d/(dt)((s)(16t^2+4t-1))

Lösung

\frac{d}{d t} ((s) (16 t^{2} + 4 t - 1))

s (32 t + 4)

Schritte zur Lösung

\frac{d}{d t} ((s) (16 t^{2} + 4 t - 1))

Entferne die Konstante:

(a \cdot f)^{'} = a \cdot f^{'}

= s \frac{d}{d t} (16 t^{2} + 4 t - 1)

Wende die Summen-/Differenzregel an:

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

= s (\frac{d}{d t} (16 t^{2}) + \frac{d}{d t} (4 t) - \frac{d}{d t} (1))

\frac{d}{d t} (16 t^{2}) = 32 t

\frac{d}{d t} (4 t) = 4

\frac{d}{d t} (1) = 0

= s (32 t + 4 - 0)

Vereinfache

= s (32 t + 4)

e x p an d - x^{2} - r ((1 - a) y + a x - t)^{2}

e x p an d y d x + (3 x - x y + 2) d y

e x p an d \frac{x ^{2} - 5 y}{y ^{2} ( 5 x - 2 y )}

e x p an d \frac{sin ( 2 ( y - 3 ) )}{2 ( y - 3 )}

e x p an d x^{(1 - n) (k - 1)}