Material zum Vorkurs P2 WiSe 2013/14

WiSe 2013/14
30.09.2013
16. Tag – Differenzieren
Lösungen zu Differenzieren I & II
Lösungen zu Differenzieren I & II
1.) a) f(x) = x 4
⇒ f ′ (x) = 4 · x 4−1 = 4x 3
)
⇒ f ′′ (x) = 4 ·
(3 · x 3−1 = 12 · x 2
)
⇒ f ′′′ (x) = 12 ·
(2x 2−1 = 24x
b) f(x) = 2x 5
)
⇒ f ′ (x) = 2 ·
(5 · x 5−1 = 10x 4
) )
⇒ f ′′ (x) = 10 ·
(4 · x 4−1 = 10 ·
(4x 3 = 40x 3
)
⇒ f ′′′ (x) = 40 ·
(3x 3−1 = 120x 2
c) f(x) = 3x 4 + 4x 2
)
⇒ f ′ (x) = 3 ·
(4x 3 + 4 · (2x) = 12x 3 + 8x
)
⇒ f ′′ (x) = 12 ·
(3x 2 + 8 = 36x 2 + 8
⇒
f ′′′ (x) = 36 · (2x) + 0 = 72x
d) f(x) = 3x 11 − 3x 7
) )
⇒ f ′ (x) = 3 ·
(11x 10 − 3 ·
(7x 6 = 33x 10 − 21x 6
) )
⇒ f ′′ (x) = 33 ·
(10x 9 − 21 ·
(6x 5 = 330x 9 − 126x 5
)
)
⇒ f ′′′ (x) = 330 ·
(9x 8 − 126 ·
(5x 4 = 2970x 8 − 630x 4
e) f(x) = 3x −2
)
⇒ f ′ (x) = 3 ·
(−2 · x −2−1 = −6x −3
)
⇒ f ′′ (x) = −6 ·
(−3 · x −3−1 = 18x −4
)
⇒ f ′′′ (x) = 18 ·
(−4 · x −4−1 = −72x −5
f) f(x) = 1
x 3
= x −3
⇒ f ′ (x) = −3x −4
)
⇒ f ′′ (x) = −3 ·
(−4x −5 = 12x −5
)
⇒ f ′′′ (x) = 12 ·
(−5x −6 = −60x −6
g) f(x) = 5x 4 + 2x −4
)
⇒ f ′ (x) = 5 ·
(4x 3
⇒ f ′′ (x) = 20 ·
)
+ 2 ·
(−4x −5 ) (−5x −6
(3x 2 )
− 8 ·
= 20x 3 − 8x −5
= 60x 2 + 40x −6
Mathematik Vorkurs P2
✞ ☎
✝124 ✆
Dipl. Math. Stefan Podworny