Matematika | Analízis » Analízis képletek - deriválás, integrálás, sorok

f (x) c xα ex ax ln x loga x sin x cos x tgx ctgx arcsin x arccos x arctgx arcctgx shx chx thx cthx arshx archx arthx arcthx Deriválás f 0 (x) 0 αxα−1 ex x a ln a 1 x 1 x ln a cos x − sin x 1 cos2 x 1 − 2 sin x 1 √ 1 − x2 1 −√ 1 − x2 1 1 + x2 1 − 1 + x2 chx shx 1 ch2 x 1 − 2 sh x 1 √ 1 + x2 1 √ 2 x −1 1 |x| < 1 1 − x2 1 |x| > 1 1 − x2 Deriválási szabályok 0 (cf ) = cf 0 0 (f ± g) = f 0 ± g 0 0 (f g) = f 0 g + f g 0 µ ¶0 f 0 g − f g0 f = g g2 0 0 (f ◦ g) = (f ◦ g) g 0 ¡ ¢0 1 f¯ = 0 ¯ f ◦f Paraméteres  megadású függvény: x = ϕ(t) ψ̇(t) f (x) : f 0 (x) = y = ψ(t) ϕ̇(t) Kiegészı́tések 1 + cos 2x 1 − cos 2x cos2 x = sin2 x = 2 2 ch2x − 1 ch2x + 1 2 2 ch x = sh x = 2 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) Integrálás F (x) xα+1 α 6= −1 α+1 ln |x| f (x) xα 1 x ex ex

ax ln a − cos x sin x ax sin x cos x 1 tgx cos2 x 1 − ctgx sin2 x shx chx chx shx 1 thx ch2 x 1 − cthx sh2 x 1 √ arcsin x 1 − x2 1 √ arshx 1 + x2 1 √ archx x2 − 1 1 arctgx 1 + x2 ¯ ¯ 1 1 ¯¯ 1 + x ¯¯ ln 1 − x2 2 ¯1 − x¯ Integrálási szabályok R F (ax + b) +C f (ax + b) dx = a α+1 R α f (x) f (x) f 0 (x) dx = +C α+1 ha α 6= −1 Z 0 f (x) dx = ln |f (x)| + C f (x) R f (g (x)) g 0 (x) dx = F (g (x)) + C R 0 u (x) v (x) dx = R = u (x) v (x) − u (x) v 0 (x) dx x t = tg helyettesı́tés: 2 2t 1 − t2 sin x = cos x = 1 + t2 1 + t2 Rb 2 V = π f (x)dx a Rb q 2 L= 1 + (f 0 (x)) dx a q Rb 2 F = 2π f (x) 1 + (f 0 (x)) dx a  x = ϕ(t) y = ψ(t) Rt2 2 V = π ψ (t)ϕ̇(t)dt t1 Rt2 q L= ϕ̇2 (t) + ψ̇ 2 (t)dt t1 q Rt2 F = 2π ψ(t) ϕ̇2 (t) + ψ̇ 2 (t)dt t1 Laplace-transzformáció f (t) f¯ (s) = L [f (t)] 1 eat s−a a sin (at) s2 + a 2 s cos (at) 2 s + a2 n! tn sn+1 a sh(at) s2 − a 2 s ch(at) 2 s − a2 eat f (t) f¯ (s − a) dn

f¯(s) tn f (t) (−1)n dsn f 0 (t) sf¯ (s) − f (0) f 00 (t) s2 f¯ (s) − sf (0) − f 0 (0) (n) f (t) sn f¯ (s) − sn−1 f (0) − . . − f (n−1) (0) t R 1¯ f (s) f (u) du s 0 f (t − a) e−as f¯ (s) Taylor-sorok ∞ P xn ex = n! n=0 x2n+1 (2n + 1)! n=0 2n ∞ P n x cos x = (−1) (2n)! n=0 ∞ ¡ ¢ P α α n (1 + x) = |x| < 1 n x ¡α¢ ¡α¢n=0 α(α−1)·.·(α−n+1) n! 0 =1 n = sin x = ∞ P n (−1) Fourier-sorok f (x) = a0 + ∞ P + (an cos (nωx) + bn sin (nωx)) n=1 1 a0 = T 2 an = T bn = 2 T a+T Z a+T Z f (x) dx a f (x) cos (nωx) dx a a+T Z f (x) sin (nωx) dx a f (x + T ) = f (x) és ω = Vektoranalı́zis Rt2 s = |ṙ| dt t1 ¯ |ṙ × r̈| G = ¯ 3¯ T = |ṙ| µ ¶ ¯ ∂ ∂ ∂ ∇= , , ∂x ∂y ∂z 2π T . ṙ r̈ r ¯¯ ¯ 2 |ṙ × r̈| ¯ ¯ gradu = ∇u divv = ∇v rotv = ∇ × v