derivative 160+3.2x-0.03x^2

Lösung

ableitung von

160 + 3.2 x - 0.03 x^{2}

- 0.06 x + 3.2

Schritte zur Lösung

\frac{d}{d x} (160 + 3.2 x - 0.03 x^{2})

Wende die Summen-/Differenzregel an:

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

= \frac{d}{d x} (160) + \frac{d}{d x} (3.2 x) - \frac{d}{d x} (0.03 x^{2})

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

\frac{d}{d x} (3.2 x) = 3.2

\frac{d}{d x} (0.03 x^{2}) = 0.06 x

= 0 + 3.2 - 0.06 x

Vereinfache

= - 0.06 x + 3.2

\frac{d}{d x} (\frac{e ^{2 x} + e ^{- 2 x}}{4})

\frac{d}{d x} (3 x^{2} - 8 x)

n = 0 \sum \infty \frac{2}{3} n (\frac{1}{3})^{n}

n or ma l y = x^{4} + 9 e^{x}, (0, 5)

x \to 81 lim (\frac{( x - 9 )}{x - 81})