(\partial)/(\partial x)(xe^{sqrt(2xy)})

Lösung

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

\frac{e ^{2 x y} x y}{2} + e^{2 x y}

Schritte zur Lösung

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

Vereinfache

x e^{2 x y} : e^{2 x y} x

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

Behandle

y

als Konstante

Wende die Produktregel an:

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

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

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

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

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

= \frac{y e ^{2 y x}}{2 x} x + 1 \cdot e^{2 x y}

Vereinfache

\frac{y e ^{2 y x}}{2 x} x + 1 \cdot e^{2 x y} : \frac{e ^{2 x y} x y}{2} + e^{2 x y}

= \frac{e ^{2 x y} x y}{2} + e^{2 x y}

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

d er i v a t i v e f (x) = \frac{1}{x ^{6}} + \frac{1}{x ^{8}}

a re a sin (x), cos (x), [\frac{π}{4}, \frac{5 π}{4}]

x^{2} y^{^{''}} - 3 x y^{^{'}} + 13 y = 0

\int arccot (9 y) d y