Ask your own question, for FREE!
Mathematics 42 Online
OpenStudy (anonymous):

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

OpenStudy (anonymous):

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

OpenStudy (anonymous):

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

OpenStudy (anonymous):

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

OpenStudy (anonymous):

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

OpenStudy (anonymous):

I don't get it haha.

OpenStudy (anonymous):

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

OpenStudy (anonymous):

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

OpenStudy (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)

OpenStudy (anonymous):

In your function you have two things mutliplied together.

OpenStudy (anonymous):

okay, let me try and figure it out. thanks

OpenStudy (anonymous):

How is going there?

OpenStudy (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.

OpenStudy (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

OpenStudy (anonymous):

please do

OpenStudy (anonymous):

Here how I would solve it

OpenStudy (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

OpenStudy (anonymous):

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

OpenStudy (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]

OpenStudy (anonymous):

You left out derivative of inside function

Can't find your answer? Make a FREE account and ask your own questions, OR help others and earn volunteer hours!

Join our real-time social learning platform and learn together with your friends!
Can't find your answer? Make a FREE account and ask your own questions, OR help others and earn volunteer hours!

Join our real-time social learning platform and learn together with your friends!