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

Lösung

ableitung von

(x + 1)^{x + 1}

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

Schritte zur Lösung

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

Wende Exponentenregel an:

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

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

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

Wende die Kettenregel an:

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

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

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

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

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

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

f (θ) = e^{θ}

\frac{d}{d x} (cos (e^{4 x}))

d er i v a t i v e f (x) = \frac{7 x ^{2} - 6 x + 1}{3 x + 2}

\frac{d ^{2}}{d x ^{2}} (3 + 5 x + 2 x^{2} + 0.4 x^{3})

y^{^{'}} = 4 x y