derivalasi - integralasi kepletek
1
Deriv´ al´ as f (x) f 0 (x) c 0 x α αx α-1 e x e x a x a x ln a ln x 1 x log a x 1 x ln a sin x cos x cos x - sin x tgx 1 cos 2 x ctgx - 1 sin 2 x arcsin x 1 √ 1 - x 2 arccos x - 1 √ 1 - x 2 arctgx 1 1+ x 2 arcctgx - 1 1+ x 2 shx chx chx shx thx 1 ch 2 x cthx - 1 sh 2 x arshx 1 √ 1+ x 2 archx 1 √ x 2 - 1 arthx 1 1 - x 2 |x| < 1 arcthx 1 1 - x 2 |x| > 1 Deriv´ al´ asi szab´ alyok (cf ) 0 = cf 0 (f ± g) 0 = f 0 ± g 0 (fg) 0 = f 0 g + fg 0 f g ¶ 0 = f 0 g - fg 0 g 2 (f ◦ g) 0 =(f 0 ◦ g) g 0 ( ¯ f ) 0 = 1 f 0 ◦ ¯ f Param´ eteresmegad´as´ u f¨ uggv´ eny: f (x): x = ϕ(t) y = ψ(t) f 0 (x)= ˙ ψ(t) ˙ ϕ(t) Kieg´ esz´ ıt´ esek sin 2 x = 1 - cos 2x 2 cos 2 x = 1 + cos 2x 2 sh 2 x = ch2x - 1 2 ch 2 x = ch2x +1 2 2 sin α sin β = cos (α - β) - cos (α + β) 2 cos α cos β = cos (α - β) + cos (α + β) 2 sin α cos β = sin (α + β) + sin (α - β) Integr´ al´ as f (x) F (x) x α x α+1 α +1 α 6= -1 1 x ln |x| e x e x a x a x ln a sin x - cos x cos x sin x 1 cos 2 x tgx 1 sin 2 x - ctgx shx chx chx shx 1 ch 2 x thx 1 sh 2 x - cthx 1 √ 1 - x 2 arcsin x 1 √ 1+ x 2 arshx 1 √ x 2 - 1 archx 1 1+ x 2 arctgx 1 1 - x 2 1 2 ln fl fl fl fl 1+ x 1 - x fl fl fl fl Integr´ al´ asi szab´ alyok R f (ax + b) dx = F (ax + b) a + C R f α (x) f 0 (x) dx = f α+1 (x) α +1 + C ha α 6= -1 Z f 0 (x) f (x) dx = ln |f (x)| + C R f (g (x)) g 0 (x) dx = F (g (x)) + C R u 0 (x) v (x) dx = = u (x) v (x) - R u (x) v 0 (x) dx t = tg x 2 helyettes´ ıt´ es: sin x = 2t 1+ t 2 cos x = 1 - t 2 1+ t 2 V = π b R a f 2 (x)dx L = b R a q 1+(f 0 (x)) 2 dx F =2π b R a f (x) q 1+(f 0 (x)) 2 dx x = ϕ(t) y = ψ(t) V = π t 2 R t1 ψ 2 (t)˙ ϕ(t)dt L = t 2 R t 1 q ˙ ϕ 2 (t)+ ˙ ψ 2 (t)dt F =2π t2 R t 1 ψ(t) q ˙ ϕ 2 (t)+ ˙ ψ 2 (t)dt Laplace-transzform´ aci´ o f (t) ¯ f (s)= L [f (t)] e at 1 s - a sin (at) a s 2 + a 2 cos (at) s s 2 + a 2 t n n! s n+1 sh(at) a s 2 - a 2 ch(at) s s 2 - a 2 e at f (t) ¯ f (s - a) t n f (t) (-1) n d n ¯ f (s) ds n f 0 (t) s ¯ f (s) - f (0) f 00 (t) s 2 ¯ f (s) - sf (0) - f 0 (0) f (n) (t) s n ¯ f (s) - s n-1 f (0) - ... ... - f (n-1) (0) t R 0 f (u) du 1 s ¯ f (s) f (t - a) e -as ¯ f (s) Taylor-sorok e x = ∞ ∑ n=0 x n n! sin x = ∞ ∑ n=0 (-1) n x 2n+1 (2n + 1)! cos x = ∞ ∑ n=0 (-1) n x 2n (2n)! (1 + x) α = ∞ ∑ n=0 ( α n ) x n |x| < 1 ( α 0 ) =1 ( α n ) = α(α-1)·...·(α-n+1) n! Fourier-sorok f (x)= a 0 + + ∞ ∑ n=1 (a n cos (nωx)+ b n sin (nωx)) a 0 = 1 T a+T Z a f (x) dx a n = 2 T a+T Z a f (x) cos (nωx) dx b n = 2 T a+T Z a f (x) sin (nωx) dx f (x + T )= f (x)´ es ω = 2π T Vektoranal´ ızis s = t 2 R t 1 | ˙ r ¯ | dt G = | ˙ r ¯ × ¨ r ¯ | | ˙ r ¯ | 3 T = ˙ r ¯ ¨ r ¯ ... r ¯ | ˙ r ¯ × ¨ r ¯ | 2 ∇ = ∂ ∂x , ∂ ∂y , ∂ ∂z ¶ gradu = ∇u divv ¯ = ∇v ¯ rotv ¯ = ∇× v ¯
Transcript of derivalasi - integralasi kepletek
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¯
