# Thread: Finishing this derivative problem (work included)

1. ## Finishing this derivative problem (work included)

Here's the problem:

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

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: $[ -9(2v - 1)^2 ] / (1+v)^4$

How did they get that??

2. 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.

3. Ahh thank you!!

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

5. 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.

6. 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]]

7. Man, don't you know that
$(a-b)^2=(b-a)^2$ .It's really easy to prove this.
So $(1-2v)^2=(2v-1)^2$ and your's and book's answer is correct.

8. Hello, janedoe!

But now I get: . $\frac{-9(1-2v)^2}{(1+v)^4}$

Their answer has $(2v-1)^2$
You both are correct!

Note that: . $(1-2v)^2 \:=\: (2v-1)^2$

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

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