(\partial)/(\partial x)(e^{2xyz-1})

Lösung

\frac{\partial}{\partial x} (e^{2 x yz - 1})

e^{2 x yz - 1} \cdot 2 yz

Schritte zur Lösung

\frac{\partial}{\partial x} (e^{2 x yz - 1})

Behandle

y, z

als Konstanten

Wende die Kettenregel an:

e^{2 x yz - 1} \frac{\partial}{\partial x} (2 x yz - 1)

= e^{2 x yz - 1} \frac{\partial}{\partial x} (2 x yz - 1)

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

= e^{2 x yz - 1} \cdot 2 yz

\frac{d}{d x} (e^{2 x} + 4)

\int e^{x} e^{x} + 5 d x

t an g e n t f (x) = \frac{1}{2 x}, at x = 9

d er i v a t i v e f (x) = (x^{4} - \frac{5}{x ^{3}})^{- 4}

\int_{0}^{3} \frac{x}{x + 1} d x