derivative f(x)=(e^x+x^pi)(x^2+x+1)

Lösung

ableitung von

f (x) = (e^{x} + x^{π}) (x^{2} + x + 1)

e^{x} x^{2} + 3 e^{x} x + π x^{π} + x^{π} + π x^{π + 1} + π x^{π - 1} + 2 x^{π + 1} + 2 e^{x}

Schritte zur Lösung

\frac{d}{d x} ((e^{x} + x^{π}) (x^{2} + x + 1))

Wende die Produktregel an:

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

f = e^{x} + x^{π}, g = x^{2} + x + 1

= \frac{d}{d x} (e^{x} + x^{π}) (x^{2} + x + 1) + \frac{d}{d x} (x^{2} + x + 1) (e^{x} + x^{π})

\frac{d}{d x} (e^{x} + x^{π}) = e^{x} + π x^{π - 1}

\frac{d}{d x} (x^{2} + x + 1) = 2 x + 1

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

Vereinfache

(e^{x} + π x^{π - 1}) (x^{2} + x + 1) + (2 x + 1) (e^{x} + x^{π}) : e^{x} x^{2} + 3 e^{x} x + π x^{π} + x^{π} + π x^{π + 1} + π x^{π - 1} + 2 x^{π + 1} + 2 e^{x}

= e^{x} x^{2} + 3 e^{x} x + π x^{π} + x^{π} + π x^{π + 1} + π x^{π - 1} + 2 x^{π + 1} + 2 e^{x}

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