Analysis I - IV - Fachhochschule Nordwestschweiz

88 Kapitel 4. Differenzialrechnung
Lösung 4.11.2.
a. f ′ (x) = 2
3 3√ x
b. f ′ (u) = 8u 1 3 + 2 3 u−4 3 −4u −3 5
c. f ′ (t) = 3 2√
t
d. f ′ (x) = 11
4
4√
x
7
e. f ′ (x) = − a
2 √ bx 3 − 1
2 √ bx
f. f ′ (u) = − 1
2 √ u 3
g. f ′ (x) = 1
2 √ x − 3√ x
2
h. f ′ (t) = √ t+ t−1
2 √ t
i. f ′ (ξ) = 6 √ ξ − 3ξ2 −a
2 √ ξ 3
j. f ′ (x) = 8x+1
3 3√ x 4 (3x2 −1)+6 3√ x 2 (4x−1)
k. f ′ (x) = − 4+ 4√ x
12 3√ x 4
l. f ′ (x) = −3√ x+4
4x 3
m. f ′ (x) =
n. f ′ (z) =
−a
√ x(a+
√ x) 2
−a
√ z(a+
√ z) 2
(1+ √ x) 2 + x−1
2 √ x 5
Lösung 4.11.3.
a. f ′ −1
(x) =
2 √ 1−x
g. f ′ 2 √ x
(x) =
2 −1+2
√
x 2 −1(1− √ x 2 −1) 2
b. f ′ (x) = −6x(1−√ 1+x 2 ) 5
√
1+x 2 h. f ′ (s) = −cos(√ 1−2s)
√ 1−2s
c. f ′ 1
(z) = √
4 z + √ z 3 i. f ′ (x) = 3+x+2√ 1+x
2(1−x) 2√ 1+x
)
−sin(
1
d. f ′ 1−x
j. f ′ (x) = 3x 2 −3x √ x 2 −1−1
(x) =
(1−x) 2
k. f ′ −2
(x) =
e. f ′ 1
(t) =
2cos 2 (t) √ 3(x−1) 3√ 4x 2 (x−1)
tan(t)
f. f ′ −1
(x) =
(1+x) √ 1−x 2