Use the Fundamental Theorem of Calculus to find the derivative of
f(x)=∫[4,x^2]((1/4)t^2−1)^15 dt
that would be
((1/4)(x^2)^2−1)^15 * 2x
= 2x(x^4/4 - 1)^15
To fully understand what Steve did, watch these two KhanAcademy clips
https://www.khanacademy.org/math/ap-calculus-ab/fundamental-theorem-of-calculus-ab/fundamental-theorem-of-calculus-tut-ab/v/fundamental-theorem-of-calculus
and
https://www.khanacademy.org/math/ap-calculus-ab/fundamental-theorem-of-calculus-ab/fundamental-theorem-of-calculus-tut-ab/v/applying-the-fundamental-theorem-of-calculus
special emphasis on the second of those two.
To find the derivative of the function f(x), we can use the Fundamental Theorem of Calculus. The theorem states that if we have a function defined by an integral of another function, then the derivative of the integral function is equal to the original function.
In this case, our function f(x) is defined as the integral of ((1/4)t^2 − 1)^15 with respect to t, where the lower limit of integration is 4 and the upper limit is x^2. To find the derivative of f(x), we can differentiate the integral function with respect to x.
Let's start by rewriting the integral function as a function of x:
f(x) = ∫[4,x^2]((1/4)t^2 − 1)^15 dt
Notice that the upper limit of integration is x^2, which depends on the variable x. To differentiate the function, we need to apply the chain rule. The chain rule states that if we have a composite function, such as g(h(x)), the derivative is given by:
(g(h(x)))' = g'(h(x)) * h'(x)
Using this rule, let's differentiate the integral function. We denote the integral function as F(x):
F(x) = ∫[4,x^2]((1/4)t^2 − 1)^15 dt
To apply the chain rule, we need to differentiate the integrand ((1/4)t^2 − 1)^15 with respect to x. Let u = (1/4)t^2 − 1, then we have:
F(x) = ∫[4,x^2] u^15 dt
To differentiate u^15 with respect to x, we can use the chain rule. Let's denote du/dx as du/dt * dt/dx:
F'(x) = (d/du)(u^15) * (du/dt * dt/dx)
Now, let's find the derivative of u^15 with respect to u:
(d/du)(u^15) = 15u^14
Next, let's find du/dt. Recall that u = (1/4)t^2 − 1:
du/dt = (d/dt)[(1/4)t^2 − 1] = (1/2)t
Finally, let's find dt/dx. Since the upper limit of integration is x^2, we can differentiate this with respect to x:
dt/dx = (d/dx)(x^2) = 2x
Now, we can substitute all the derivatives back into the expression for F'(x):
F'(x) = 15u^14 * (1/2)t * 2x
Simplifying this expression, we have:
F'(x) = 15u^14tx
Substituting back u = (1/4)t^2 − 1, we get:
F'(x) = 15((1/4)t^2 − 1)^14tx
Therefore, the derivative of f(x) is given by:
f'(x) = 15((1/4)x^2 − 1)^14xt
Note that this is the derivative of the original function f(x), and we obtained it by applying the Fundamental Theorem of Calculus and using the chain rule.