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

1 Answer
Jul 2, 2016

- 1/x^2 arctan(3/x)

Explanation:

FTC part 2 says that

if F(x) = int_a^x dt qquad f(t)

then (dF)/dx = f(x)

here you want

d/dx int_4^{1/x} dt qquad arctan(3t)

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

int_4^{1/x} = int_a^{1/x} - int_a^4

so

d/dx int_4^{1/x} \ arctan(3t) \ dt

= d/dx ( color{green}{int_a^{1/x} dt \ arctan(3t)} - color{red}{ int_a^4 dt \ arctan(3t)} )

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(u) = int_a^u dt \ f(t)

then (dF)/(du) = f(u)

if then u = u(x) = 1/x then

(dF)/(dx) = (dF)/(du) * (du)/dx

= f(u) * (du)/dx

So we can say that

= d/(du) ( int_a^{u} dt \ arctan(3t) )

= arctan(3u) = arctan(3/x)

and

= d/dx (int_a^{1/x} dt \ arctan(3t) )

arctan(3/x) * d/dx (1/x)

- 1/x^2 arctan(3/x)