deriv.
h(t)=(tˆ4-1)ˆ3(tˆ3+1)ˆ4
product rule and chain rule:
h = uv
h' = u'v + uv'
if u=w^n, u' = n w^(n-1) w'
so, whatcha got?
(3t^4-3)^2(4t^6+4t^3)^4+(t^4-1)^3(12t^4+12t)^3
but i'm not sure if this is a simple as it gets.
To find the derivative of the function h(t) = (t^4 - 1)^3(t^3 + 1)^4, we can use the product rule and chain rule of differentiation.
First, let's simplify the function using the power rule for exponents:
h(t) = [(t^4)^3 - 1^3][(t^3)^4 + 1^4]
= (t^12 - 1)(t^12 + 1)
= t^24 - 1
Now, let's find the derivative of h(t) using the power rule:
h'(t) = 24t^(24-1)
= 24t^23
Therefore, the derivative of h(t) is h'(t) = 24t^23.