# How do you calculate the derivative of intarctan t dt  from [2,1/x]?

Aug 14, 2015

$\frac{d}{\mathrm{dx}} {\int}_{2}^{\frac{1}{x}} \arctan t \mathrm{dt} = - \frac{\arctan \left(\frac{1}{x}\right)}{x} ^ 2$

#### Explanation:

$\frac{d}{\mathrm{dx}} {\int}_{2}^{\frac{1}{x}} \arctan t \mathrm{dt} = \arctan \left(\frac{1}{x}\right) \frac{d}{\mathrm{dx}} \left(\frac{1}{x}\right)$
$= - \frac{\arctan \left(\frac{1}{x}\right)}{x} ^ 2$

Aug 14, 2015

#### Explanation:

Fundamental Theorem of Calculus, Part 1 tells us that if $f$ is continuous on interval $\left[a , b\right]$, if $g$ is defined by

$g \left(x\right) = {\int}_{a}^{x} f \left(t\right) \mathrm{dt} \text{ }$ for $x \in \left[a , b\right]$

then ($g$ is continuous on $\left[a , b\right]$)
and ($g$ is differentiable on $\left(a , b\right)$)

and $g ' \left(x\right) = f \left(x\right)$ (which is what we need here).

In this problem we have a composition, so we need the chain rule:

$g \left(x\right) = {\int}_{2}^{u} \arctan t \mathrm{dt}$ with $u = \frac{1}{x}$

So the chain rule gives us:

$g ' \left(x\right) = \arctan u \frac{\mathrm{du}}{\mathrm{dx}}$

In this case:

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

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