Question

Find the d/dx of the integral of e^t^3 dt from 3 to x^2

Answer 1

What have you tried with this problem so far? Any ideas on what we might use?

Answer 2

I tried the FFT, but I am having trouble solving it because of the log.

Answer 3

First fundamental theorem? That sounds like a good idea. And if you could, it'd be great to see the work up to where you got stuck!

Answer 4

F'(x^2)-F'(3)
This is where I am stuck....2xe(^-x^3)-(e(^-3^2))

Answer 5

Why you stuck?

Answer 6

That setup looks good. And then you take the derivative of F(x^2) - F(3).
d/dx ( F(x^2) - F(3) )
We take the derivative by terms first.
d/dx( F(x^2) - d/dx( F(3) )
But notice that the second term is just a constant. We plug in a value for the variable, so F(3) is just some number. What is the derivative of a constant?

Answer 7

The derivative of a constant is zero, so d/dx(Fx^2)-d/dx(F(3))=2x-0?

Answer 8

We're getting closer.
We are left with d/dx( F(x^2) ) after making that second part 0. And I agree that we need chain rule. But I think we just need to be a bit more careful.
d/dx(x^2) * f(x^2)
This time f(x^2) is from the integrand, we plug in x^2 directly for t here: \(f(t) = e^{t^3} \)

Answer 9

So f(x^2)=e(^x^2)^3) or e^x^6

Answer 10

Yes! We end up with 2x * e^(x^6) now, making sure our chain rule is accounted for. :)

Answer 11

Thank you so much. I was really over-thinking this problem.

Answer 12

Glad to help! :)