i) Berechnen Sie die Ableitung der folgenden Funktionen. A) b ( v ) = 7 v 5 {\displaystyle \ b(v)=7\ v^{5}\quad } B) x ( y ) = y 9 {\displaystyle \ x(y)=y^{9}\quad } C) f ( x ) = 5 12 x 12 {\displaystyle \ f(x)={\frac {5}{12}}x^{12}} D) a ( c ) = b c d {\displaystyle \ a(c)=b\ c^{d}} E) v ( t ) = t 3 4 {\displaystyle \ v(t)={\sqrt[{4}]{t^{3}}}\quad } F) f ( x ) = 5 x 5 {\displaystyle \ f(x)={\frac {5}{x^{5}}}\quad } G) V ( h ) = 1 h 5 4 {\displaystyle \ V(h)={\sqrt[{4}]{\frac {1}{h^{5}}}}\quad } H) H ( x ) = 27 x 7 3 {\displaystyle \ H(x)={\sqrt[{3}]{\frac {27}{x^{7}}}}\quad } I) m ( n ) = 3125 n 15 5 {\displaystyle \ m(n)={\sqrt[{5}]{\frac {3125}{n^{15}}}}\ } . ii) Berechnen Sie den Wert der Funktion und der Ableitung . an der Stelle 2 bei Aufgaben A, B, C, D, G und I
i)
A) b ( v ) = 7 v 5 ⇒ b ′ ( v ) = 35 v 4 {\displaystyle \ b(v)=7\ v^{5}\ \Rightarrow \ b'(v)=35\ v^{4}\ \quad } B) x ( y ) = y 9 ⇒ x ′ ( y ) = 9 y 8 {\displaystyle \ x(y)=y^{9}\ \Rightarrow \ x'(y)=9\ y^{8}\ \quad }
C) f ( x ) = 5 12 x 12 ⇒ f ′ ( x ) = 5 x 11 {\displaystyle \ f(x)={\frac {5}{12}}x^{12}\ \Rightarrow \ f'(x)={5}x^{11}\ \quad } D) a ( c ) = b c d ⇒ a ′ ( c ) = b d c d − 1 {\displaystyle \ a(c)=b\ c^{d}\ \Rightarrow \ a'(c)=b\ d\ c^{d-1}}
E) v ( t ) = t 3 4 = t 3 4 ⇒ v ′ ( t ) = 3 4 t 3 4 − 1 = 3 4 t − 1 4 ( = 3 4 t 4 ) {\displaystyle \ v(t)={\sqrt[{4}]{t^{3}}}\ =t^{\frac {3}{4}}\ \ \Rightarrow \ v'(t)={\frac {3}{4}}t^{{\frac {3}{4}}-1}\ ={\frac {3}{4}}t^{-{\frac {1}{4}}}\left(={\frac {3}{4\ {\sqrt[{4}]{t}}}}\right)\quad }
F) f ( x ) = 5 x 5 = 5 x − 5 ⇒ f ′ ( x ) = 5 ⋅ ( − 5 ) x − 5 − 1 = − 25 x − 6 {\displaystyle \ f(x)={\frac {5}{x^{5}}}=5\ x^{-5}\ \ \Rightarrow \ f'(x)=5\cdot (-5)\ x^{-5-1}\ =-25\ x^{-6}\quad }
G) V ( h ) = 1 h 5 4 = h − 5 4 = h − 5 4 ⇒ V ′ ( h ) = − 5 4 h − 5 4 − 1 = − 5 4 h − 9 4 ( = − 5 4 h 9 4 ) {\displaystyle \ V(h)={\sqrt[{4}]{\frac {1}{h^{5}}}}\ ={\sqrt[{4}]{h^{-5}}}\ =h^{-{5 \over 4}}\ \Rightarrow \ V'(h)=-{5 \over 4}h^{-{\frac {5}{4}}-1}\ =-{5 \over 4}h^{-{\frac {9}{4}}}\left(=-{\frac {5}{4\ {\sqrt[{4}]{h^{9}}}}}\right)}
H) H ( x ) = 27 x 7 3 = 27 3 x − 7 3 = 3 x − 7 3 ⇒ {\displaystyle \ H(x)={\sqrt[{3}]{\frac {27}{x^{7}}}}\ ={\sqrt[{3}]{27}}{\sqrt[{3}]{x^{-7}}}=3\ x^{-{\frac {7}{3}}}\ \Rightarrow \ }
H ′ ( x ) = 3 ⋅ ( − 7 3 ) x − 7 3 − 1 = − 7 x − 10 3 ( = − 7 x 10 3 ) {\displaystyle \qquad \qquad H'(x)=3\cdot \left(-{\frac {7}{3}}\right)x^{-{\frac {7}{3}}-1}=-7\ x^{-{\frac {10}{3}}}\left(=-{\frac {7}{\sqrt[{3}]{x^{10}\ }}}\right)}
I) m ( n ) = 3125 n 15 5 = 3125 5 n − 15 5 = 5 n − 15 5 = 5 n − 3 ⇒ {\displaystyle \ m(n)={\sqrt[{5}]{\frac {3125}{n^{15}}}}\ ={\sqrt[{5}]{3125}}\ {\sqrt[{5}]{n^{-15}}}=5\ n^{-{\frac {15}{5}}}=5\ n^{-3}\ \Rightarrow \ }
m ′ ( n ) = 5 ⋅ ( − 3 ) n − 3 − 1 = − 15 n − 4 {\displaystyle \qquad \qquad m'(n)=5\cdot \left(-3\right)n^{-3-1}=-15\ n^{-4}} .
ii)
A) b ( 2 ) = 7 ⋅ 2 5 = 224 b ′ ( 2 ) = 35 ⋅ 2 4 = 560 {\displaystyle \ b(2)=7\cdot 2^{5}=224\ \ \ b'(2)=35\cdot 2^{4}=560\ \quad } B) x ( 2 ) = 2 9 = 512 x ′ ( 2 ) = 9 ⋅ 2 8 = 2304 {\displaystyle \ x(2)=2^{9}=512\ \ \ x'(2)=9\cdot 2^{8}=2304\ \quad }
C) f ( 2 ) = 5 12 ⋅ 2 12 = 1706 , 6 ˙ f ′ ( x ) = 5 2 11 = 20480 {\displaystyle \ f(2)={\frac {5}{12}}\cdot 2^{12}=1706{,}{\dot {6}}\ \ \ f'(x)={\frac {5}{2}}^{11}=20480\ \quad } D) a ( 2 ) = b ⋅ 2 d a ′ ( 2 ) = b ⋅ d ⋅ 2 d − 1 {\displaystyle \ a(2)=b\cdot 2^{d}\ \ \ a'(2)=b\cdot d\cdot 2^{d-1}}
G) V ( 2 ) = 2 − 5 4 ≈ 0 , 42 V ′ ( 2 ) = − 5 4 ⋅ 2 1 4 ≈ − 1 , 05 {\displaystyle \ V(2)=2^{-{5 \over 4}}\approx 0{,}42\ \ \ V'(2)=-{5 \over 4}\cdot 2^{\frac {1}{4}}\approx -1{,}05}
I) m ( 2 ) = 5 ⋅ 2 − 3 = 0,625 m ′ ( 2 ) = − 15 ⋅ 2 − 4 = − 0,937 5 {\displaystyle \ m(2)=5\cdot 2^{-3}=0{,}625\ \ \ m'(2)=-15\cdot 2^{-4}=-0{,}9375}
B) 
D) 
![{\displaystyle \ v(t)={\sqrt[{4}]{t^{3}}}\ =t^{\frac {3}{4}}\ \ \Rightarrow \ v'(t)={\frac {3}{4}}t^{{\frac {3}{4}}-1}\ ={\frac {3}{4}}t^{-{\frac {1}{4}}}\left(={\frac {3}{4\ {\sqrt[{4}]{t}}}}\right)\quad }](https://wikimedia.org/api/rest_v1/media/math/render/svg/74a2f0af4b0eb82f780e5c028f25bad960ffc921)
![{\displaystyle \ V(h)={\sqrt[{4}]{\frac {1}{h^{5}}}}\ ={\sqrt[{4}]{h^{-5}}}\ =h^{-{5 \over 4}}\ \Rightarrow \ V'(h)=-{5 \over 4}h^{-{\frac {5}{4}}-1}\ =-{5 \over 4}h^{-{\frac {9}{4}}}\left(=-{\frac {5}{4\ {\sqrt[{4}]{h^{9}}}}}\right)}](https://wikimedia.org/api/rest_v1/media/math/render/svg/7eebca1e4afd23dcee131a0d1893a79301cf7c5e)
![{\displaystyle \ H(x)={\sqrt[{3}]{\frac {27}{x^{7}}}}\ ={\sqrt[{3}]{27}}{\sqrt[{3}]{x^{-7}}}=3\ x^{-{\frac {7}{3}}}\ \Rightarrow \ }](https://wikimedia.org/api/rest_v1/media/math/render/svg/2fc4cabc121b43763b96ee2fe5b9d05f641124da)
![{\displaystyle \qquad \qquad H'(x)=3\cdot \left(-{\frac {7}{3}}\right)x^{-{\frac {7}{3}}-1}=-7\ x^{-{\frac {10}{3}}}\left(=-{\frac {7}{\sqrt[{3}]{x^{10}\ }}}\right)}](https://wikimedia.org/api/rest_v1/media/math/render/svg/f3a507dc0b34b8e800aba57b3f77f00e2737e608)
![{\displaystyle \ m(n)={\sqrt[{5}]{\frac {3125}{n^{15}}}}\ ={\sqrt[{5}]{3125}}\ {\sqrt[{5}]{n^{-15}}}=5\ n^{-{\frac {15}{5}}}=5\ n^{-3}\ \Rightarrow \ }](https://wikimedia.org/api/rest_v1/media/math/render/svg/1608ef0bcd097dd53166fe4f78e4e2c732823f06)
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B) 
D) 
![{\displaystyle \ v(t)={\sqrt[{4}]{t^{3}}}\quad }](https://wikimedia.org/api/rest_v1/media/math/render/svg/d526346eb3a2a98cbf0efb74449697145f89a41f)
![{\displaystyle \ V(h)={\sqrt[{4}]{\frac {1}{h^{5}}}}\quad }](https://wikimedia.org/api/rest_v1/media/math/render/svg/adbec42ff2d23596366bdd416a557e93d4761e91)
![{\displaystyle \ H(x)={\sqrt[{3}]{\frac {27}{x^{7}}}}\quad }](https://wikimedia.org/api/rest_v1/media/math/render/svg/daa4a072dbb975f068d96688396d3124b06c53b3)
![{\displaystyle \ m(n)={\sqrt[{5}]{\frac {3125}{n^{15}}}}\ }](https://wikimedia.org/api/rest_v1/media/math/render/svg/34212041b1a93b5ff6998c3760946d914eb6c7f0)
