derivative f(x)=-(x^3+x-1)/2

Lösung

ableitung von

f (x) = - \frac{x ^{3} + x - 1}{2}

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

Schritte zur Lösung

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

Entferne die Konstante:

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

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

Wende die Summen-/Differenzregel an:

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

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

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

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

\frac{d}{d x} (1) = 0

= - \frac{1}{2} (3 x^{2} + 1 - 0)

Vereinfache

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

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

\frac{d}{d x} (\frac{1}{a} e^{a x})

\int_{0}^{\frac{π}{2}} sin^{2} (x) d x

\int_{1}^{e^{3}} x^{3} ln (x) d x

y^{^{''}} - 2 y^{^{'}} - 3 y = 4 x - 5 + 6 x e^{2 x}