Thread: Finishing this derivative problem (work included)

Can someone please help me finish this derivative problem?
Here's the problem:



Here's my work so far:

f '(v) = 3[ (1-2v) / (1+v)] ^2 times [(1+v)(-2) - (1-2v)(v) / (1+v)^2 ]
f '(v) = [3(1-2v)^2] / (1+v)2 times [-2 - 2v - v + 2v^2] / (1+v)^2 ]
f '(v) = " times (-2-3v+2v^2) / (1+v)^2
f '(v) = [3(1-2v)^2][-2-3v+2v^2] / [1+v]^4


..thats as far as I got, but the answer is:

How did they get that??

Originally Posted by janedoe

[snip]

Here's my work so far:

f '(v) = 3[ (1-2v) / (1+v)] ^2 times [(1+v)(-2) - (1-2v)(1) / (1+v)^2 ]

[/snip]

See my minor correction in red.

Ahh thank you!!

But now I get: -9(1-2v)^2 / (1+v)^4

their answer is 2v-1 instead ... ?

is the answer: 3*[(1-2v)^2*(v-3)]/[1+v)^4

no its



but I keep getting 1-2v in stead of 2v-1 and so does my friend....so we think it's the book's error.

this is my solution

= [(1-2v)/(1+v)]^3
= 3*[[(1-2v)/(1+v)]^2]*[(1-2v+2+2v)/[(1+v)^2]]
= 9*[[(1-2v)^2]/[(1+v)^4]]

Man, don't you know that
.It's really easy to prove this.
So and your's and book's answer is correct.

Hello, janedoe!

But now I get: .

Their answer has

You both are correct!

Note that: .

But I have no idea why they switched that binomial.
.

I don't think there's a - sign, my answer is 9, not -9