derivative of x^{x+1}

Lösung

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

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

Schritte zur Lösung

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

Wende Exponentenregel an:

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

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

= \frac{d}{d x} (e^{(x + 1) l n (x)})

Wende die Kettenregel an:

e^{(x + 1) l n (x)} \frac{d}{d x} ((x + 1) ln (x))

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

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

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

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

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

s l o p e (- 20, - 4), (- 12, - 10)

n = 1 \sum \infty \frac{2}{25} \cdot (\frac{1}{5})^{n}

\frac{d y}{d x} - 2 x y^{2} = 0

4 y^{^{''}} + 20 y^{^{'}} + 25 y = 0

\int e^{x} 8 - e^{x} d x