# How do you find the derivative of F(x)=cose^tdt from [pi, lnx]?

Apr 28, 2017

The Fundamental Theorem of Calculus, part 1, tells us that:

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

But, if we have an interval limit that is a function of $x$, then we need also add on the chain rule:

$\frac{d}{\mathrm{dx}} \left(\setminus {\int}_{a}^{u \left(x\right)} f \left(t\right) \setminus \mathrm{dt}\right) = f \left(u \left(x\right)\right) \cdot u '$

So:

$\frac{d}{\mathrm{dx}} \left({\int}_{\pi}^{\ln x} \cos \left({e}^{t}\right) \setminus \mathrm{dt}\right)$

$= \cos {e}^{\ln x} \cdot \left(\ln x\right) '$

$= \frac{\cos x}{x}$