d/(dy)(2xe^{xy}+x^2ye^{xy})

Lösung

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

3 e^{x y} x^{2} + e^{x y} x^{3} y

Schritte zur Lösung

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

Behandle

x

als Konstante

Wende die Summen-/Differenzregel an:

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

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

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

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

= 2 x^{2} e^{x y} + x^{2} (e^{x y} + x e^{x y} y)

Vereinfache

2 x^{2} e^{x y} + x^{2} (e^{x y} + x e^{x y} y) : 3 e^{x y} x^{2} + e^{x y} x^{3} y

= 3 e^{x y} x^{2} + e^{x y} x^{3} y

\frac{\partial}{\partial y} (y^{2} - cos (x y))

\int 2 tan (x) d x

\frac{d}{d x} (- \frac{1}{( 2 x + 8 ) ^{\frac{3}{2}}})

\int + \frac{1}{x ^{2}} d x

\frac{d}{d t} (cos (2 t))