d/(dq)((C)(7+2q-3q^2))

Lösung

\frac{d}{d q} ((C) (7 + 2 q - 3 q^{2}))

C (- 6 q + 2)

Schritte zur Lösung

\frac{d}{d q} ((C) (7 + 2 q - 3 q^{2}))

Behandle

C

als Konstante

Entferne die Konstante:

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

= C \frac{d}{d q} (7 + 2 q - 3 q^{2})

Wende die Summen-/Differenzregel an:

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

= C (\frac{d}{d q} (7) + \frac{d}{d q} (2 q) - \frac{d}{d q} (3 q^{2}))

\frac{d}{d q} (7) = 0

\frac{d}{d q} (2 q) = 2

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

= C (0 + 2 - 6 q)

Vereinfache

= C (- 6 q + 2)

e x p an d \frac{x}{( x - 3 ) ( x + 1 )}

e x p an d - \frac{5 ( y - 1 )}{13 ( y + 1 )}

e x p an d f (f (f (- 2.8) + 3, 5) - 1)

e x p an d e^{- x^{2}} (1 + 2 x) - x

e x p an d (1 + 3 t) t^{2} d t