derivative y= 1/(x+\frac{1){x+1}}

Lösung

ableitung von

y = \frac{1}{x + \frac{1}{x + 1}}

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

Schritte zur Lösung

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

Wende Exponentenregel an:

\frac{1}{a} = a^{- 1}

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

Wende die Kettenregel an:

- \frac{1}{( x + \frac{1}{x + 1} ) ^{2}} \frac{d}{d x} (x + \frac{1}{x + 1})

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

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

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

Vereinfache

- \frac{1}{( x + \frac{1}{x + 1} ) ^{2}} (1 - \frac{1}{( x + 1 ) ^{2}}) : - \frac{x ^{2} + 2 x}{( x ( x + 1 ) + 1 ) ^{2}}

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

\frac{d}{d x} ((3 x)^{- 4})

d er i v a t i v e 8 e^{x} + x

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\int \frac{1}{64 e ^{- 2 x} + e ^{2 x}} d x

\int sin (3 t - 3 x) d x