derivative f(x)=9x^8e^x+e^xx^9

Lösung

ableitung von

f (x) = 9 x^{8} e^{x} + e^{x} x^{9}

e^{x} x^{9} + 18 e^{x} x^{8} + 72 e^{x} x^{7}

Schritte zur Lösung

\frac{d}{d x} (9 x^{8} e^{x} + e^{x} x^{9})

Wende die Summen-/Differenzregel an:

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

= \frac{d}{d x} (9 x^{8} e^{x}) + \frac{d}{d x} (e^{x} x^{9})

\frac{d}{d x} (9 x^{8} e^{x}) = 9 (8 x^{7} e^{x} + e^{x} x^{8})

\frac{d}{d x} (e^{x} x^{9}) = e^{x} x^{9} + 9 x^{8} e^{x}

= 9 (8 x^{7} e^{x} + e^{x} x^{8}) + e^{x} x^{9} + 9 x^{8} e^{x}

Vereinfache

9 (8 x^{7} e^{x} + e^{x} x^{8}) + e^{x} x^{9} + 9 x^{8} e^{x} : e^{x} x^{9} + 18 e^{x} x^{8} + 72 e^{x} x^{7}

= e^{x} x^{9} + 18 e^{x} x^{8} + 72 e^{x} x^{7}

\int_{0}^{1} x^{2} ln (x) d x

\int_{0}^{N} e^{- s t} e^{- 3 t} d t

\frac{d}{d x} (\frac{1}{0})

12 x^{2} y + y^{^{'}} = 5 x^{2}

d er i v a t i v e f (x) = cos (2 x)