# How do you use the second fundamental theorem of Calculus to find the derivative of given int dt/(2+3sqrtt) from [0,x]?

Aug 25, 2016

$\frac{d}{\mathrm{dx}} {\int}_{0}^{x} \frac{1}{2 + 3 \sqrt{t}} \setminus \mathrm{dt} = \frac{1}{2 + 3 \sqrt{x}}$

#### Explanation:

By the 2nd FTC:

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

So

$\frac{d}{\mathrm{dx}} {\int}_{0}^{x} \frac{1}{2 + 3 \sqrt{t}} \setminus \mathrm{dt} = \frac{1}{2 + 3 \sqrt{x}}$