derivative f(t)=6t^3-54t^2+70

Lösung

ableitung von

f (t) = 6 t^{3} - 54 t^{2} + 70

18 t^{2} - 108 t

Schritte zur Lösung

\frac{d}{d t} (6 t^{3} - 54 t^{2} + 70)

Wende die Summen-/Differenzregel an:

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

= \frac{d}{d t} (6 t^{3}) - \frac{d}{d t} (54 t^{2}) + \frac{d}{d t} (70)

\frac{d}{d t} (6 t^{3}) = 18 t^{2}

\frac{d}{d t} (54 t^{2}) = 108 t

\frac{d}{d t} (70) = 0

= 18 t^{2} - 108 t + 0

Vereinfache

= 18 t^{2} - 108 t

\frac{d}{d x} (\frac{1}{3 x ^{3} + 1})

\frac{d}{d x} (\frac{x ^{2} + 1}{x ^{2} + x + 1})

\frac{\partial}{\partial x} (e^{- \frac{1}{2} x y})

\int - 2 x + 6 d x

y^{^{'}} + \frac{1}{x + 7} y = x^{- 2}, y (1) = 4