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

2 Answers
Aug 14, 2015

Answer:

# d/dxint_2^(1/x) arctan t dt = -(arctan (1/x))/x^2 #

Explanation:

# d/dxint_2^(1/x) arctan t dt = arctan (1/x) d/dx(1/x) #
# = -(arctan (1/x))/x^2 #

Aug 14, 2015

Answer:

Explanation:

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

#g(x) = int_a^x f(t) dt" "# for #x in [a,b]#

then (#g# is continuous on #[a,b]#)
and (#g# is differentiable on #(a,b)#)

and #g'(x) = f(x)# (which is what we need here).

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

#g(x) = int_2^u arctan t dt# with #u = 1/x#

So the chain rule gives us:

#g'(x) = arctanu (du)/dx#

In this case:

#g'(x) = arctan(1/x) (-1)/x^2#

# = - arctan(1/x)/x^2#