Question

h(t)=(t^4-1)^3) (t^3+1)^4
the chain rule. help please.

Answer 1

I will just do the first term

1) When you are taking chain rule always takes derivative of outside function first

Answer 2

h(t)=(t^4-1)^3 (t^3+1)^4

Answer 3

2) derivative of outside function is
3(t^4-1)^2

derivative of inside function is
4t^3

multiply inside and outside derivative

Answer 4

This is a chain rule problem and a product rule. So be careful.. if you don't get it reply.

Answer 5

I don't get it haha.

Answer 6

Yes I know but he just wanted help with chain rule so I used the first term as example

Answer 7

i'm actually a she. Since it is a product rule how would i find in the outside and inside function?

Answer 8

You will have to use both product rule and chain rule to solve whole problem.

The product rule is = F(x)'g(x)+g(x)'(F(x)

Answer 9

In your function you have two things mutliplied together.

Answer 10

okay, let me try and figure it out. thanks

Answer 11

How is going there?

Answer 12

okay, so i got the derivative of the first one, which was the same as you got. and i ended up with a long equation.

Answer 13

Yes it's pretty messy if you don't simplify it.
i can tell you the answer if you think you got it or you're stumped

Answer 14

please do

Answer 15

Here how I would solve it

Answer 16

applying product rule
(t^4-1)^3)* 4(t^3+1)^3 * 3t^2 + (t^3+1)^4 *3(t^4-1)^2 *4t^3
From here it is just algebra

Answer 17

If It is not a word problem, you can leave it here

Answer 18

im so confused! okay so after i take the derivative of the outside functions, what do i do? i ended up with 3(t^4-1)^2 (t^3+1)^4 + (t^4-1)^2 [4(t^3+1)^3]

Answer 19

You left out derivative of inside function