# How do you use the second fundamental theorem of Calculus to find the derivative of given int arctan(3t) dt from [4,1/x]?

Jul 2, 2016

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

#### Explanation:

FTC part 2 says that

if $F \left(x\right) = {\int}_{a}^{x} \mathrm{dt} q \quad f \left(t\right)$

then $\frac{\mathrm{dF}}{\mathrm{dx}} = f \left(x\right)$

here you want

$\frac{d}{\mathrm{dx}} {\int}_{4}^{\frac{1}{x}} \mathrm{dt} q \quad \arctan \left(3 t\right)$

we can start by separating out the integration limits so they match the theorem

${\int}_{4}^{\frac{1}{x}} = {\int}_{a}^{\frac{1}{x}} - {\int}_{a}^{4}$

so

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

$= \frac{d}{\mathrm{dx}} \left(\textcolor{g r e e n}{{\int}_{a}^{\frac{1}{x}} \mathrm{dt} \setminus \arctan \left(3 t\right)} - \textcolor{red}{{\int}_{a}^{4} \mathrm{dt} \setminus \arctan \left(3 t\right)}\right)$

the red term is constant and can be ignored. in fact we could have specified that a = 4 and continued with the basic formulation of FTC pt 2

the green term is not exactly in the form required but the FTC pt 2 can be expressed in terms of a new variable u as

if $F \left(u\right) = {\int}_{a}^{u} \mathrm{dt} \setminus f \left(t\right)$

then $\frac{\mathrm{dF}}{\mathrm{du}} = f \left(u\right)$

if then $u = u \left(x\right) = \frac{1}{x}$ then

$\frac{\mathrm{dF}}{\mathrm{dx}} = \frac{\mathrm{dF}}{\mathrm{du}} \cdot \frac{\mathrm{du}}{\mathrm{dx}}$

$= f \left(u\right) \cdot \frac{\mathrm{du}}{\mathrm{dx}}$

So we can say that

$= \frac{d}{\mathrm{du}} \left({\int}_{a}^{u} \mathrm{dt} \setminus \arctan \left(3 t\right)\right)$

$= \arctan \left(3 u\right) = \arctan \left(\frac{3}{x}\right)$

and

$= \frac{d}{\mathrm{dx}} \left({\int}_{a}^{\frac{1}{x}} \mathrm{dt} \setminus \arctan \left(3 t\right)\right)$

$\arctan \left(\frac{3}{x}\right) \cdot \frac{d}{\mathrm{dx}} \left(\frac{1}{x}\right)$

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