derivative f(x)=-x^3+2x-e^{-x}-3

Lösung

ableitung von

f (x) = - x^{3} + 2 x - e^{- x} - 3

- 3 x^{2} + e^{- x} + 2

Schritte zur Lösung

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

Wende die Summen-/Differenzregel an:

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

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

\frac{d}{d x} (x^{3}) = 3 x^{2}

\frac{d}{d x} (2 x) = 2

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

\frac{d}{d x} (3) = 0

= - 3 x^{2} + 2 - (- e^{- x}) - 0

Vereinfache

= - 3 x^{2} + e^{- x} + 2

in v erse l a pl a ce \frac{3}{\frac{s + 2}{s}}

\frac{\partial}{\partial x} (2 x^{4} + 5 y^{2} + x y)

\int \frac{3}{x} + 10 sec^{2} (x) + 40 e^{x} d x

\frac{d}{d x} (e^{x y} cos (x + y))

y^{^{'}} - \frac{3}{x ^{2}} y = \frac{1}{x ^{2}}