(\partial)/(\partial y)(2x+y-1)

Lösung

\frac{\partial}{\partial y} (2 x + y - 1)

1

Schritte zur Lösung

\frac{\partial}{\partial y} (2 x + y - 1)

Behandle

x

als Konstante

Wende die Summen-/Differenzregel an:

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

= \frac{\partial}{\partial y} (2 x) + \frac{\partial}{\partial y} (y) - \frac{\partial}{\partial y} (1)

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

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

\frac{\partial}{\partial y} (1) = 0

= 0 + 1 - 0

Vereinfache

= 1

\frac{d}{d x} (ln (tan (\frac{x}{2})))

x \to - 2 lim (\frac{1 - x ^{2}}{x + 2})

in v erse l a pl a ce \frac{e ^{- 5 s}}{s + 1}

x \to - 2 + lim (\frac{2 x - 1}{x + 2})

y^{^{''}} + 4 y = sin^{2} (x)