h(t)= cubed root t(t^2=4) using the product rule. how do i set it up
To solve the problem using the product rule, you first need to determine the two functions you will be differentiating. In this case, you have the function h(t) and you can break it down into two functions:
1) f(t) = ³√t (the cube root of t)
2) g(t) = t² - 4
Now, let's set up the differentiation using the product rule. The product rule states that the derivative of the product of two functions f(t) and g(t) is given by:
d/dt [f(t) * g(t)] = f'(t) * g(t) + f(t) * g'(t)
Applying this to the functions f(t) and g(t) we have:
d/dt [h(t)] = d/dt [f(t) * g(t)] = f'(t) * g(t) + f(t) * g'(t)
Now, let's find the derivatives:
- f'(t) = d/dt [³√t]
- g'(t) = d/dt [t² - 4]
To calculate f'(t), we need to apply the chain rule since we have a composite function. The chain rule states that if we have a composite function h(u), where u = g(t), the derivative of h(u) with respect to t is given by:
d/dt [h(u)] = d/du [h(u)] * d/dt [g(t)]
So, applying the chain rule to our function f(t) = ³√t, we have:
f'(t) = d/du [³√u] * d/dt [t]
= (1/3) * (u^(-2/3)) * 1 / dt
= (1/3) * (t^(-2/3)) * 1
= t^(-2/3) / 3
Now, let's calculate g'(t):
g'(t) = d/dt [t² - 4]
= (d/dt [t²]) - (d/dt [4])
= 2t - 0
= 2t
Finally, we can substitute the derivatives back into the product rule equation:
d/dt [h(t)] = f'(t) * g(t) + f(t) * g'(t)
= (t^(-2/3) / 3) * (t² - 4) + (³√t) * (2t)
That's the setup to differentiate the function h(t) = ³√t(t² - 4) using the product rule.