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

Lösung

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

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

Schritte zur Lösung

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

Wende Exponentenregel an:

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

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

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

Wende die Kettenregel an:

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

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

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

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

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

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

y^{^{''}} + 2 y^{^{'}} = 2 x + 7 - e^{- 2 x}

\int_{- 1}^{1} (x^{2} - 1)^{2} d x

\int \frac{x - 1}{x} d x

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

\frac{d}{d x} (e^{x} (3 x + 1)^{3})