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

Question

Christopher Allers · Accepted Answer

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

#d/dx (\int _{a}^{x} f(t) \ dt) = f(x)#

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

#d/dx (\int _{a}^{u(x)} f(t) \ dt) = f(u(x)) cdot u'#

So:

#d/dx (int_pi^(ln x) cos(e^t) \ dt )#

#= cose^(ln x) cdot (ln x)'#

#= (cos x)/x#

Charles Anctil · Answer

undefined To find the derivative of $F(x) = \int_{\pi}^{\ln(x)} \cos(e^t) \, dt$ with respect to $x$, we can apply the Fundamental Theorem of Calculus and the Chain Rule.

$$\frac{d}{dx} F(x) = \frac{d}{dx} \left( \int_{\pi}^{\ln(x)} \cos(e^t) \, dt \right)$$

By the Fundamental Theorem of Calculus, if $F(x) = \int_{g(x)}^{h(x)} f(t) \, dt$, then $F'(x) = f(h(x)) \cdot h'(x) - f(g(x)) \cdot g'(x)$.

In our case, $f(t) = \cos(e^t)$, $g(x) = \pi$, and $h(x) = \ln(x)$. So, we substitute these values into the formula:

$$F'(x) = \cos(e^{\ln(x)}) \cdot \frac{d}{dx} \ln(x) - \cos(e^{\pi}) \cdot \frac{d}{dx} \pi$$

$$= \cos(x) \cdot \frac{1}{x} - \cos(e^{\pi}) \cdot 0$$

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

Therefore, the derivative of $F(x)$ is $\frac{\cos(x)}{x}$.