# How do you use the second fundamental theorem of Calculus to find the derivative of given int (cos(t^3)) from [cosx, 7x]?

Jul 9, 2016

$7 \cos {\left(7 x\right)}^{3} + \sin x \cos \left({\cos}^{3} x\right)$

#### Explanation:

FTC part 2 tells us that

$\textcolor{b l u e}{\frac{d}{\mathrm{du}} {\int}_{a}^{u} \setminus f \left(t\right) \setminus \mathrm{dt} = f \left(u\right)}$ where $a = c o n s t$

if $u = u \left(x\right)$, we can add in the chain rule to conclude that

$\frac{d}{\mathrm{dx}} {\int}_{a}^{u} \setminus f \left(t\right) \setminus \mathrm{dt} = f \left(u\right) \cdot \frac{\mathrm{du}}{\mathrm{dx}}$

we can also further conclude from FTC that

$\frac{d}{\mathrm{du}} {\int}_{u}^{a} \setminus f \left(t\right) \setminus \mathrm{dt} = - \frac{d}{\mathrm{du}} {\int}_{a}^{u} \setminus f \left(t\right) \setminus \mathrm{dt} = - f \left(u\right)$

if we put that all together we get from the FTC and the chain rule:

$\frac{d}{\mathrm{dx}} {\int}_{u}^{v} \setminus f \left(t\right) \setminus \mathrm{dt} = f \left(v\right) \cdot \frac{\mathrm{dv}}{\mathrm{dx}} - f \left(u\right) \cdot \frac{\mathrm{du}}{\mathrm{dx}}$

so here we have

$f \left(t\right) = \cos {t}^{3}$

$u = \cos x , u ' = - \sin x$
$v = 7 x , v ' = 7$

and so we pattern match to get

$7 \cos {\left(7 x\right)}^{3} + \sin x \cos \left({\cos}^{3} x\right)$