# How do you use Part 1 of the Fundamental Theorem of Calculus to find the derivative of the function h(x)=int_4^(1/x) arctan(3t) dt?

May 6, 2015

$h \left(x\right) = {\int}_{4}^{\frac{1}{x}} \arctan \left(3 t\right) \mathrm{dt}$

If we think of $g \left(x\right)$ as $g \left(x\right) = {\int}_{4}^{x} \arctan \left(3 t\right) \mathrm{dt}$,
then the function $h \left(x\right) = g \left(\frac{1}{x}\right)$.

To find the derivative with respect to $x$, we need the chain rule along with Part 1 of the Fundamental Theorem of Calculus.

By the chain rule:
$h ' \left(x\right) = g ' \left(\frac{1}{x}\right) \cdot \frac{d}{\mathrm{dx}} \left(\frac{1}{x}\right)$

By FTC 1,

$g ' \left(\frac{1}{x}\right) = \arctan \left(\frac{3}{x}\right)$ . of course $\frac{d}{\mathrm{dx}} \left(\frac{1}{x}\right) = - \frac{1}{x} ^ 2$.

So we have:

$h ' \left(x\right) = \arctan \left(\frac{3}{x}\right) \cdot - \frac{1}{x} ^ 2 = - \frac{1}{x} ^ 2 \arctan \left(\frac{3}{x}\right)$

or

$h ' \left(x\right) = - \arctan \frac{\frac{3}{x}}{x} ^ 2$.

Note that, the end result is that the derivative w.r.t. $x$ of

$h \left(x\right) = {\int}_{a}^{u} f \left(t\right) \mathrm{dt}$ is $h ' \left(x\right) = f \left(u\right) \frac{\mathrm{du}}{\mathrm{dx}}$