# 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?

Aug 22, 2015

We need FTC 1 and the chain rule.

#### Explanation:

$h \left(x\right) = {\int}_{7}^{{x}^{2}} \sqrt{6 + {r}^{3}} \mathrm{dx}$ does not quite fit the form:

$g \left(x\right) = {\int}_{a}^{b} f \left(t\right) \mathrm{dt}$ whose derivative is $f \left(x\right)$

But we can use the chain rule here;

The $h \left(x\right)$ can be written as

$h \left(u\right) = {\int}_{7}^{u} \sqrt{6 + {r}^{3}} \mathrm{dx}$ where $u = {x}^{2}$

So

$\frac{d}{\mathrm{dx}} \left(h\right) = \frac{\mathrm{dh}}{\mathrm{du}} \cdot \frac{\mathrm{du}}{\mathrm{dx}}$

$= \sqrt{6 + {u}^{3}} \frac{\mathrm{du}}{\mathrm{dx}}$

$= \sqrt{6 + {\left({x}^{2}\right)}^{3}} \cdot \left(2 x\right)$

$= 2 x \sqrt{6 + {x}^{6}}$

Summary

For
$h \left(x\right) = {\int}_{a}^{g \left(x\right)} f \left(t\right) \mathrm{dt}$ $\text{ }$(With the right kind of $f$ and domain of $h$)

We have

$h ' \left(x\right) = f \left(g \left(x\right)\right) \cdot g ' \left(x\right)$ $\text{ }$ (Which is one way of writing the chain rule.)