(\partial)/(\partial y)(3x-x^2y^2+2x^3y)

Lösung

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

2 x^{3} - 2 x^{2} y

Schritte zur Lösung

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

Behandle

x

als Konstante

Wende die Summen-/Differenzregel an:

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

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

\frac{\partial}{\partial y} (3 x) = 0

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

\frac{\partial}{\partial y} (2 x^{3} y) = 2 x^{3}

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

Vereinfache

= 2 x^{3} - 2 x^{2} y

\int \frac{1}{3 + 4 x ^{2}} d x

d er i v a t i v e y = e^{x + 2}

n = 4 \sum \infty (\frac{3}{4})^{- n}

\frac{d}{d x} (\frac{15 x ^{7} + 27 x ^{5}}{3 x})

y = x^{3 x}