derivative of e^{(2x}(x^3+6x-1)ln(x))

Lösung

\frac{d}{d x} (e^{(2 x)} (x^{3} + 6 x - 1) ln (x))

2 e^{2 x} x^{3} ln (x) + 3 e^{2 x} x^{2} ln (x) + 12 e^{2 x} x ln (x) + 4 e^{2 x} ln (x) + \frac{e ^{2 x} ( x ^{3} + 6 x - 1 )}{x}

Schritte zur Lösung

\frac{d}{d x} (e^{(2 x)} (x^{3} + 6 x - 1) ln (x))

Wende die Produktregel an:

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

f = e^{2 x}, g = (x^{3} + 6 x - 1) ln (x)

= \frac{d}{d x} (e^{2 x}) (x^{3} + 6 x - 1) ln (x) + \frac{d}{d x} ((x^{3} + 6 x - 1) ln (x)) e^{2 x}

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

\frac{d}{d x} ((x^{3} + 6 x - 1) ln (x)) = ln (x) (3 x^{2} + 6) + \frac{x ^{3} + 6 x - 1}{x}

= e^{2 x} \cdot 2 (x^{3} + 6 x - 1) ln (x) + (ln (x) (3 x^{2} + 6) + \frac{x ^{3} + 6 x - 1}{x}) e^{2 x}

Vereinfache

e^{2 x} \cdot 2 (x^{3} + 6 x - 1) ln (x) + (ln (x) (3 x^{2} + 6) + \frac{x ^{3} + 6 x - 1}{x}) e^{2 x} : 2 e^{2 x} x^{3} ln (x) + 3 e^{2 x} x^{2} ln (x) + 12 e^{2 x} x ln (x) + 4 e^{2 x} ln (x) + \frac{e ^{2 x} ( x ^{3} + 6 x - 1 )}{x}

= 2 e^{2 x} x^{3} ln (x) + 3 e^{2 x} x^{2} ln (x) + 12 e^{2 x} x ln (x) + 4 e^{2 x} ln (x) + \frac{e ^{2 x} ( x ^{3} + 6 x - 1 )}{x}

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

d er i v a t i v e \frac{x}{8}

in v erse l a pl a ce {\frac{( s - 1 )}{s + 1}}

\frac{\partial}{\partial y} (e^{x y} + x^{2} z)

d er i v a t i v e \frac{x ^{4}}{1 - x ^{3}}