# How do you use Part 1 of the Fundamental Theorem of Calculus to find the derivative of the function h(x) = int arctan t dt from [2,1/x]?

May 14, 2015

The theorem says:

Given the function:

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

then:

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

So if:

$y = {\int}_{0}^{\frac{1}{x}} \arctan t \mathrm{dt}$,

then

$y ' = \arctan \left(\frac{1}{x}\right) \cdot \left(- \frac{1}{x} ^ 2\right) - \arctan 2 \cdot 0 =$

$= - \arctan \frac{\frac{1}{x}}{x} ^ 2$.