(\partial)/(\partial x)((z)(e^{xyz}-x-y))

Lösung

\frac{\partial}{\partial x} ((z) (e^{x yz} - x - y))

z (e^{z x y} zy - 1)

Schritte zur Lösung

\frac{\partial}{\partial x} ((z) (e^{x yz} - x - y))

Behandle

z, y

als Konstanten

Entferne die Konstante:

(a \cdot f)^{'} = a \cdot f^{'}

= z \frac{\partial}{\partial x} (e^{x yz} - x - y)

Wende die Summen-/Differenzregel an:

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

= z (\frac{\partial}{\partial x} (e^{x yz}) - \frac{\partial}{\partial x} (x) - \frac{\partial}{\partial x} (y))

\frac{\partial}{\partial x} (e^{x yz}) = e^{x yz} zy

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

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

= z (e^{x yz} zy - 1 - 0)

Vereinfache

= z (e^{z x y} zy - 1)

e x p an d \frac{x ^{3} - x}{( x + 2 ) ( x - 2 )}

e x p an d \frac{2 ( - 3 x ^{2} + 1 )}{( x ^{2} + 1 ) ^{3}}

e x p an d - \frac{1}{8} (x + 2)^{2} (x - 4)^{2}

s im pl i f y (1 - \frac{1}{x ^{2}}) \cdot (2 + x)

e x p an d f (12 x - 6) x^{2} d x