5.3 The Fundamental Theorem of Calculus/15

Use part 1 of the FTC to find the derivative of the function: $y=\int _{0}^{tan(x)}{\sqrt {t+{\sqrt {t}}}}\,dt$

${\begin{aligned}y=\int _{0}^{tan(x)}{\sqrt {t+{\sqrt {t}}}}\,dt={\sqrt {tan(x)+{\sqrt {t}}an(x)}}\cdot \sec ^{2}(x)\end{aligned}}$

FTC 1: ${\frac {d}{dx}}\int _{a(x)}^{b(x)}f(t)\,dt=b^{\prime }{(x)}\cdot \,f(b(x))-\,a^{\prime }{(x)}\cdot \,f(a(x))$ In this problem $a^{\prime }{(x)}=0$, so when it is multiplied by $f(a(x))$ it will result in 0.