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 $h'(x) = f(u) (du)/dx$

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) dth(x)=int_4^(1/x) arctan(3t) dt?