derivative f(x)=3x^2e^x+e^x(x^3+9)

Lösung

ableitung von

f (x) = 3 x^{2} e^{x} + e^{x} (x^{3} + 9)

e^{x} x^{3} + 6 e^{x} x^{2} + 6 e^{x} x + 9 e^{x}

Schritte zur Lösung

\frac{d}{d x} (3 x^{2} e^{x} + e^{x} (x^{3} + 9))

Wende die Summen-/Differenzregel an:

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

= \frac{d}{d x} (3 x^{2} e^{x}) + \frac{d}{d x} (e^{x} (x^{3} + 9))

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

\frac{d}{d x} (e^{x} (x^{3} + 9)) = e^{x} (x^{3} + 9) + 3 x^{2} e^{x}

= 3 (2 x e^{x} + e^{x} x^{2}) + e^{x} (x^{3} + 9) + 3 x^{2} e^{x}

Vereinfache

3 (2 x e^{x} + e^{x} x^{2}) + e^{x} (x^{3} + 9) + 3 x^{2} e^{x} : e^{x} x^{3} + 6 e^{x} x^{2} + 6 e^{x} x + 9 e^{x}

= e^{x} x^{3} + 6 e^{x} x^{2} + 6 e^{x} x + 9 e^{x}

y \frac{d y}{d x} = \frac{4}{5} x

t an g e n t f (x) = 4 sec (x), at x = \frac{7 π}{6}

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

t an g e n t y = (x + 2) (x - 1) (x - 3)

\int t sec^{2} (9 t) d t