derivative of (Ax+Be^x)

Lösung

\frac{d}{d x} ((A x + B) e^{x})

A e^{x} + e^{x} (A x + B)

Schritte zur Lösung

\frac{d}{d x} ((A x + B) e^{x})

Behandle

A, B

als Konstanten

Wende die Produktregel an:

(f \cdot g)^{'} = f^{'} \cdot g + f \cdot g^{'}

f = A x + B, g = e^{x}

= \frac{d}{d x} (A x + B) e^{x} + \frac{d}{d x} (e^{x}) (A x + B)

\frac{d}{d x} (A x + B) = A

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

= A e^{x} + e^{x} (A x + B)

y^{^{'}} + y = e^{9 t}, y (0) = - 1

x \to 2 lim (\frac{x + 3}{x + 2})

\frac{\partial}{\partial y} (x y^{3} + 4 y e^{x} - 2 y^{2})

\frac{\partial}{\partial y} (ln ((\frac{y}{x})^{2}))

t an g e n t f (x) = - 3 x^{3} - 2 x^{2} - x + 3, at x = - 1