d/(dq)(-0.1q^2+1.2q+46.4)

Lösung

\frac{d}{d q} (- 0.1 q^{2} + 1.2 q + 46.4)

- 0.2 q + 1.2

Schritte zur Lösung

\frac{d}{d q} (- 0.1 q^{2} + 1.2 q + 46.4)

Wende die Summen-/Differenzregel an:

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

= - \frac{d}{d q} (0.1 q^{2}) + \frac{d}{d q} (1.2 q) + \frac{d}{d q} (46.4)

\frac{d}{d q} (0.1 q^{2}) = 0.2 q

\frac{d}{d q} (1.2 q) = 1.2

\frac{d}{d q} (46.4) = 0

= - 0.2 q + 1.2 + 0

Vereinfache

= - 0.2 q + 1.2

n = 1 \sum \infty (\frac{9}{8})^{n}

im pl i c i t \frac{d y}{d x}, x y + 3 = y

\int \frac{( x ^{7} )}{1 + x ^{16}} d x

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d er i v a t i v e 2 4 - x