Mathematrix: Aufgabenbeispiele/ Ableitung und Grenzwerten

01

 Mit Hilfe von Grenzwerten berechnen Sie die Ableitung der Funktion ${\displaystyle y=ax^{4}}$!

${\displaystyle {\frac {d\ (y)}{dx}}={\frac {d\ (ax^{4})}{dx}}={\underset {\Delta x\rightarrow 0}{\lim }}{\frac {\Delta \ (ax^{4})}{\Delta x}}={\underset {\Delta x\rightarrow 0}{\lim }}{\frac {a(x+\Delta x)^{4}-a\Delta x^{4}}{\Delta x}}=}$

${\displaystyle {\underset {\Delta x\rightarrow 0}{\lim }}{\frac {a(x^{4}+4x^{3}\Delta x+6x^{2}\Delta x^{2}+4x\Delta x^{3}+\Delta x^{4})-ax^{4}}{\Delta x}}=}$

${\displaystyle {\underset {\Delta x\rightarrow 0}{\lim }}{\frac {{\cancel {\ \ ax^{4}\ \ }}+4ax^{3}\Delta x+6ax^{2}\Delta x^{2}+4ax\Delta x^{3}+a\Delta x^{4}-{\cancel {\ \ ax^{4}\ \ }}}{\Delta x}}=}$

${\displaystyle {\underset {\Delta x\rightarrow 0}{\lim }}{\frac {{\cancel {\ \ \Delta x\ \ }}(4ax^{3}+6ax^{2}\Delta x+4ax\Delta x^{2}+a\Delta x^{3})}{\cancel {\ \ \Delta x\ \ }}}=}$

${\displaystyle {\underset {\Delta x\rightarrow 0}{\lim }}(4ax^{3}+6ax^{2}{\cancelto {0}{\ \ \Delta x\ \ }}+4ax{\cancelto {0}{\ \ \Delta x^{2}\ \ }}+a{\cancelto {0}{\ \ \Delta x^{3}\ \ }})=4ax^{3}}$

Also: ${\displaystyle \qquad \qquad {\frac {d\ (ax^{4})}{dx}}=4ax^{3}}$

02

 Mit Hilfe von Grenzwerten berechnen Sie die Ableitung der Funktion ${\displaystyle y={\frac {2}{5}}x^{5}}$!

${\displaystyle {\frac {d\ (y)}{dx}}={\frac {d\ ({\frac {2}{5}}x^{5})}{dx}}={\underset {\Delta x\rightarrow 0}{\lim }}{\frac {\Delta \ ({\frac {2}{5}}x^{5})}{\Delta x}}={\underset {\Delta x\rightarrow 0}{\lim }}{\frac {{\frac {2}{5}}(x+\Delta x)^{5}-{\frac {2}{5}}\Delta x^{5}}{\Delta x}}=}$

${\displaystyle {\underset {\Delta x\rightarrow 0}{\lim }}{\frac {{\frac {2}{5}}(x^{5}+5x^{4}\Delta x+10x^{3}\Delta x^{2}+10x^{2}\Delta x^{3}+5x\Delta x^{4}+\Delta x^{5})-{\frac {2}{5}}x^{5}}{\Delta x}}=}$

${\displaystyle {\underset {\Delta x\rightarrow 0}{\lim }}{\frac {{\cancel {\ \ {\frac {2}{5}}x^{5}\ \ }}+{\frac {2}{5}}5x^{4}\Delta x+{\frac {2}{5}}10x^{3}\Delta x^{2}+{\frac {2}{5}}10x^{2}\Delta x^{3}+{\frac {2}{5}}5x\Delta x^{4}+{\frac {2}{5}}\Delta x^{5}-{\cancel {\ \ {\frac {2}{5}}x^{5}\ \ }}}{\Delta x}}=}$

${\displaystyle {\underset {\Delta x\rightarrow 0}{\lim }}{\frac {{\frac {2}{5}}(5x^{4}\Delta x+10x^{3}\Delta x^{2}+10x^{2}\Delta x^{3}+5x\Delta x^{4}+\Delta x^{5})}{\Delta x}}=}$

${\displaystyle {\underset {\Delta x\rightarrow 0}{\lim }}{\frac {{\frac {2}{5}}{\cancel {\ \ \Delta x\ \ }}(5x^{4}+10x^{3}\Delta x+10x^{2}\Delta x^{2}+5x\Delta x^{3}+\Delta x^{4})}{\cancel {\ \ \Delta x\ \ }}}=}$

${\displaystyle {\underset {\Delta x\rightarrow 0}{\lim }}{\frac {2}{5}}(5x^{4}+10x^{3}{\cancelto {0}{\ \ \Delta x\ \ }}+10x^{2}{\cancelto {0}{\ \ \Delta x^{2}\ \ }}+5x{\cancelto {0}{\ \ \Delta x^{3}\ \ }}+{\cancelto {0}{\ \ \Delta x^{4}\ \ }})={\frac {2}{5}}5x^{4}=2x^{4}}$

Also: ${\displaystyle \qquad \qquad {\frac {d\ ({\frac {2}{5}}x^{5})}{dx}}={\frac {2}{5}}\cdot 5x^{4}=2x^{4}}$

03

 Mit Hilfe von Grenzwerten berechnen Sie die Ableitung der Funktion ${\displaystyle y=a\ x}$!

${\displaystyle {\frac {d\ (y)}{dx}}={\frac {d\ (ax)}{dx}}={\underset {\Delta x\rightarrow 0}{\lim }}{\frac {\Delta \ (ax)}{\Delta x}}={\underset {\Delta x\rightarrow 0}{\lim }}{\frac {a(x+\Delta x)-a\ x}{\Delta x}}={\underset {\Delta x\rightarrow 0}{\lim }}{\frac {a{\cancel {\ \ \Delta x\ \ }}}{\cancel {\ \ \Delta x\ \ }}}=a}$

Also: ${\displaystyle \qquad \qquad {\frac {d\ (ax)}{dx}}=a}$