Use part 1 of the FTC to find the derivative of the function: y = ∫ 0 t a n ( x ) t + t d t {\displaystyle y=\int _{0}^{tan(x)}{\sqrt {t+{\sqrt {t}}}}\,dt}
y = ∫ 0 t a n ( x ) t + t d t = t a n ( x ) + t a n ( x ) ⋅ sec 2 ( x ) {\displaystyle {\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: d d x ∫ a ( x ) b ( x ) f ( t ) d t = b ′ ( x ) ⋅ f ( b ( x ) ) − a ′ ( x ) ⋅ f ( a ( x ) ) {\displaystyle {\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 ′ ( x ) = 0 {\displaystyle a^{\prime }{(x)}=0} , so when it is multiplied by f ( a ( x ) ) {\displaystyle f(a(x))} it will result in 0.
, so when it is multiplied by 
it will result in 0.