Question

what is the derivative of 2x(x-2)^3?

Answer 1

if you don't want to expand (which is not that hard) and use the power rule, you can use the product rule

Answer 2

How would I use the power rule? Do i distribute the 2x?

Answer 3

expanding gives \(2 x^4-12 x^3+24 x^2-16 x\) and that is easy to find the derivative of

Answer 4

but you can use
\[(fg)'=f'g+g'f\] with \[f(x)=2x,f'(x)=2,g(x)=(x-2)^3,g'(x)=3(x-2)^2\]

Answer 5

Would you mind walking me through that? I'm not quite sure how you got the end product.

Answer 6

you mean how i got this\[2x(x-2)^3=2 x^4-12 x^3+24 x^2-16 x\]??

Answer 7

Yea, that would be awesome.

Answer 8

\[2x(x-2)^3=2x(x-2)(x-2)(x-2)\] is a start
then either grind it til you find it, or recall that \((a-b)^3=a^3-b^3=a^3-3 a^2 b+3 a b^2-b^3\)

Answer 9

typo there, i meant \[(a-b)^3=a^3-3 a^2 b+3 a b^2-b^3\]

Answer 10

so you get
\[(x-2)^3=x^3+6x^3+12x-8\]

Answer 11

then multiply everything by \(2x\) and get the answer i wrote above

Answer 12

Oh, I see. Thank you so much. Breaking down the problem makes it much simpler.

Answer 13

yw
btw you can still use the product rule, then you don't have to do all that multiplication

Answer 14

\[f(x)=2x(x-2)^3= 2x^4-12x^3+24x^2-16x f'(x)=8x^3-36x^2+48x-16\]