1. fundamental theorem of calculus

compute the following
(a) d/dx[ integrate(x to e^(x^2)) (ln(t))dt]
(b) d/dx[ int (2 to e^(x^2) )(ln(t))dt]

2. Originally Posted by gracy
compute the following
(a) d/dx[ integrate(x to e^(x^2)) (ln(t))dt]
$\displaystyle \frac{d}{dx} \int_x^{e^{x^2}} dt \, ln(t)$

= $\displaystyle \frac{d}{dx} \left ( t \cdot ln(t) - t \right )|_x^{e^{x^2}}$

= $\displaystyle \frac{d}{dx} \left ( e^{x^2} \cdot x^2 - e^{x^2} - x \cdot ln(x) + x \right )$

= $\displaystyle 2xe^{x^2} + 2x^3e^{x^2} - 2xe^{x^2} - ln(x) - x \cdot \frac{1}{x} + 1$

= $\displaystyle 2x^3e^{x^2} - ln(x)$

-Dan

3. Originally Posted by gracy
(b) d/dx[ int (2 to e^(x^2) )(ln(t))dt]
Again I am so angry that you do not mention a function on what domain.
But since,
$\displaystyle e^{x^2}>0$ this function is on the entire real domain.

You have,
$\displaystyle f(x)=\int_2^{e^{x^2}}\ln t dt$

You can consider this function as a composition of functions.
$\displaystyle g(x)=\int_2^x \ln tdt,\, x>0$
$\displaystyle h(x)=e^{x^2}$
Then,
$\displaystyle f =g\circ h$
Since, $\displaystyle h$ is differenciable everywhere and $\displaystyle g$ is differenciable for $\displaystyle x>0$ for that is the fundamental theorem. Thus,
$\displaystyle f'=h' g'\circ h$
Thus,
$\displaystyle f'=2xe^{x^2} \ln (e^{x^2})=2x^3e^{x^2}$ for $\displaystyle x>0$.

4. Originally Posted by gracy
(b) d/dx[ int (2 to e^(x^2) )(ln(t))dt]
$\displaystyle \frac{d}{dx} \int_2^{e^{x^2}} dt \, ln(t)$

= $\displaystyle \frac{d}{dx}(t \cdot ln(t) - t)|_2^{e^{x^2}}$

= $\displaystyle \frac{d}{dx} \left ( e^{x^2} \cdot x^2 - e^{x^2} - 2 \cdot ln(2) + 2 \right )$

= $\displaystyle 2xe^{x^2} + 2x^3e^{x^2} - 2xe^{x^2}$

= $\displaystyle 2x^3e^{x^2}$

-Dan