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

Lösung

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

- 6 y

Schritte zur Lösung

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

Behandle

x

als Konstante

Wende die Summen-/Differenzregel an:

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

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

\frac{\partial}{\partial y} (2 x^{2}) = 0

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

\frac{\partial}{\partial y} (2 x) = 0

= 0 - 6 y - 0

Vereinfache

= - 6 y

\frac{\partial}{\partial x} (sin (x) + cos (- 6 y))

\frac{\partial}{\partial x} (\frac{5 x y ^{2}}{x ^{2} + y ^{4}})

in v erse l a pl a ce \frac{20}{s \cdot ( s + 10 )}

t \to 7 lim \frac{( t - 7 ) ^{3}}{t - 7}

n = 0 \sum \infty \frac{1}{10 n + 1}