Calculo II Derivadas
Enviado por 1324569 • 17 de Enero de 2022 • Apuntes • 1.343 Palabras (6 Páginas) • 90 Visitas
Calculo II
Andr´es Gerardo Cruz Diaz January 2022
Derivadas
f (x) = k −→ f′(x) = 0
f (x) = xn −→ f′(x) = nxn−1
f (x) = ax −→ f′(x) = ax ln a
f (x) = ex −→ f′(x) = ex
[pic 1]
Integrales
∫ dx −→ x + c[pic 2]
[pic 3]
[pic 4]
f (x) = loga x −→ f′(x) = 1
∫ xndx −→ xn+1 + c
f (x) = ln x −→ f′(x) = 1
∫ 1 −→ ln |x| + c
x[pic 5]
f (x) = sin x −→ ′[pic 6]
f (x) = cos x[pic 7][pic 8]
f (x) = cos x f′(x) = sin x
f (x) = tan x f′(x) = sec2 x
f (x) = cot x f′(x) = csc2 x
f (x) = sec x f′(x) = sec x tan x[pic 9]
f (x) = csc x f′(x) = csc x cot x[pic 10]
f (x) = sin−1 x f′(x) = 1 [pic 11]
1−x2 [pic 12][pic 13][pic 14]
∫ cos x −→ sin x + c
∫ tan xdx −→ − ln (cos x) + c[pic 15]
∫ sec xdx −→ ln (sec x + tan x) + c[pic 16]
∫ sec2 x −→ tan x + c
[pic 17]
f (x) = cos−1 x −→ f′(x) = −√ 1
[pic 18]
∫ csc2 xdx −→ − cot x + c
[pic 19]
f (x) = tan x −→ f (x) = 1+x2 ∫
f (x) = cot−1 x −→ f′(x) = − 1 2[pic 20][pic 21][pic 22]
[pic 23]
∫ csc x cot x −→ − csc x + c
[pic 24][pic 25]
f (x) = sec−1 x −→ f′(x) =[pic 26][pic 27]
1+x
√1 [pic 28][pic 29]
1 dx sin−1
∫ 1−x2[pic 30]
x + c
f (x) = csc−1 x f′(x) = 1 [pic 31][pic 32]
x2 −1
1+x2
√ 1 dx −→ sec−1 x + c[pic 33][pic 34][pic 35]
...