# How do you use the Fundamental Theorem of Calculus to find the derivative of int sqrt(3t+ sqrt(t)) dt from 5 to tanx?

Apr 28, 2018

$= {\sec}^{2} x \setminus \sqrt{3 \tan x + \sqrt{\tan x}}$

#### Explanation:

FTC, Pt 1: if $F \left(u\right) = {\int}_{a}^{u} \setminus f \left(t\right) \setminus \mathrm{dt}$, then $F ' \left(u\right) = f \left(u\right)$.

With u(x) = tan x, and $F = F \left(u \left(x\right)\right)$, it follows from the chain rule that:

$\frac{\mathrm{dF}}{\mathrm{dx}} = F ' \left(u\right) \setminus u ' \left(x\right)$

So:

$\frac{d}{\mathrm{dx}} \left({\int}_{5}^{\tan \left(x\right)} \sqrt{3 t + \sqrt{t}} \setminus \setminus \mathrm{dt}\right)$

$= \sqrt{3 \tan x + \sqrt{\tan x}} \setminus \frac{d}{\mathrm{dx}} \left(\tan x\right)$

$= {\sec}^{2} x \setminus \sqrt{3 \tan x + \sqrt{\tan x}}$