need some help
if f(x) = x^2
t^2dt
x
then
f'(x)=
f'(1)=
Is this what you're asking?
If $\displaystyle f(x) = \int_x^{x^2}{t^2\,dt}$
what is $\displaystyle f'(x)$ and $\displaystyle f'(1)$?
If so... evaluate the integral...
$\displaystyle f(x) = \left[\frac{1}{3}t^3\right]_x^{x^2}$
$\displaystyle = \frac{1}{3}(x^2)^3 - \frac{1}{3}x^3$
$\displaystyle = \frac{1}{3}x^6 - \frac{1}{3}x^3$.
Now take the derivative...
$\displaystyle f'(x) = 2x^5 - x^2$.
Now let $\displaystyle x = 1$
$\displaystyle f'(1) = 2\cdot 1^5 - 1^2$
$\displaystyle = 1$.