# How do you use Part 1 of the Fundamental Theorem of Calculus to find the derivative of the function of y=int sqrt(2t+sqrt(t))dt from 5 to tanx?

Apr 30, 2015

To find the derivative of the function of $y = {\int}_{5}^{\tan} x \sqrt{2 t + \sqrt{t}} \mathrm{dt}$, we have to pair Part 1 of the Fundamental Theorem of Calculus with the Chain Rule.

FTC Part I tells us that for $g \left(x\right) = {\int}_{a}^{x} f \left(t\right) \mathrm{dt}$, the derivative with respect to $x$ is $g ' \left(x\right) = f \left(x\right)$

Now we combine this with the chain rule:

For the function $g \left(u\right) = {\int}_{a}^{u} f \left(t\right) \mathrm{dt}$, the derivative with respect to $x$ is

$\frac{d}{\mathrm{dx}} \left(g \left(u\right)\right) = g ' \left(u\right) \frac{\mathrm{du}}{\mathrm{dx}} = f \left(u\right) \frac{\mathrm{du}}{\mathrm{dx}}$.

So here's the answer to this question:

For $y = {\int}_{5}^{\tan} x \sqrt{2 t + \sqrt{t}} \mathrm{dt}$, the derivative with respect to $x$ is

$\frac{\mathrm{dy}}{\mathrm{dx}} = \sqrt{2 \tan x + \sqrt{\tan x}} \cdot {\sec}^{2} x$.