derivative 4xe^{x^2-1}

Lösung

ableitung von

4 x e^{x^{2} - 1}

4 (e^{x^{2} - 1} + 2 e^{x^{2} - 1} x^{2})

Schritte zur Lösung

\frac{d}{d x} (4 x e^{x^{2} - 1})

Entferne die Konstante:

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

= 4 \frac{d}{d x} (x e^{x^{2} - 1})

Wende die Produktregel an:

(f \cdot g)^{'} = f^{'} \cdot g + f \cdot g^{'}

f = x, g = e^{x^{2} - 1}

= 4 (\frac{d x}{d x} e^{x^{2} - 1} + \frac{d}{d x} (e^{x^{2} - 1}) x)

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

\frac{d}{d x} (e^{x^{2} - 1}) = e^{x^{2} - 1} \cdot 2 x

= 4 (1 \cdot e^{x^{2} - 1} + e^{x^{2} - 1} \cdot 2 xx)

Vereinfache

4 (1 \cdot e^{x^{2} - 1} + e^{x^{2} - 1} \cdot 2 xx) : 4 (e^{x^{2} - 1} + 2 e^{x^{2} - 1} x^{2})

= 4 (e^{x^{2} - 1} + 2 e^{x^{2} - 1} x^{2})

d er i v a t i v e f (x) = \frac{2}{3 x} - \frac{1}{6 x}

t a y l or sin (x), π

\frac{d P}{d t} = P (850 - P)

\frac{d y}{d x} = - y

y^{^{'}} + \frac{5}{x} y = 0