derivative of 0.001x^3-0.14x^2+7x+160

Lösung

\frac{d}{d x} (0.001 x^{3} - 0.14 x^{2} + 7 x + 160)

0.003 x^{2} - 0.28 x + 7

Schritte zur Lösung

\frac{d}{d x} (0.001 x^{3} - 0.14 x^{2} + 7 x + 160)

Wende die Summen-/Differenzregel an:

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

= \frac{d}{d x} (0.001 x^{3}) - \frac{d}{d x} (0.14 x^{2}) + \frac{d}{d x} (7 x) + \frac{d}{d x} (160)

\frac{d}{d x} (0.001 x^{3}) = 0.003 x^{2}

\frac{d}{d x} (0.14 x^{2}) = 0.28 x

\frac{d}{d x} (7 x) = 7

\frac{d}{d x} (160) = 0

= 0.003 x^{2} - 0.28 x + 7 + 0

Vereinfache

= 0.003 x^{2} - 0.28 x + 7

t a y l or [(z - 1)^{3} (z + 1)^{2}] - 1

\int \frac{t ^{2} - 3}{- t ^{3} + 9 t + 1} d t

t an g e n t 3 x

\frac{d}{d x} (co x)

\int \frac{x ^{3} + 1}{3 x ^{2}} d x