Question

How do you use Part 1 of the Fundamental Theorem of Calculus to find the derivative of the function `h(x) = int sqrt(6 + r^3) dx` from 7 to `x^2`?

Answer 1

We need FTC 1 and the chain rule.

Explanation:

`h(x) = int_7^(x^2) sqrt(6 + r^3) dx` does not quite fit the form:

`g(x) = int_a^b f(t) dt` whose derivative is `f(x)`

But we can use the chain rule here;

The `h(x)` can be written as

`h(u) = int_7^u sqrt(6 + r^3) dx` where `u = x^2`

So

`d/dx(h) = (dh)/(du)* (du)/dx`

`= sqrt(6 + u^3) (du)/dx`

`= sqrt(6 + (x^2)^3)*(2x)`

`= 2xsqrt(6 + x^6)`

Summary

For
`h(x) = int_a^(g(x)) f(t) dt` `" "`(With the right kind of `f` and domain of `h`)

We have

`h'(x) = f(g(x))*g'(x)` `" "` (Which is one way of writing the chain rule.)