Derivalasf (x) f ′ (x)

c 0xα αxα−1

ex ex

ax ax ln a

ln x1x

loga x1

x ln asin x cos x

cos x − sin x

tgx1

cos2 x

ctgx − 1sin2 x

arcsin x1√

1− x2

arccosx − 1√1− x2

arctgx1

1 + x2

arcctgx − 11 + x2

shx chx

chx shx

thx1

ch2x

cthx − 1sh2x

arshx1√

1 + x2

archx1√

x2 − 1

arthx1

1− x2|x| < 1

arcthx1

1− x2|x| > 1

Derivalasi szabalyok

(cf)′ = cf ′

(f ± g)′ = f ′ ± g′

(fg)′ = f ′g + fg′(f

g

)′=

f ′g − fg′

g2

(f ◦ g)′ = (f ′ ◦ g) g′(f)′ =

1f ′ ◦ f

Parameteres megadasu fuggveny:

f(x) :

x = ϕ(t)

y = ψ(t)f ′(x) =

ψ(t)ϕ(t)

Kiegeszıtesek

sin2 x =1− cos 2x

2cos2 x =

1 + cos 2x

2

sh2x =ch2x− 1

2ch2x =

ch2x + 12

2 sin α sin β = cos (α− β)− cos (α + β)

2 cos α cosβ = cos (α− β) + cos (α + β)

2 sinα cos β = sin (α + β) + sin (α− β)

Integralasf (x) F (x)

xα xα+1

α + 1α 6= −1

1x

ln |x|ex ex

ax ax

ln asin x − cosx

cosx sin x1

cos2 xtgx

1sin2 x

− ctgx

shx chx

chx shx1

ch2xthx

1sh2x

− cthx

1√1− x2

arcsin x

1√1 + x2

arshx

1√x2 − 1

archx

11 + x2

arctgx

11− x2

12

ln∣∣∣∣1 + x

1− x

∣∣∣∣Integralasi szabalyok

∫f (ax + b) dx =

F (ax + b)a

+ C

∫fα (x) f ′ (x) dx =

fα+1 (x)α + 1

+ C

ha α 6= −1∫f ′ (x)f (x)

dx = ln |f (x)|+ C∫

f (g (x)) g′ (x) dx = F (g (x)) + C∫u′ (x) v (x) dx =

= u (x) v (x)− ∫u (x) v′ (x) dx

t = tgx

2helyettesıtes:

sin x =2t

1 + t2cosx =

1− t2

1 + t2

V = πb∫

a

f2(x)dx

L =b∫

a

√1 + (f ′(x))2dx

F = 2πb∫

a

f(x)√

1 + (f ′(x))2dx

x = ϕ(t)

y = ψ(t)

V = πt2∫t1

ψ2(t)ϕ(t)dt

L =t2∫t1

√ϕ2(t) + ψ2(t)dt

F = 2πt2∫t1

ψ(t)√

ϕ2(t) + ψ2(t)dt

Laplace-transzformaciof (t) f (s) = L [f (t)]

eat 1s− a

sin (at)a

s2 + a2

cos (at)s

s2 + a2

tnn!

sn+1

sh(at)a

s2 − a2

ch(at)s

s2 − a2

eatf (t) f (s− a)

tnf (t) (−1)n dnf(s)dsn

f ′ (t) sf (s)− f (0)f ′′ (t) s2f (s)− sf (0)− f ′ (0)f (n) (t) snf (s)− sn−1f (0)− . . .

. . .− f (n−1) (0)t∫0

f (u) du1sf (s)

f (t− a) e−asf (s)Taylor-sorok

ex =∞∑

n=0

xn

n!

sin x =∞∑

n=0(−1)n x2n+1

(2n + 1)!

cos x =∞∑

n=0(−1)n x2n

(2n)!

(1 + x)α =∞∑

n=0

(αn

)xn |x| < 1

(α0

)= 1

(αn

)= α(α−1)·...·(α−n+1)

n!

Fourier-sorok

f (x) = a0+

+∞∑

n=1(an cos (nωx) + bn sin (nωx))

a0 =1T

a+T∫

a

f (x) dx

an =2T

a+T∫

a

f (x) cos (nωx) dx

bn =2T

a+T∫

a

f (x) sin (nωx) dx

f (x + T ) = f (x) es ω =2π

T

Vektoranalızis

s =t2∫t1

|r| dt

G =|r¯× r||r|3 T =

r¯r¯...r¯|r

¯× r|2

∇ =(

∂

∂x,

∂

∂y,

∂

∂z

)

gradu = ∇u

divv¯

= ∇v¯

rotv¯

= ∇× v¯