# How do you calculate the derivative of int(cos(t^3)+t) from x=4 to x=sinx?

Aug 10, 2015

The derivative is $\left(\cos \left({\sin}^{3} \left(x\right)\right) + \sin \left(x\right)\right) \cdot \cos \left(x\right)$

$= \cos \left(x\right) \cos \left({\sin}^{3} \left(x\right)\right) + \cos \left(x\right) \sin \left(x\right)$

#### Explanation:

One form of the Fundamental Theorem of Calculus says that if $f$ is continuous on some interval $\left[c , d\right]$ and $F \left(x\right) = {\int}_{a}^{x} f \left(t\right) \setminus \mathrm{dt}$ for some number $a$ in $\left[c , d\right]$, then $F ' \left(x\right) = f \left(x\right)$ for all $x$ in $\left[c , d\right]$. In other words, you can use the integral in a theoretical/abstract way to "construct" an antiderivative of $f$ (that doesn't mean you can find a "nice" formula for $F$...it might not be an "elementary function ").

If, in addition, you have a function $g$ that is differentiable on the interval $\left[c , d\right]$, and if you create another function $h$ by the (composition) formula $h \left(x\right) = F \left(g \left(x\right)\right) = {\int}_{a}^{g \left(x\right)} f \left(t\right) \setminus \mathrm{dt}$, then the Chain Rule implies that $h ' \left(x\right) = F ' \left(g \left(x\right)\right) \cdot g ' \left(x\right) = f \left(g \left(x\right)\right) \cdot g ' \left(x\right)$.

In the present problem, $f \left(t\right) = \cos \left({t}^{3}\right) + t$ and $g \left(x\right) = \sin \left(x\right)$. Since $g ' \left(x\right) = \cos \left(x\right)$, we get the answer:

$h ' \left(x\right) = \frac{d}{\mathrm{dx}} \left({\int}_{a}^{\sin \left(x\right)} \left(\cos \left({t}^{3}\right) + t\right) \setminus \mathrm{dt}\right)$

$= \left(\cos \left({\sin}^{3} \left(x\right)\right) + \sin \left(x\right)\right) \cdot \cos \left(x\right)$

$= \cos \left(x\right) \cos \left({\sin}^{3} \left(x\right)\right) + \cos \left(x\right) \sin \left(x\right)$