# How do you use part 1 of the fundamental theorem of calculus to find the derivative of g(x)= int cos(sqrt( t))dt 6 to x?

May 14, 2015

The theorem says:

Given the function:

$y = {\int}_{h \left(x\right)}^{g} \left(x\right) f \left(t\right) \mathrm{dt}$

then:

$y ' = f \left(g \left(x\right)\right) \cdot g ' \left(x\right) - f \left(h \left(x\right)\right) \cdot h ' \left(x\right)$.

So:

$g ' \left(x\right) = \cos \sqrt{x} \cdot 1 - \cos \sqrt{6} \cdot 0 = \cos \sqrt{x}$.