# How do you calculate the derivative of intsqrt(3t+ sqrt(t)) dt from [5, tanx]?

May 26, 2015

Fundamental Theorem of Calculus, Part 1.

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

This question is the chain rule version:

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

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

$\frac{d}{\mathrm{dx}} {\int}_{5}^{\tan x} f \left(t\right) \mathrm{dt} = f \left(\tan x\right) \cdot {\sec}^{2} x$

So for this problems we get:

$\sqrt{3 \tan x + \sqrt{\tan x}} \cdot {\sec}^{2} x$, or, written more clearly:

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