Question

Calculus help?

Answer 1

\[\large f(x)=\sqrt[3]{(x^2-1)^2}\] Find f'(x) and state its domain.

Answer 2

\[f(x)=\sqrt[3]{(x^2-1)^2}=(x^2-1)^{2/3}\]

Answer 3

\[f'(x)=\frac{\text d }{\text dx}(x^2−1)^{2/3}\]

Answer 4

\[\frac{\text d }{\text d x}\left(f(x)^n\right)=nf^{n-1}\]
\[\frac{\text d }{\text d x}\left(f \circ g\right)=f'(g)\cdot g'\]

Answer 5

So\[\large f'(x)=(x^2-1)^{\frac23}+\frac23(x^2-1)^{-\frac13}(2x)x\]?

Answer 6

where did that plus sign come from?

Answer 7

Product rule, I thought.

Answer 8

there is no product

Answer 9

i have the chain rule \((f \circ g)'\)

Answer 10

Oh shoot\[\large f(x)=x \sqrt[3]{(x^2-1)^2}\]This is the correct function, sorry. I think I got the derivative, but I'm stuck in finding the domain.

Answer 11

well now the question has changed you almost had the right solution

Answer 12

using the product rule was right

Answer 13

and you actually had it more than almost right you have it exactly right

Answer 14

i cant see any numbers to exclude form the domain

Answer 15

Well, I have x≠±1. But the answer sheet also says -1.4<x<1.4. I don't get that part :(

Answer 16

im getting x cannot be \([-1,1]\)

Answer 17

That's what I have too. But that's incomplete apparently

Answer 18

but i mean
x cannot be -1, any where in between up 1

Answer 19

actually it can be between. i think you are right and the answer has not got the right answer

Answer 20

answer sheet is wrong*

Answer 21

ok thanks :) yeah maybe