Answer
$y' = -\frac{ tan(\sqrt x)}{2}$
Work Step by Step
We are told to find the derivative using Part 1 of the Fundamental Theorem of Calculus given the expression: $ y =\int_{\sqrt x} ^{\pi/4} θtan(θ)dθ$
Swap upper and lower bounds: (this makes the expression negative)
$ y =-\int^{\sqrt x} _{\pi/4} θtan(θ)dθ$
Since the upper bound is a function we need to use the chain rule:
$u = \sqrt x$
$u' = \frac{1}{2\sqrt x}$
$ y =-\int^{u} _{\pi/4} θtan(θ)dθ$
Using the FTC
$F(u) =\int_0^uf(θ)dθ$
$F'(u) = f(u)$
Thus,
$y'= -utan(u)u'$
$y' = -\sqrt x tan(\sqrt x) \frac{1}{2\sqrt x}$
$y' = -\frac{\sqrt x tan(\sqrt x)}{2\sqrt x}$
$y' = -\frac{ tan(\sqrt x)}{2}$
