Mit Hilfe von Grenzwerten berechnen Sie die Ableitung der Funktion y = a x 4 {\displaystyle y=ax^{4}} !
d ( y ) d x = d ( a x 4 ) d x = lim Δ x → 0 Δ ( a x 4 ) Δ x = lim Δ x → 0 a ( x + Δ x ) 4 − a Δ x 4 Δ x = {\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}}=}
lim Δ x → 0 a ( x 4 + 4 x 3 Δ x + 6 x 2 Δ x 2 + 4 x Δ x 3 + Δ x 4 ) − a x 4 Δ 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}}=}
lim Δ x → 0 a x 4 + 4 a x 3 Δ x + 6 a x 2 Δ x 2 + 4 a x Δ x 3 + a Δ x 4 − a x 4 Δ 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}}=}
lim Δ x → 0 Δ x ( 4 a x 3 + 6 a x 2 Δ x + 4 a x Δ x 2 + a Δ x 3 ) Δ 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\ \ }}}=}
lim Δ x → 0 ( 4 a x 3 + 6 a x 2 Δ x 0 + 4 a x Δ x 2 0 + a Δ x 3 0 ) = 4 a x 3 {\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: d ( a x 4 ) d x = 4 a x 3 {\displaystyle \qquad \qquad {\frac {d\ (ax^{4})}{dx}}=4ax^{3}}
Mit Hilfe von Grenzwerten berechnen Sie die Ableitung der Funktion y = 2 5 x 5 {\displaystyle y={\frac {2}{5}}x^{5}} !
d ( y ) d x = d ( 2 5 x 5 ) d x = lim Δ x → 0 Δ ( 2 5 x 5 ) Δ x = lim Δ x → 0 2 5 ( x + Δ x ) 5 − 2 5 Δ x 5 Δ x = {\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}}=}
lim Δ x → 0 2 5 ( x 5 + 5 x 4 Δ x + 10 x 3 Δ x 2 + 10 x 2 Δ x 3 + 5 x Δ x 4 + Δ x 5 ) − 2 5 x 5 Δ 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}}=}
lim Δ x → 0 2 5 x 5 + 2 5 5 x 4 Δ x + 2 5 10 x 3 Δ x 2 + 2 5 10 x 2 Δ x 3 + 2 5 5 x Δ x 4 + 2 5 Δ x 5 − 2 5 x 5 Δ 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}}=}
lim Δ x → 0 2 5 ( 5 x 4 Δ x + 10 x 3 Δ x 2 + 10 x 2 Δ x 3 + 5 x Δ x 4 + Δ x 5 ) Δ 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}}=}
lim Δ x → 0 2 5 Δ x ( 5 x 4 + 10 x 3 Δ x + 10 x 2 Δ x 2 + 5 x Δ x 3 + Δ x 4 ) Δ 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\ \ }}}=}
lim Δ x → 0 2 5 ( 5 x 4 + 10 x 3 Δ x 0 + 10 x 2 Δ x 2 0 + 5 x Δ x 3 0 + Δ x 4 0 ) = 2 5 5 x 4 = 2 x 4 {\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: d ( 2 5 x 5 ) d x = 2 5 ⋅ 5 x 4 = 2 x 4 {\displaystyle \qquad \qquad {\frac {d\ ({\frac {2}{5}}x^{5})}{dx}}={\frac {2}{5}}\cdot 5x^{4}=2x^{4}}
Mit Hilfe von Grenzwerten berechnen Sie die Ableitung der Funktion y = a x {\displaystyle y=a\ x} !
d ( y ) d x = d ( a x ) d x = lim Δ x → 0 Δ ( a x ) Δ x = lim Δ x → 0 a ( x + Δ x ) − a x Δ x = lim Δ x → 0 a Δ x Δ x = a {\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: d ( a x ) d x = a {\displaystyle \qquad \qquad {\frac {d\ (ax)}{dx}}=a}
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