Apply the Fundamental Theorem of Calculus: $g'(x) = f(x)$.

Apply the Fundamental Theorem of Calculus and the Chain Rule: $g'(x) = f(u(x)) cdot u'(x)$.

First find $g'(x)$ using the Fundamental Theorem of Calculus, then evaluate $g'(c)$.

First, switch the limits: $F(x) = -\int_{b}^{u(x)} f(t) dt$. Then, $F'(x) = -f(u(x)) cdot u'(x)$.

Split the integral: $g(x) = \int_{0}^{v(x)} f(t) dt - \int_{0}^{u(x)} f(t) dt$. Then, $g'(x) = f(v(x))v'(x) - f(u(x))u'(x)$.

Apply the Fundamental Theorem of Calculus and the chain rule: $\cos(x^3) cdot 3x^2$.

Apply the Fundamental Theorem of Calculus: $F'(x) = x^2 + 1$.

First find $g'(x) = \sqrt{x^2} cdot 2x = |x| cdot 2x$. Then, $g'(2) = \sqrt{2^2} cdot 2(2) = 8$.

Switch the bounds and apply the Fundamental Theorem of Calculus: $-\int_{0}^{x} e^{t^2} dt$, so the derivative is $-e^{x^2}$.

Switch the bounds: $h(x) = -\int_{3}^{\sin(x)} t^3 dt$. Apply the Fundamental Theorem of Calculus and the chain rule: $h'(x) = -(\sin(x))^3 \cos(x)$.

$\frac{d}{dx}left[int_{a}^{x}f(t)dt ight]=f(x)$

$\int_{a}^{b}f(x) , dx = -\int_{b}^{a}f(x) , dx$

$\frac{d}{dx}left[int_{a}^{g(x)}f(t)dt ight]=f(g(x)) cdot g'(x)$

$F(x) = \int_{a}^{x} f(t) , dt$

A function defined by a definite integral where one of the limits of integration is a variable.

Differentiation and integration; it shows they are inverse processes.

The function inside the integral that is being integrated.

An integral with upper and lower limits, resulting in a numerical value.

A function whose derivative is the original function.

The upper limit of integration in a definite integral.

The lower limit of integration in a definite integral.

A rule for differentiating composite functions.

The instantaneous rate of change of a function with respect to its variable.

The process of finding the area under a curve.