Question

Derive f ' (x) for f(x) = 2x^2+2x/(1+2x)^2

Answer 1

\[2x ^{2}+2x \div(1+2x)^{2}\]

Answer 2

use the difference rule.

\[(\frac{ u }{v })' = \frac{ \left( (u' * v) + v' * u\right) }{ v ^{2} }\]

Answer 3

In other words, let the numerator equal to u and the denominator equal to v and then solve.

Answer 4

I did. I'm having trouble simplifying it. Here's where I am stuck \[(1+2x)^{2}(2)(2x+1)-2x(x+1)2(1+2x)(2)\div(1+2x)^{4}\]

Answer 5

do you know chain rule?

Answer 6

you can avoid using quotiont rule and turn it into a product rule problem as:
\[f(x)=2x^2+2x(1+2x)^{-2}\]

Answer 7

interesting... I do know the chain rule. I'm having trouble simplifying it though.

Answer 8

(1+2x)2(2)(2x+1)−2x(x+1)2(1+2x)(2)÷(1+2x)4 Once I'm here I don't know where to go next.

Answer 9

there is always more than 1 way to eat a pie

Answer 10

especially in math

Answer 11

\[
f'(x)=4x+\left[(2)(1+2x)^{-2}+(2x)(-2)(1+2x)^{-3}(2)\right]
\]
simplify and thats the answer using my method!

Answer 12

can you re-write your step legibly?

Answer 13

I did but I think its wrong... Let me show you...

Answer 14

\[4\div(1+2x)^{5}\]

Answer 15

that small?
here using quotient rule:
\[
f'(x)=4x+\frac{(2)(1+2x)^2-(2x)(2)(1+2x)^1(2)}{(1+2x)^4}
\]

Answer 16

is it 4/(1+2x)^3?

Answer 17

how?
\[
f'(x)=4x+\frac{2(1+2x)-(2x)(4)}{(1+2x)^3}\\
f'(x)=4x+\frac{2(1-2x)}{(1+2x)^3}
\]
i do not think this can be simplified further

Answer 18

what happens to the (2x)(4)?

Answer 19

the numerator of the second term got simplified