(\partial)/(\partial x)(-2xe^{-x^2})

Lösung

\frac{\partial}{\partial x} (- 2 x e^{- x^{2}})

- 2 (e^{- x^{2}} - 2 e^{- x^{2}} x^{2})

Schritte zur Lösung

\frac{\partial}{\partial x} (- 2 x e^{- x^{2}})

Entferne die Konstante:

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

= - 2 \frac{\partial}{\partial x} (x e^{- x^{2}})

Wende die Produktregel an:

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

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

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

\frac{\partial}{\partial x} (x) = 1

\frac{\partial}{\partial x} (e^{- x^{2}}) = - 2 e^{- x^{2}} x

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

Vereinfache

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

= - 2 (e^{- x^{2}} - 2 e^{- x^{2}} x^{2})

d er i v a t i v e sec^{2} (x)

2 y^{^{''}} + 17 y^{^{'}} - 9 y = 0

\frac{d}{d y} (x y + y z + x z)

y^{^{'}} = 3 e^{x - y}

f (x) = tan (4 x)