Question

Calc derivatives would f(x) = e^[(4x-1)^2] just be f'(x) = e^[(4x-1)^2] ?

Answer 1

No You need to apply the chain rule

Answer 2

let me work this out for a second

Answer 3

ok

Answer 4

wait, I am lost :(

Answer 5

I would start by doing (4x-1)^(2) (e) ?

Answer 6

i mean (4x-1)^(2) (e)^[(4x-1)^(2) -1]

Answer 7

\[\frac{d}{du}e^u = e^u \cdot u' \]

Answer 8

no its not like that,use this rule \[\huge (e^{f(x)})'=f'(x)e^{f(x)}\]

Answer 9

so u need to derive \[(4x-1)^2\]

Answer 10

ok, 2(4x-1)

Answer 11

use chain rule

Answer 12

\[\frac{d}{dx} e^{(4x-1)^2}=e^{(4x-1)^2}\cdot \frac{d}{dx}(4x-1)^2\]

Answer 13

Don't forget to find the derivative inside of (4x-1) when using the chain rule

Answer 14

Oh, thanks. I almost forgot

Answer 15

Ok, so would the answer be
\[e ^{4x-1^{2}} (8x+2)\]

Answer 16

Euh not quite

Answer 17

What did i miss? Is it just simplification?

Answer 18

where did + come from

Answer 19

the (8x+2) factor is wrong. And you forgot the parentheses in the exponential (4x-1)^2

Answer 20

so we shud have \[(2)(4x-1)(4)=8(4x-1)\]

Answer 21

^yup

Answer 22

Oh, i see it. I skipped a step

Answer 23

Wait, why is it (2)(4x−1)(4) and not (2)(4x−1) + (4) ?

Answer 24

never mind, I see. I mixed up my rules

Answer 25

Thank you both. I don't know who to give the medal to, so I will give it to the lower rank to help them out, but thank you both!!!

Answer 26

I'll give a medal too haha