AP Calculus BC: Limits, Derivatives, Integrals, & Theorems

d/du [ |u| ]

u(u’) / |u| if u ≠ 0

d/dx [ log_a u ]

u’ / (ln a)(u)

d/du [ a^u ]

(ln a)(a^u )(u')

d/dx [ cot u ]

-csc² u (u')

d/dx [ sec u ]

(sec u tan u) (u')

d/du [ csc u ]

-(csc u cot u) (u')

d/dx [ arcsin u ]

(u') / √(1 - u²)

d/dx [ arccos u ]

-(u') / √(1 - u²)

d/dx [ arctan u ]

(u') / (1 + u²)

d/dx [ arccot u ]

-(u') / (1 + u²)

d/dx [ arcsec u ]

(u') / |u| √(u² - 1)

d/dx [ arccsc u ]

-(u') / |u| √(u² - 1)

d/dx [ sin u ]

cos(u) (u')

d/dx [ cos u ]

- sin(u) (u')

d/dx [ tan u ]

sec²(u) (u')

The Mean Value Theorem

states that if a function is continuous on a closed interval [ a, b ] and differentiable on the open interval ( a, b ), then there exists at least one point “c” in the open interval ( a, b ) where the derivative equals the average rate of change over that interval (difference quotient of a and b).

Rolle’s Theorem

states that if a function is continuous on a closed interval [ a, b ] and differentiable on the open interval ( a, b ), and f(a) = f(b), then there exists at least one point "c" in ( a, b ) where the derivative is zero.

Intermediate Value Theorem

This theorem states that for any value “k” between f(a) and f(b) for a continuous function on [a, b], there exists at least one point “c” in (a, b) such that f(c) equals that value “k”. Must use the following statement: “f(x) is continuous and f(a) < k < f(b). By the intermediate Value Theorem, f(c) = k for at least one value of “c” between “a” and “b”.

∫ kf(u) du

k ∫ f(u) du

∫ du

u + C

∫ uⁿ du

uⁿ⁺¹ / n+1 + C

∫ 1/u du

ln |u| + C

∫ e^u du

e^u + C

∫ sin u du

-cos u + C

∫ cos u du

sin u + C

∫ tan u du

-ln |cos u| + C

∫ cot u du

ln |sin u| + C

∫ sec u du

ln |sec u + tan u| + C

∫ csc u du

-ln |csc u + cot u| + C

∫ sec² u du

tan u + C

∫ csc² u du

-cot u + C

∫ sec u tan u du

sec u + C

∫ csc u cot u du

-csc u + C

∫ 1 / √(a² - u²) du

arcsin (u/a) + C

∫ 1 / (a² + u²) du

(1/a) arctan(u/a) + C

∫ 1 / u√(u² - a²) du

(1/a) arcsec(|u|/a) + C

∫ a^u du

(a^u / ln a) + C

∫ ln x dx

x ln(x) - x + C

Let lim_x→c f(x) = L // lim_x→c kf(x) =

kL

Let lim_x→c f(x) = L and lim_x→c g(x) = M // lim_x→c f(x) ± g(x) =

L ± M

Let lim_x→c f(x) = L and lim_x→c g(x) = M // lim_x→c f(x)g(x) =

LM

Let lim_x→c f(x) = L and lim_x→c g(x) = M , lim_x→c f(x)/g(x) =

L/M, M ≠ 0.

Let lim_x→c f(x) = L // lim_x→c f(x)^k =

L^k