## Calculus 8th Edition

$y' = -\frac{ tan(\sqrt x)}{2}$
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, $y'= -utan(u)u'$ $y' = -\sqrt x tan(\sqrt x) \frac{1}{2\sqrt x}$ $y' = -\frac{\sqrt x tan(\sqrt x)}{2\sqrt x}$ $y' = -\frac{ tan(\sqrt x)}{2}$