derivative f(x)=(x^6-4x^2+2)ln(x)

Lösung

ableitung von

f (x) = (x^{6} - 4 x^{2} + 2) ln (x)

ln (x) (6 x^{5} - 8 x) + \frac{x ^{6} - 4 x ^{2} + 2}{x}

Schritte zur Lösung

\frac{d}{d x} ((x^{6} - 4 x^{2} + 2) ln (x))

Wende die Produktregel an:

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

f = x^{6} - 4 x^{2} + 2, g = ln (x)

= \frac{d}{d x} (x^{6} - 4 x^{2} + 2) ln (x) + \frac{d}{d x} (ln (x)) (x^{6} - 4 x^{2} + 2)

\frac{d}{d x} (x^{6} - 4 x^{2} + 2) = 6 x^{5} - 8 x

\frac{d}{d x} (ln (x)) = \frac{1}{x}

= (6 x^{5} - 8 x) ln (x) + \frac{1}{x} (x^{6} - 4 x^{2} + 2)

\frac{1}{x} (x^{6} - 4 x^{2} + 2) = \frac{x ^{6} - 4 x ^{2} + 2}{x}

= ln (x) (6 x^{5} - 8 x) + \frac{x ^{6} - 4 x ^{2} + 2}{x}

\int \frac{2}{3} t^{\frac{3}{2}} - 8 sin (t) + C d t

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

y^{^{'}} = x y (y + 2)

n = 1 \sum \infty \frac{9}{n ^{3}}

n = 0 \sum \infty 4 (\frac{3}{4})^{n}