(\partial)/(\partial y)(2y^3-3y^2-12y+6)

Lösung

\frac{\partial}{\partial y} (2 y^{3} - 3 y^{2} - 12 y + 6)

6 y^{2} - 6 y - 12

Schritte zur Lösung

\frac{\partial}{\partial y} (2 y^{3} - 3 y^{2} - 12 y + 6)

Wende die Summen-/Differenzregel an:

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

= \frac{\partial}{\partial y} (2 y^{3}) - \frac{\partial}{\partial y} (3 y^{2}) - \frac{\partial}{\partial y} (12 y) + \frac{\partial}{\partial y} (6)

\frac{\partial}{\partial y} (2 y^{3}) = 6 y^{2}

\frac{\partial}{\partial y} (3 y^{2}) = 6 y

\frac{\partial}{\partial y} (12 y) = 12

\frac{\partial}{\partial y} (6) = 0

= 6 y^{2} - 6 y - 12 + 0

Vereinfache

= 6 y^{2} - 6 y - 12

\int_{0}^{1} 8 ln (x) d x

n = 1 \sum \infty \frac{8}{( n + 1 ) !}

ma c l a u r in 1 - \frac{x}{3}

\int \frac{1 + tan ( x )}{1 - tan ( x )} d x

x \to \infty lim ((1 - \frac{1}{4 x})^{8 x})