Question

Can someone know if I've set up this problem right, I'm trying to find the derivative of this function using product rule (not chain rule )

h(t) = t squared ( 1 - t^2)

Answer 1

So far I've gotten this:
(1/2 t ^ -1/2) (1 - t ^ 2) + (t ^ 1/2) ( - t)

Answer 2

what do you mean by "not chain rule"

Answer 3

sorry that t ^ 2 should be square rooted

Answer 4

Here is the derivative.
h'(t) = 2t - 4t^3

Answer 5

take a deep breathe, and do us all a favor; re write this so it resembles some sort of coherent mathematical notation :)

Answer 6

alright hold on

Answer 7

\[h(t) = \sqrt{t}(1-t ^{2})\]

Answer 8

no matter how you do this; other than distributing the sqrt(t) thru the (..) you are going to have to implement the "chain rule" ....

Answer 9

teh product rule is simply:

rl = r'l + rl' ; where right is left and left is right :) just looks better
to me like this ...

r = sqrt(t) ; r' = 1/2(sqrt(t))

l = (1 - t^2) ; l' = -2t .. and thats where the chain rule would have
been :)

Answer 10

r = sqrt(t) ; r' = 1/2(sqrt(t))
l' = -2t l = (1 - t^2)
-------------------------
\(-2t\sqrt{t}\) + \(\cfrac{(1-t^2)}{2\sqrt{t}}\)

Answer 11

simplify as needed