# How do you find the derivative of F(x) = int sqrt(1+sec(3t)) dt?

Oct 24, 2015

Without knowing the limits of integration all we can say if the following.

#### Explanation:

For $F \left(x\right) = {\int}_{f \left(x\right)}^{g} \left(x\right) \sqrt{1 + \sec \left(3 t\right)} \mathrm{dt}$

$F ' \left(x\right) = \sqrt{1 + \sec \left(3 g \left(x\right)\right)} g ' \left(x\right) - \sqrt{1 + \sec \left(3 f \left(x\right)\right)} f ' \left(x\right)$

So

Example 1
For $F \left(x\right) = {\int}_{- \frac{\pi}{8}}^{x} \sqrt{1 + \sec \left(3 t\right)} \mathrm{dt}$

The Fundamental Theorem of Calculus gives us

F'(x) = sqrt(1+sec(3x)

Example 2

For $F \left(x\right) = {\int}_{0}^{x} ^ 2 \sqrt{1 + \sec \left(3 t\right)} \mathrm{dt}$

The Fundamental Theorem of Calculus gives us

$F ' \left(x\right) = 2 x \sqrt{1 + \sec \left(3 {x}^{2}\right)}$

Example 3

For $F \left(x\right) = {\int}_{x}^{x} ^ 2 \sqrt{1 + \sec \left(3 t\right)} \mathrm{dt}$

The Fundamental Theorem of Calculus gives us

$F ' \left(x\right) = 2 x \sqrt{1 + \sec \left(3 {x}^{2}\right)} - \sqrt{1 + \sec \left(3 x\right)}$