# How do you use the second fundamental theorem of Calculus to find the derivative of given int (cos(t^2)+t) dt from [-5, sinx]?

May 17, 2016

By the FTC and the Chain Rule, the derivative is $\left(\cos \left({\sin}^{2} \left(x\right)\right) + \sin \left(x\right)\right) \cdot \cos \left(x\right)$.

#### Explanation:

The "second" Fundamental Theorem of Calculus (I just call it part of the FTC) says $\frac{d}{\mathrm{dx}} \left({\int}_{a}^{x} f \left(t\right) \setminus \mathrm{dt}\right) = f \left(x\right)$ when $f$ is continuous on the interval between $a$ and $x$.

Let $F \left(x\right) = {\int}_{- 5}^{x} \left(\cos \left({t}^{2}\right) + t\right) \setminus \mathrm{dt}$ and $g \left(x\right) = \sin \left(x\right)$. The given function for this question is $h \left(x\right) = F \left(g \left(x\right)\right)$. The Chain Rule implies that $h ' \left(x\right) = F ' \left(g \left(x\right)\right) \cdot g ' \left(x\right)$. Since $F ' \left(x\right) = \cos \left({x}^{2}\right) + x$ and $g ' \left(x\right) = \cos \left(x\right)$, it follows that

$h ' \left(x\right) = \frac{d}{\mathrm{dx}} \left({\int}_{- 5}^{\sin \left(x\right)} \left(\cos \left({t}^{2}\right) + t\right) \setminus \mathrm{dt}\right)$

$= \left(\cos \left({\sin}^{2} \left(x\right)\right) + \sin \left(x\right)\right) \cdot \cos \left(x\right)$.