5.3 The Fundamental Theorem of Calculus/15

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

FTC 1: ${\frac {d}{dx}}\int _{a(x)}^{b(x)}f(t)\,dt=b^{\prime }{(x)}\cdot \,f(b(x))-\,a^{\prime }{(x)}\cdot \,f(a(x))$

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

In this problem $a^{\prime }{(x)}=0$, so when it is multiplied by $f(a(x))$ it will result in 0 and doesn't need to be added.