anonymous
  • anonymous
h(t)=(t^4-1)^3) (t^3+1)^4 the chain rule. help please.
Mathematics
chestercat
  • chestercat
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anonymous
  • anonymous
I will just do the first term 1) When you are taking chain rule always takes derivative of outside function first
anonymous
  • anonymous
h(t)=(t^4-1)^3 (t^3+1)^4
anonymous
  • 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
  • anonymous
This is a chain rule problem and a product rule. So be careful.. if you don't get it reply.
anonymous
  • anonymous
I don't get it haha.
anonymous
  • anonymous
Yes I know but he just wanted help with chain rule so I used the first term as example
anonymous
  • anonymous
i'm actually a she. Since it is a product rule how would i find in the outside and inside function?
anonymous
  • 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
  • anonymous
In your function you have two things mutliplied together.
anonymous
  • anonymous
okay, let me try and figure it out. thanks
anonymous
  • anonymous
How is going there?
anonymous
  • 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
  • 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
  • anonymous
please do
anonymous
  • anonymous
Here how I would solve it
anonymous
  • 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
  • anonymous
If It is not a word problem, you can leave it here
anonymous
  • 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
  • anonymous
You left out derivative of inside function

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