# How do you use Part 1 of the Fundamental Theorem of Calculus to find the derivative of the function y = int sin^3 t dt from [e^x, 0]?

Apr 20, 2015

The theorem says:

Given the function:

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

than:

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

So, since $y = {\int}_{{e}^{x}}^{0} {\sin}^{3} t \mathrm{dt}$:

$y ' = {\sin}^{3} 0 \cdot 0 - {\sin}^{3} {e}^{x} \cdot {e}^{x} = - {e}^{x} {\sin}^{3} {e}^{x}$.