derivative of 1/(x+\frac{1{x+1}})

Lösung

\frac{d}{d x} (\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{\partial}{\partial x} (ln (sin (x y)))

l a pl a ce t r an s f or m sin^{2} (4 t)

\frac{\partial}{\partial x} (2 \frac{y}{x})

x \to 0 lim (e^{4 x})

l a pl a ce t r an s f or m 2 x^{2} - 3 x + 4