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- anonymous

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

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- anonymous

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

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- anonymous

I will just do the first term
1) When you are taking chain rule always takes derivative of outside function first

- anonymous

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

- anonymous

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

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- anonymous

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

- anonymous

I don't get it haha.

- anonymous

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

- anonymous

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

- anonymous

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)

- anonymous

In your function you have two things mutliplied together.

- anonymous

okay, let me try and figure it out. thanks

- anonymous

How is going there?

- anonymous

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.

- anonymous

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

- anonymous

please do

- anonymous

Here how I would solve it

- anonymous

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

- anonymous

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

- anonymous

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]

- anonymous

You left out derivative of inside function

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