# How do you use the Fundamental Theorem of Calculus to find the derivative of int (cos(t^4) + t) dt from -4 to sinx?

Oct 19, 2017

Use the Chain Rule along with the Fundamental Theorem of Calculus to get the answer $\left(\cos \left({\sin}^{4} \left(x\right)\right) + \sin \left(x\right)\right) \cdot \cos \left(x\right)$

#### Explanation:

For a function $F \left(x\right) = {\int}_{a}^{x} f \left(t\right) \setminus \mathrm{dt}$, the Fundamental Theorem of Calculus implies that, if $f$ is continuous, then $F ' \left(x\right) = f \left(x\right)$.

If $f \left(t\right) = \cos \left({t}^{4}\right) + t$, the function to differentiate in this problem is $h \left(x\right) = {\int}_{- 4}^{\sin \left(x\right)} f \left(t\right) \setminus \mathrm{dt}$. If $F \left(x\right) = {\int}_{- 4}^{x} f \left(t\right) \setminus \mathrm{dt}$, then $h \left(x\right) = F \left(\sin \left(x\right)\right)$.

By the Chain Rule, $h ' \left(x\right) = F ' \left(\sin \left(x\right)\right) \cdot \frac{d}{\mathrm{dx}} \left(\sin \left(x\right)\right)$

$= \left(\cos \left({\sin}^{4} \left(x\right)\right) + \sin \left(x\right)\right) \cdot \cos \left(x\right)$.

(In general, the Chain Rule says that, under appropriate assumptions about differentiability, $\frac{d}{\mathrm{dx}} \left(f \left(g \left(x\right)\right)\right) = f ' \left(g \left(x\right)\right) \cdot g ' \left(x\right)$).