ISM – Differentialrechnung – Kapitel 3.2 Ableitungen
Lösungen zum Aufgabenblatt
Aufgabe 1a
\[\begin{array}{l}
f\left( x \right) = 5\\
f’\left( x \right) = 0
\end{array}\]
Aufgabe 1b
\[\begin{array}{l}
f\left( x \right) = x\\
f'\left( x \right) = 1
\end{array}\]
Aufgabe 1c
\[\begin{array}{l}
f\left( x \right) = 3{x^2} + 4x - 2\\
f'\left( x \right) = 6x + 4
\end{array}\]
Aufgabe 1d
\[\begin{array}{l}
f\left( x \right) = \sqrt x = {x^{\frac{1}{2}}}\\
f'\left( x \right) = \frac{1}{2}{x^{ - \frac{1}{2}}} = \frac{1}{2} \cdot \frac{1}{{{x^{\frac{1}{2}}}}} = \frac{1}{{2\sqrt x }}
\end{array}\]
Aufgabe 1e
\[\begin{array}{l}
f\left( x \right) = \frac{2}{{{x^3}}} = 2{x^{ - 3}}\\
f'\left( x \right) = - 6{x^{ - 4}} = - \frac{6}{{{x^4}}}
\end{array}\]
Aufgabe 1f
\[\begin{array}{l}
f\left( x \right) = 5 \cdot \sin \left( x \right)\\
f'\left( x \right) = 5\cos \left( x \right)
\end{array}\]
Aufgabe 2a
\[\begin{array}{l}
f\left( x \right) = \frac{1}{2}{\left( {{x^2} - 4x} \right)^4}\\
f'\left( x \right) = \left( {2x - 4} \right) \cdot \frac{1}{2} \cdot 4 \cdot {\left( {{x^2} - 4x} \right)^3} = 2\left( {2x - 4} \right) \cdot {\left( {{x^2} - 4x} \right)^3}
\end{array}\]
Aufgabe 2b
\[\begin{array}{l}
f\left( x \right) = \sqrt {5{x^3} - 4x + 1} \\
f'\left( x \right) = \left( {15{x^2} - 4} \right) \cdot \frac{1}{{2\sqrt {5{x^3} - 4x + 1} }} = \frac{{15{x^2} - 4}}{{2\sqrt {5{x^3} - 4x + 1} }}
\end{array}\]
Aufgabe 2c
\[\begin{array}{l}
f\left( x \right) = \frac{3}{{{x^2} - x}} = 3 \cdot {\left( {{x^2} - x} \right)^{ - 1}}\\
f'\left( x \right) = \left( {2x - 1} \right) \cdot 3 \cdot \left( { - 1} \right) \cdot {\left( {{x^2} - x} \right)^{ - 2}}\\
= - \frac{{3\left( {2x - 1} \right)}}{{{{\left( {{x^2} - x} \right)}^2}}}
\end{array}\]
Aufgabe 2d
\[\begin{array}{l}
f\left( x \right) = \frac{1}{4}{e^{ - 0,3x - 4}}\\
f'\left( x \right) = - 0,3 \cdot \frac{1}{4} \cdot {e^{ - 0,3x - 4}}\\
= - \frac{3}{{40}}{e^{ - 0,3x - 4}}
\end{array}\]
Aufgabe 2e
\[\begin{array}{l}
f\left( x \right) = 4\ln \left( {4{x^4} - 3x} \right)\\
f'\left( x \right) = \left( {16{x^3} - 3} \right) \cdot \frac{4}{{4{x^4} - 3x}}\\
= \frac{{64{x^3} - 12}}{{4{x^4} - 3x}}
\end{array}\]
Aufgabe 2f
\[\begin{array}{l}
f\left( x \right) = - \frac{1}{\pi }\sin \left( {{x^2} - \pi x + 1} \right)\\
f'\left( x \right) = - \frac{1}{\pi } \cdot \left( {2x - \pi } \right) \cdot \cos \left( {{x^2} - \pi x + 1} \right)
\end{array}\]
Aufgabe 3a
\[\begin{array}{l}
f\left( x \right) = 2x \cdot \sqrt x \\
f'\left( x \right) = 2 \cdot \sqrt x + 2x \cdot \frac{1}{{2\sqrt x }}\\
= 2\sqrt x + \frac{x}{{\sqrt x }} = 3\sqrt x
\end{array}\]
Aufgabe 3b
\[\begin{array}{l}
f\left( x \right) = \left( {\frac{1}{2}{x^2} - 4x} \right) \cdot \sin \left( x \right)\\
f'\left( x \right) = \left( {x - 4} \right) \cdot \sin \left( x \right) + \left( {\frac{1}{2}{x^2} - 4x} \right) \cdot \cos \left( x \right)
\end{array}\]
Aufgabe 3c
\[\begin{array}{l}
f\left( x \right) = - \frac{4}{{{x^2}}} \cdot \ln \left( x \right) = - 4x{}^{ - 2} \cdot \ln \left( x \right)\\
f'\left( x \right) = 8{x^{ - 3}} \cdot \ln \left( x \right) - 4{x^{ - 2}} \cdot \frac{1}{x}\\
= \frac{{8\ln \left( x \right)}}{{{x^3}}} - \frac{4}{{{x^2}}} \cdot \frac{1}{x}\\
= \frac{{8\ln \left( x \right) - 4}}{{{x^3}}}
\end{array}\]
Aufgabe 3d
\[\begin{array}{l}
f\left( x \right) = \frac{x}{{x - 1}}\\
f'\left( x \right) = \frac{{1 \cdot \left( {x - 1} \right) - x \cdot 1}}{{{{\left( {x - 1} \right)}^2}}}\\
= - \frac{1}{{{{\left( {x - 1} \right)}^2}}}
\end{array}\]
Aufgabe 3e
\[\begin{array}{l}
f\left( x \right) = \frac{{{x^2} - x}}{{2x + 1}}\\
f'\left( x \right) = \frac{{\left( {2x - 1} \right) \cdot \left( {2x + 1} \right) - \left( {{x^2} - x} \right) \cdot 2}}{{{{\left( {2x + 1} \right)}^2}}}\\
= \frac{{4{x^2} - 1 - 2{x^2} + 2x}}{{{{\left( {2x + 1} \right)}^2}}}\\
= \frac{{2{x^2} + 2x - 1}}{{{{\left( {2x + 1} \right)}^2}}}
\end{array}\]
Aufgabe 3f
\[\begin{array}{l}
f\left( x \right) = - \frac{{2{x^3} - {x^2}}}{{4{x^2} - 2x}} = \frac{{ - 2{x^3} + {x^2}}}{{4{x^2} - 2x}} = \frac{{ - {x^2} \cdot \left( {2x - 1} \right)}}{{2x \cdot \left( {2x - 1} \right)}} = \frac{{ - {x^2}}}{{2x}} = - \frac{1}{2}x\\
f'\left( x \right) = - \frac{1}{2}
\end{array}\]
Aufgabe 4a
\[\begin{array}{l}
f\left( x \right) = {e^{2x\sqrt {4x + 1} }}\\
f'\left( x \right) = \left( {2\sqrt {4x + 1} + 2x \cdot \frac{1}{{2\sqrt {4x + 1} }} \cdot 4} \right){e^{2x\sqrt {4x + 1} }}\\
= \left( {2\sqrt {4x + 1} + \frac{{4x}}{{\sqrt {4x + 1} }}} \right){e^{2x\sqrt {4x + 1} }}
\end{array}\]
Aufgabe 4b
\[\begin{array}{l}
f\left( x \right) = \frac{{\ln \left( {{x^2} + 1} \right)}}{{{x^2} - 4}} = \ln \left( {{x^2} + 1} \right) \cdot {\left( {{x^2} - 4} \right)^{ - 1}}\\
f'\left( x \right) = \frac{1}{{{x^2} + 1}} \cdot 2x \cdot {\left( {{x^2} - 4} \right)^{ - 1}} + \ln \left( {{x^2} + 1} \right) \cdot 2x \cdot \left( { - 1} \right) \cdot {\left( {{x^2} - 4} \right)^{ - 2}}\\
= \frac{{2x}}{{\left( {{x^2} + 1} \right) \cdot \left( {{x^2} - 4} \right)}} - \frac{{2x\ln \left( {{x^2} + 1} \right)}}{{{{\left( {{x^2} - 4} \right)}^2}}}
\end{array}\]
Aufgabe 4c
\[\begin{array}{l}
f\left( x \right) = \sin {\left( {\frac{\pi }{x}} \right)^4}\\
f'\left( x \right) = 4\sin {\left( {\frac{\pi }{x}} \right)^3} \cdot \left( { - \frac{\pi }{{{x^2}}}} \right) \cdot \cos \left( {\frac{\pi }{x}} \right)\\
= \frac{{ - 4\pi \cdot \sin {{\left( {\frac{\pi }{4}} \right)}^3} \cdot \cos \left( {\frac{\pi }{x}} \right)}}{{{x^2}}}
\end{array}\]
Aufgabe 4d
\[\begin{array}{l}
f\left( x \right) = x \cdot \ln \left( {\frac{1}{x}} \right)\\
f'\left( x \right) = 1 \cdot \ln \left( {\frac{1}{x}} \right) + x \cdot \frac{1}{{\frac{1}{x}}} \cdot \left( { - \frac{1}{{{x^2}}}} \right)\\
= \ln \left( {\frac{1}{x}} \right) - x \cdot x \cdot \frac{1}{{{x^2}}}\\
= \ln \left( {\frac{1}{x}} \right) - \frac{{{x^2}}}{{{x^2}}}\\
= \ln \left( {\frac{1}{x}} \right) - 1
\end{array}\]
Aufgabe 4e
\[\begin{array}{l}
f\left( x \right) = \frac{{{e^{ - {x^2} + x}}}}{{{x^2}}} = {e^{ - {x^2} + x}} \cdot {x^{ - 2}}\\
f'\left( x \right) = \left( { - 2x + 1} \right) \cdot {e^{ - {x^2} + x}} \cdot {x^{ - 2}} + {e^{ - {x^2} + x}} \cdot \left( { - 2} \right) \cdot {x^{ - 3}}\\
= \frac{{\left( { - 2x + 1} \right) \cdot {e^{ - {x^2} + x}}}}{{{x^2}}} - \frac{{2{e^{ - {x^2} + x}}}}{{{x^3}}}\\
= \frac{{x \cdot \left( { - 2x + 1} \right) \cdot {e^{ - {x^2} + x}}}}{{{x^3}}} - \frac{{2{e^{ - {x^2} + x}}}}{{{x^3}}}\\
= \frac{{x \cdot \left( { - 2x + 1} \right) \cdot {e^{ - {x^2} + x}} - 2{e^{ - {x^2} + x}}}}{{{x^3}}}\\
= \frac{{\left( { - 2{x^2} + x - 2} \right)}}{{{x^3}}} \cdot {e^{ - {x^2} + x}}
\end{array}\]
Aufgabe 4f
\[\begin{array}{l}
f\left( x \right) = {\left( {2x - 1} \right)^4} \cdot {e^{ - 3{x^2} + 6x}}\\
f'\left( x \right) = 2 \cdot 4 \cdot {\left( {2x - 1} \right)^3} \cdot {e^{ - 3{x^2} + 6x}} + {\left( {2x - 1} \right)^4} \cdot \left( { - 6x + 6} \right){e^{ - 3{x^2} + 6x}}\\
= \left[ {8 \cdot {{\left( {2x - 1} \right)}^3} + {{\left( {2x - 1} \right)}^4} \cdot \left( { - 6x + 6} \right)} \right]{e^{ - 3{x^2} + 6x}}
\end{array}\]