(x^x)^'

Lösung

(x^{x})^{^{'}}

x^{x} (ln (x) + 1)

Schritte zur Lösung

(x^{x})^{'}

Wende Exponentenregel an:

a^{b} = e^{b l n (a)}

x^{x} = e^{x l n (x)}

= (e^{x l n (x)})^{^{'}}

Wende die Kettenregel an:

e^{x l n (x)} (x ln (x))^{'}

= e^{x l n (x)} (x ln (x))^{'}

(x ln (x))^{'} = ln (x) + 1

= e^{x l n (x)} (ln (x) + 1)

e^{x l n (x)} = x^{x}

= x^{x} (ln (x) + 1)

x \to 2 - lim (ln (4 - 2 x))

\int tan^{2} (t) sec^{3} (t) d t

n = 0 \sum \infty 5^{- n}

\frac{d}{d x} (ln (1 + 8 x))

x \to 1 + lim (\frac{1}{x ^{2} + 1})