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

Let $F \left(x\right) = \setminus {\int}_{2}^{x} \setminus \sqrt{5 t + \setminus \sqrt{t}} \setminus \mathrm{dt}$ and $g \left(x\right) = \setminus \tan \left(x\right)$. The Fundamental Theorem of Calculus implies that $F ' \left(x\right) = \setminus \sqrt{5 x + \setminus \sqrt{x}}$. The derivative of the tangent function is $g ' \left(x\right) = \setminus {\sec}^{2} \left(x\right)$.
Since $\setminus {\int}_{2}^{\setminus \tan \left(x\right)} \setminus \sqrt{5 t + \setminus \sqrt{t}} \setminus \mathrm{dt} = F \left(g \left(x\right)\right)$,
$\frac{d}{\mathrm{dx}} \left(\setminus {\int}_{2}^{\setminus \tan \left(x\right)} \setminus \sqrt{5 t + \setminus \sqrt{t}} \setminus \mathrm{dt}\right) = F ' \left(g \left(x\right)\right) \cdot g ' \left(x\right)$
$= \setminus \sqrt{5 \setminus \tan \left(x\right) + \setminus \sqrt{\setminus \tan \left(x\right)}} \cdot \setminus {\sec}^{2} \left(x\right)$.