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#?

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Answer 1 · 2015-05-06T16:13:19

#h(x)=int_4^(1/x) arctan(3t) dt#

If we think of #g(x)# as #g(x) = int_4^x arctan(3t) dt#,
then the function #h(x) = g(1/x)#.

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'(x) = g'(1/x) * d/dx(1/x)#

By FTC 1,

# g'(1/x) = arctan(3/x)# . of course #d/dx(1/x) = -1/x^2#.

So we have:

#h'(x) = arctan(3/x) * -1/x^2 =-1/x^2 arctan(3/x)#

or

#h'(x) = -arctan(3/x) /x^2#.

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

#h(x) = int_a^u f(t) dt# is #h'(x) = f(u) (du)/dx#