# How do you calculate the derivative of int(6t + sqrt(t)) dt from [1, tanx]?

Jun 20, 2015

#### Explanation:

$y = {\int}_{1}^{u} f \left(t\right) \mathrm{dt}$ then $\frac{\mathrm{dy}}{\mathrm{du}} = f \left(u\right)$

In this case we want $\frac{\mathrm{dy}}{\mathrm{dx}}$ so we need the chain rule: $\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{\mathrm{dy}}{\mathrm{du}} \frac{\mathrm{du}}{\mathrm{dx}}$

$y = {\int}_{1}^{\tan} x \left(6 t + \sqrt{t}\right) \mathrm{dt}$

$\frac{\mathrm{dy}}{\mathrm{dx}} = {\underbrace{\left[6 \left(\tan x\right) + \sqrt{\tan x}\right]}}_{\frac{\mathrm{dy}}{\mathrm{du}}} \cdot {\underbrace{{\sec}^{2} x}}_{\frac{\mathrm{du}}{\mathrm{dx}}}$

$= \left(6 \tan x + \sqrt{\tan x}\right) {\sec}^{2} x$