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# Find F'(x)?

If F(x)=∫ sqrt(1+t^3)dt from 1 to x^2, find F'(x).

The Second Fundamental Theorem of Calculus states:

d/dx (∫{a,x} f(t) dt) = f(x)

Using the chain rule:

d/dx (∫{a,u(x)} f(t) dt) = (f(x))(du/dx)

So if F(x) = ∫{1,x^2} √(1+t^3) dt, then

F'(x) = (√(1+t^3))(2x) = 2x√(1+t^3)
Use chain rule

F'(x) = dF(x)/dx = (dF(x)/dx2)(dx2/dx) = 2x (dF(x)/dx2)  (1)

When you differentiate the integral with variable upper limit it gives you the function being integrated.

Thus (because the lower limit doesn't play any role being a constant number)

dF(x)/dx2 = √(1 + t3)  where t = x2                         (2)

F'(x) = 2x√(1+x6)

Comment. There is no need to integrate and then differentiate by x.

If you want to type x6 instead of x^6 simply type x6, highlight only 6 and click on the icon (x2).

Good luck!

X^2                                         X^2
F( X) =  ∫     √( 1 + t^3) dt = √ (1 + t^3 ) du/dx       = 2X √(1+ X^6) - ( 1- 1) = 2X √( 1 + X^6)
1                                              1

Note: Integration and differentiation are inverse of each other.