POCHODNE WZORY
( c )′ = 0 ( xα )′ = α ⋅ xα −1 ( x )′ =
1
2 x
'
1 1 = − x x2
(a x )′ = a x ln a ( e x )′ = e x (ln x)′ = 1x
x ∈ R , a ∈ R+ \ {1}
x ∈ R+
1 x ∈ R , a ∈ R+ \ {1} x ln a (sin x)′ = cos x (cos x)′ = − sin x (tgx)′ = 12 x ≠ π2 + kπ , k ∈ C cos x (ctgx)′ = − 12 x ≠ kπ , k ∈ C sin x (log a x)′ =
Arkadiusz Lisak
1
TWIERDZENIE. Jeżeli funkcje f(x) i g(x) są różniczkowalne na zbiorze X, to dla każdego x∈X
(cf (x))’ = c f ’(x) (f (x) ± g(x))’ = f’‘(x) ± g’(x) (f (x) ⋅ g(x))’ = f ’(x)g(x) + f (x)g’(x) '
f ( x) f ' ( x) g ( x) − f ( x) g ' ( x) = g ( x) [g ( x)]2
[ f (g (x )) ] '= f ' (t ) ⋅ g ' (x ) , gdzie
t = t (x )
Przykład. 2 2 1 1 1 4 + cos x − 4 = + cos x − 1. x + sin x − 4 ln x '= 3
(
32 x
x
x
3 x
)
2. 3x ⋅ x 2 '= 3x ln 3 ⋅ x 2 + 3x ⋅ 2 x = 3x ⋅ x ⋅ (ln 3 ⋅ x + 2) 2 cos x ' − 2 sin x ⋅ (3 − x ) − 2 cos x ⋅ (− 1) − 2 sin x ⋅ (3 − x ) + 2 cos x = 3. = 2 2
(3 − x )
3− x
x x − 3 x2 '= 4. 2 x 5
5
3 x2
−
2 x3
x2 5
(3 − x )
2 3 1 2 −1 3 x 2 − x 3 ⋅ x2 − x 2 − x 3 ⋅ 2x 3 2 ' = = 4 x
5
3 2 2 3 7 7 7 3 3 3 x − x − 2x 2 + 2x 3 − − − − 1 4 3 −2 2 −3 2 3 = = x − x − 2x 2 + 2x 3 = − x 2 + x 3 2 3 2 3 x4
lub 2 3 x x − 3 x2 x 2 − x 3 '= 2 x2 x
2 3 4 7 3 x2 x3 −1 − − − 1 4 ' ' ' 3 2 2 = − 2 = x − x = − x + x 3 2 2 3 x x
1 x3 − 1 x3 − 1 ' x = e x ⋅ 3x 2 + 2 5. e x
Arkadiusz Lisak
2