Question

How to differentiate this:
y = (x^3 - 3x^2 + 1)^9

Answer 1

product and chain rules

Answer 2

\[\frac{d}{dz}(z)^9|_{z=x^3-3x^2+1} \cdot (x^3-3x^2+1)'\]

Answer 3

y = 9(x^3 - 3x^2 + 1)^8 * (x^3 - 3x^2 + 1)

Answer 4

?

Answer 5

Well you got the outside derivative right
but you still didn't take derivative of the inside

Answer 6

\[y'=\frac{d}{dz}(z)^9|_{z=x^3-3x^2+1} \cdot (x^3-3x^2+1)' \\ y'=9(x^3-3x^2+1)^8 \cdot (x^3-3x^2+1)'\]

Answer 7

Okay. 9(3x^2 - 6x)(x^3 - 3x^2 +1)^8

Answer 8

looks tons better

Answer 9

Then do we do this:

Answer 10

a = (3x^2 - 6x)
b = (x^3 - 3x^2 +1)^8

a^1 x b - a x b^1

Answer 11

when i said tons better means it was good
it was right

Answer 12

what are you doing now?

Answer 13

oh okay. nevermind.

Answer 14

thanks. :)

Answer 15

unless you want to find the second derivative ?

Answer 16

because when I look at it I think that is what you want to do now
since you were applying product rule and we have a product
and I guess by f^1 that is f'