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

Apr 13, 2015

The theorem says:

Given the function:

$y = {\int}_{h \left(x\right)}^{g} \left(x\right) f \left(t\right) \mathrm{dt}$

than:

$y ' = f \left(g \left(x\right)\right) \cdot g ' \left(x\right) - f \left(h \left(x\right)\right) \cdot h ' \left(x\right)$.

So, in our case:

$y ' = \sqrt{2 \tan x + \sqrt{\tan} x} \cdot \frac{1}{\cos} ^ 2 x - \sqrt{2 \cdot 2 + \sqrt{2} \cdot} 0 \Rightarrow$

$y ' = \sqrt{2 \tan x + \sqrt{\tan} x} \cdot \frac{1}{\cos} ^ 2 x$.

This is because $\frac{1}{\cos} ^ 2 x$ is the derivative of $\tan x$ and $0$ is the derivative of $2$.