Question

Find the derivative with and without using the chain rule. f(x) = (x^2 + 1)^3

Answer 1

\[f(x)=(x^2+1)^3=(x^2+1)(x^2+1)(x^2+1)\]
\[f'(x)=6x(x^2+1)^2\]

Answer 2

used only product rule

Answer 3

is that using the chain rule?

Answer 4

no. i said i only used product rule.

Answer 5

okay

Answer 6

One, do you understand how to find the derivative `with` using chain rule?
You would just take the derivative as is.

Answer 7

not 100%

Answer 8

okay i'll try doing with the chain rule

Answer 9

Ok :)

Answer 10

okay can i post it here and correct me if i i mistake somewhere

Answer 11

k

Answer 12

\[d/dx ((x^2 + 1)^3) = 3(x^2 + 1)^2 d/dx(x^2 + 1) = 3(d/dx(x^2)(x^2 - 1)^2 = 3(2x)(x^2 = 1)^2 = 6x(x^2 + 1)^2 \]

Answer 13

@zepdrix

Answer 14

I'm not quite sure why it changed from x^2+1 to x^2-1. Was that a typo?

Answer 15

i see that it didn't show up all of it but here is the last part 3(2x)(x^2 + 1)^2 = 6x(x^2 + 1)^2

Answer 16

yes typo for that one

Answer 17

\[\large f'(x)=3(x^2+1)^2\frac{d}{dx}(x^2+1)\]\[\large f'(x)=3(x^2+1)^2(2x) \qquad \qquad f'(x)=6x(x^2+1)^2\]

Ok yah looks good :)

Answer 18

thanks!

Answer 19

`without` chain rule is going to be a little tricky. They want you multiply out your binomial :O oh boy.

Answer 20

ohh so the one @ingenuus showed is nto by without chain rule?

Answer 21

not*

Answer 22

the one ingenuus showed its using the product rule

Answer 23

That's another method for doing it without chain rule.
Which involved the product rule with `3 terms`.
Which is a little bit tricky if you've never seen it before.

Product rule with 2 terms is probably what you're used to seeing.
I was referring to a different method, multiplying out all the brackets as Wio showed you the last time you posted this. :D

Answer 24

Ohhhh okay got you