derivative f(t)=80t+t^2-1/48 t^3

Lösung

ableitung von

f (t) = 80 t + t^{2} - \frac{1}{48} t^{3}

80 + 2 t - \frac{t ^{2}}{16}

Schritte zur Lösung

\frac{d}{d t} (80 t + t^{2} - \frac{1}{48} t^{3})

Vereinfache

80 t + t^{2} - \frac{1}{48} t^{3} : 80 t + t^{2} - \frac{t ^{3}}{48}

= \frac{d}{d t} (80 t + t^{2} - \frac{t ^{3}}{48})

Wende die Summen-/Differenzregel an:

(f \pm g)^{'} = f^{'} \pm g^{'}

= \frac{d}{d t} (80 t) + \frac{d}{d t} (t^{2}) - \frac{d}{d t} (\frac{t ^{3}}{48})

\frac{d}{d t} (80 t) = 80

\frac{d}{d t} (t^{2}) = 2 t

\frac{d}{d t} (\frac{t ^{3}}{48}) = \frac{t ^{2}}{16}

= 80 + 2 t - \frac{t ^{2}}{16}

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