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Prostepochodne
Działanianapochodnych
Wzory: Przykłady:
( c ) 0 =0 (2) 0 =0
(100) 0 =0
Wzory: Przykłady:
( f + g ) 0 = f 0 + g 0
( x 2 + x 3 ) 0 =( x 2 ) 0 +( x 3 ) 0 = 2 x +3 x 2
( f g ) 0 = f 0 g 0
( x 4 x ) 0 =( x 4 ) 0 ( x ) 0 =4 x 3 1
( ax ) 0 = a ( x ) 0 =1
(3 x ) 0 =3
( c · f ) 0 = c · f 0
(5 x 3 ) 0 =5 · ( x 3 ) 0 =5 · 3 x 2 =15 x 2
( x n ) 0 = nx n 1
( x 3 ) 0 =3 x 2
( x 5 ) 0 =5 x 4
( f · g ) 0 = f 0 g + fg 0
( x 2 p
x ) 0 =( x 2 ) 0 p
x + x 2 (
p
x ) 0 =
= 2 x p x + x 2 · 1
2 p x ==2 x p x + x 2
2 p x
a
x
0
= a
x 2
1 x 0 = 1 x 2
3 x
0 = 3 x 2
f
g
0
= f 0 g fg 0
g 2
x 2
p x
0
= ( x 2 ) 0 p x x 2 ( p x ) 0
( p x ) 2 = 2 x p x x 2 1
2 p x
x =
p x 0 = 1
2 p x
=
x (2 p x x 1
x =2 p x x
2 p x )
2 p x
( e x ) 0 = e x
Pochodnafunkcjizło »one j
(2 x +1) 3 0 =( y 3 ) 0 =3 y 3 1 · y 0 =3 y 2 · y 0 =
( a x ) 0 = a x ln a
[ f ( y )] 0 = f 0 · y 0
(ln x ) 0 = 1
x
=3(2 x +1) 2 · 2=6(2 x +1) 2
gdzie y =2 x +1 ,y 0 =2
p x 3 2 x 0
(log a x ) 0 = 1
x ln a
=( p y ) 0 = 1
2 p y · y 0 =
(log 2 x ) 0 = 1
x ln2
(log 5 x ) 0 = 1
x ln5
= 1
2
p
x 3 2 x · (3 x 2 2)= 3 x 2 2
p
2
x 3 2 x
Pochodnefunkcjitrygonometrycznych
(sin x ) 0 =cos x (tg x ) 0 = 1
gdzie y = x 3 2 x,y 0 =3 x 2 2
cos 2 x =1+tg 2 x
e x 2 0
(cos x ) 0 = sin x (ctg x ) 0 = 1
=( e y ) 0 = e y · y 0 = e x 2 · 2 x =2 xe x 2
sin 2 x = (1+ctg 2 x )
gdzie y = x 2 ,y 0 =2 x
(arcsin x ) 0 = 1
(arctg x ) 0 = 1
1+ x 2
p
1 x 2
(arccos x ) 0 = 1
p
1 x 2
(arcctg x ) 0 = 1
1+ x 2
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