1)/(x+1)]^2

and finally, dy/dx = 6*[(x-1)^2]/(x+1)^4
>>



U got it first .........just a small mistake in his ans.

not yours............lol
 
Thank you, Vespasian! That helped greatly. After I found out what my problem was in my edit there I was still hung up on the last step. I did out the (x + 1) ^ 2 part on the bottom which I shouldn't have and it had me confused. Thanks!

Oh and yeah there was a typo in my first message where I had my book's answer. It actually does match yours, Vespasian.
 
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