# How do you use the second fundamental theorem of Calculus to find the derivative of given int sect tant dt from [0, x^3]?

Apr 11, 2016

You have two choices for how to do this.

#### Explanation:

Method 1
${\int}_{0}^{{x}^{3}} \sec t \tan t \mathrm{dt} = F \left({x}^{3}\right) - F \left(0\right)$ where $F$ is an antiderivative of $\sec t \tan t$.

Now use the chain rule to differentiate $F \left({x}^{3}\right)$ with respect to $x$

We get $\sec {x}^{3} \tan {x}^{3} \left(3 {x}^{2}\right)$

Method 2

(Actually evaluate the definite integral.)

Since $\frac{d}{\mathrm{dt}} \left(\sec t\right) = \sec t \tan t$, we get $\sec t$ is an antiderivative of $\sec t \tan t$.

So, applying the second fundamental theorem of calculus, we find

int_0^(x^3) sect tant dt = sect]_0^(x^3) = secx^3-sec0

Now we differentiate to answer the question:

$\frac{d}{\mathrm{dx}} \left(\sec \left({x}^{3}\right)\right) = \sec {x}^{3} \tan {x}^{3} \left(3 {x}^{2}\right)$ (rearrange to taste).