derivative of ae^{2x}+be^x

Lösung

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

a e^{2 x} \cdot 2 + b e^{x}

Schritte zur Lösung

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

Wende die Summen-/Differenzregel an:

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

= \frac{d}{d x} (a e^{2 x}) + \frac{d}{d x} (b e^{x})

\frac{d}{d x} (a e^{2 x}) = a e^{2 x} \cdot 2

\frac{d}{d x} (b e^{x}) = b e^{x}

= a e^{2 x} \cdot 2 + b e^{x}

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

\int 10 sin^{3} (x) cos^{7} (x) d x

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

in v erse l a pl a ce \frac{s}{s ^{4} + 1}

d er i v a t i v e \frac{y}{x}