# How do you use the Fundamental Theorem of Calculus to find the derivative of int sqrt(6t + sqrt (t)) dt from 1 to tanx?

Aug 14, 2016

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

#### Explanation:

FTC part 1 tells us that ;

$\frac{d}{\mathrm{du}} {\int}_{a}^{u} f \left(t\right) \setminus \mathrm{dt} = f \left(u\right)$

and by the chain rule, if $u = u \left(x\right)$ then

$\frac{d}{\mathrm{dx}} \left({\int}_{a}^{u \left(x\right)} f \left(t\right) \setminus \mathrm{dt}\right)$

$\frac{d}{\mathrm{du}} \left({\int}_{a}^{u \left(x\right)} f \left(t\right) \setminus \mathrm{dt}\right) \cdot \frac{\mathrm{du}}{\mathrm{dx}}$

$= f \left(u\right) \frac{\mathrm{du}}{\mathrm{dx}}$

so $\frac{d}{\mathrm{dx}} \left({\int}_{1}^{\tan x} \sqrt{6 t + \sqrt{t}} \mathrm{dt}\right)$

$= \sqrt{6 \tan x + \sqrt{\tan x}} \frac{d}{\mathrm{dx}} \left(\tan x\right)$

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