# How do you use Part 1 of the Fundamental Theorem of Calculus to find the derivative of the function y=int_("[e"^x",5]") 5(sin(t))^5 dt?

Jul 30, 2015

If $F \left(x\right) = {\int}_{{e}^{x}}^{5} 5 {\sin}^{5} \left(t\right) \mathrm{dt}$, then $F ' \left(x\right) = - 5 {\sin}^{5} \left({e}^{x}\right) {e}^{x}$.

#### Explanation:

Let us first review the first part of the Fundamental Theorem, I will use the formulation used in Analysis I by Terence Tao:

Let $a < b$ be real numbers, and let $f : \left[a , b\right] \to \mathbb{R}$ be a Riemann integrable function. Let $F : \left[a , b\right] \to \mathbb{R}$ be the function
$F \left(x\right) = {\int}_{a}^{x} f \left(t\right) \mathrm{dt}$.
Then $F$ is continuous. Furthermore, if ${x}_{0} \in \left[a , b\right]$ and $f$ continuous in ${x}_{0}$, then $F$ is differentiable in ${x}_{0}$, and $F ' \left({x}_{0}\right) = f \left({x}_{0}\right)$.

So now we want to know the derivative of
$F \left(x\right) = {\int}_{{e}^{x}}^{5} 5 {\sin}^{5} \left(t\right) \mathrm{dt}$.
We note that if we define $G \left(x\right) = {\int}_{x}^{5} 5 {\sin}^{5} \left(t\right) \mathrm{dt}$, then $F \left(x\right) = G \left({e}^{x}\right)$. So in order to find the derivative of $F$, we use the chain rule, so:
$F ' \left(x\right) = G ' \left({e}^{x}\right) \frac{d}{\mathrm{dx}} {e}^{x} = G ' \left({e}^{x}\right) {e}^{x}$.

Now we need to know the derivative of $G$. We note that in the theorem, the variable $x$ serves as an upper limit of integration, while in $G$ it serves as a lower limit. This changes the way we can find the derivative slightly, as I will show next.

Suppose you have such a function $f$ as given in the theorem, then let $H , G : \left[a , b\right] \to \mathbb{R}$ such that:
$H \left(x\right) = {\int}_{a}^{x} f \left(t\right) \mathrm{dt}$ and $G \left(x\right) = {\int}_{x}^{b} f \left(t\right) \mathrm{dt}$.
Now note that $\left(H + G\right) \left(x\right) = H \left(x\right) + G \left(x\right) = {\int}_{a}^{b} f \left(t\right) \mathrm{dt}$.
Suppose $f$ continuous in ${x}_{0} \in \left[a , b\right]$, then $H ' \left({x}_{0}\right) = f \left({x}_{0}\right)$, using the fundamental theorem. Furthermore $\left(H + G\right) \left(x\right)$ is constant, so $\left(H + G\right) ' \left(x\right) = 0$. Since $G \left(x\right) = \left(H + G\right) \left(x\right) - H \left(x\right)$:
$G ' \left({x}_{0}\right) = \left(H + G\right) ' \left({x}_{0}\right) - H ' \left({x}_{0}\right) = 0 - f \left({x}_{0}\right) = - f \left({x}_{0}\right)$.
So $G ' \left({x}_{0}\right) = - f \left({x}_{0}\right)$.

Applying this to $G \left(x\right) = {\int}_{x}^{5} 5 {\sin}^{5} \left(t\right) \mathrm{dt}$, we note that $5 {\sin}^{5} \left(t\right)$ is continuous.
Therefore if $x \le 5$, it serves as a lower limit, so $G ' \left(x\right) = - 5 {\sin}^{5} \left(x\right)$.
If $x > 5$, we use the fact that ${\int}_{x}^{5} 5 {\sin}^{5} \left(t\right) \mathrm{dt} = - {\int}_{5}^{x} 5 {\sin}^{5} \left(t\right) \mathrm{dt}$. Since $x$ now serves as an upper limit, we may use the original theorem and once again see $G ' \left(x\right) = - 5 {\sin}^{5} \left(x\right)$.

We already saw $F ' \left(x\right) = G ' \left({e}^{x}\right) {e}^{x}$, so $F ' \left(x\right) = - 5 {\sin}^{5} \left({e}^{x}\right) {e}^{x}$.