Question

Find the derivative of the function.
h(t) = (t4 - 1)^5 (t3 + 1)^8

5(t^4 - 1)^4 (4 * t^3)
8(t^3+1)^7(3*t^2)
h'(t) = (t^4 -1)^5 (8(t^3+1)^7(3*t^2))+(t^3+1)^8(5(t^4-1)^4(4 *t^3)

Now I am confused with the problem from this point on how to get the answer, the video lecture skips exactly which steps to factor to get the right answer.

Answer 1

\(h'(t) = \color {red}{((t^4-1)^5)' }*(t^3+1)^8+(t^4-1)^5*\color{blue}{((t^3+1)^8)'}\)

Answer 2

\(h'(t) = \color {red}{(5(t^4-1)^4)(4t^3) }*(t^3+1)^8+(t^4-1)^5*\color{blue}{(8(t^3+1)^7(3t^2)}\)

Answer 3

factor out : \(h'(t) = 2t^2(t^4-1)^4(t^3+1)^7[10t(t^3+1)+12(t^4-1)]\)

Answer 4

The part I am confused with your second answer to your third answer which is the solution. I am confused how it goes to 2t^2 in front and how you get 10t and 12 to the back of the problem.

Answer 5

from 2, the first term: number: 5*4 =20, right?
the second term: number: 8*3 = 24,
instead of factor 4 out, I just take 2 out, you should take out 2 more , hehehe. my mistake

Answer 6

\(h'(t) = \color {red}{(5(t^4-1)^4)(4t^3) }*(t^3+1)^8\\~~~~~~~~~~~~~~+(t^4-1)^5*\color{blue}{(8(t^3+1)^7(3t^2)}\)
rearrange:
\(h'(t) = \color {red}{(20(t^4-1)^4)(t^3) }*(t^3+1)^8\\+~~~~~~~~~~24(t^4-1)^5*\color{blue}{((t^2)(t^3+1)^7}\)

Answer 7

Factor the smallest term out to get my third line

Answer 8

Okay thanks, I finally see what you did in the problem. thanks.