derivative of arctan(x/(1+x))

Lösung

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

\frac{1}{2 x ^{2} + 2 x + 1}

Schritte zur Lösung

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

Wende die Kettenregel an:

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

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

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

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

Vereinfache

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

= \frac{1}{2 x ^{2} + 2 x + 1}

\int \frac{( 4 x ^{3} - 3 x ^{2} - 8 x )}{6 x} d x

x \to 0 lim (x^{2} sin (\frac{1}{x}))

\frac{d}{d x} ((5 x^{6} + 2 x^{3})^{4})

\int_{0}^{l n (3)} \frac{4 e ^{x}}{1 + e ^{2 x}} d x

d er i v a t i v e x (9^{- 2 x})