derivative of 3x^3+2x^2-1x+5

Lösung

\frac{d}{d x} (3 x^{3} + 2 x^{2} - 1 x + 5)

9 x^{2} + 4 x - 1

Schritte zur Lösung

\frac{d}{d x} (3 x^{3} + 2 x^{2} - 1 \cdot x + 5)

Vereinfache

3 x^{3} + 2 x^{2} - 1 \cdot x + 5 : 3 x^{3} + 2 x^{2} - x + 5

= \frac{d}{d x} (3 x^{3} + 2 x^{2} - x + 5)

Wende die Summen-/Differenzregel an:

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

= \frac{d}{d x} (3 x^{3}) + \frac{d}{d x} (2 x^{2}) - \frac{d x}{d x} + \frac{d}{d x} (5)

\frac{d}{d x} (3 x^{3}) = 9 x^{2}

\frac{d}{d x} (2 x^{2}) = 4 x

\frac{d x}{d x} = 1

\frac{d}{d x} (5) = 0

= 9 x^{2} + 4 x - 1 + 0

Vereinfache

= 9 x^{2} + 4 x - 1

\int_{0}^{3} e^{- x^{2}} d x

x \to π lim (ln (cos (x) - 3))

\int y^{2} z + 2 x z^{2} d x

d er i v a t i v e \frac{2}{x}

\int (- 4 sin (2 t)) d t