derivative of c_{1}e^{-4x}+c_{2}xe^{-4x}

Lösung

\frac{d}{d x} (c_{1} e^{- 4 x} + c_{2} x e^{- 4 x})

- 4 c_{1} e^{- 4 x} + c_{2} (e^{- 4 x} - 4 e^{- 4 x} x)

Schritte zur Lösung

\frac{d}{d x} (c_{1} e^{- 4 x} + c_{2} x e^{- 4 x})

Wende die Summen-/Differenzregel an:

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

= \frac{d}{d x} (c_{1} e^{- 4 x}) + \frac{d}{d x} (c_{2} x e^{- 4 x})

\frac{d}{d x} (c_{1} e^{- 4 x}) = - 4 c_{1} e^{- 4 x}

\frac{d}{d x} (c_{2} x e^{- 4 x}) = c_{2} (e^{- 4 x} - 4 e^{- 4 x} x)

= - 4 c_{1} e^{- 4 x} + c_{2} (e^{- 4 x} - 4 e^{- 4 x} x)

\frac{\partial}{\partial y} (x^{4} + y^{4})

\int \frac{1}{3} sin (x) d x

\int 21 tan^{3} (θ) d θ

y^{^{'}} - \frac{4}{x} y = 4 x^{5}

\frac{d}{d x} (\frac{x}{csch ( x )})