# How do you use Part 1 of the Fundamental Theorem of Calculus to find the derivative of the function y = int (6+v^8)^3 dv from 1 to cos(x)?

Jul 7, 2015

The answer is $- {\left(6 + {\cos}^{8} \left(x\right)\right)}^{3} \cdot \sin \left(x\right)$
Given a function $f$ continuous on an interval $\left[a , b\right]$ (with $a < b$) and a number $c \setminus \in \left[a , b\right]$, the Fundamental Theorem of Calculus ("Part 1"...actually, some books call it "Part 2") says that $\frac{d}{\mathrm{dx}} \left(\setminus {\int}_{c}^{x} f \left(v\right) \setminus \mathrm{dv}\right) = f \left(x\right)$ for all $x \setminus \in \left[a , b\right]$. Thought of another way, the function $F$ defined by the equation $F \left(x\right) = \setminus {\int}_{c}^{x} f \left(v\right) \setminus \mathrm{dv}$ is an antiderivative of $f$ on the interval $\left[a , b\right]$ (also, $F \left(c\right) = 0$).
For the problem at hand, let $F \left(x\right) = \setminus {\int}_{1}^{x} {\left(6 + {v}^{8}\right)}^{3} \setminus \mathrm{dv}$ and let $g \left(x\right) = \cos \left(x\right)$ so that we must find d/dx(F(g(x)). This also requires the Chain Rule, which gives $\frac{d}{\mathrm{dx}} \left(F \left(g \left(x\right)\right)\right) = F ' \left(g \left(x\right)\right) \cdot g ' \left(x\right)$. But $F ' \left(x\right) = {\left(6 + {x}^{8}\right)}^{3}$ and $g ' \left(x\right) = - \sin \left(x\right)$. The final answer is therefore:
$\frac{d}{\mathrm{dx}} \left(\setminus {\int}_{1}^{\setminus \cos \left(x\right)} {\left(6 + {v}^{8}\right)}^{3} \setminus \mathrm{dv}\right) = - {\left(6 + {\cos}^{8} \left(x\right)\right)}^{3} \cdot \sin \left(x\right)$