derivative of arccos((1-x/(1+x)))

Lösung

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

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

Schritte zur Lösung

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

Wende die Kettenregel an:

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

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

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

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

Vereinfache

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

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

\int \frac{3 x ^{2}}{3 x ^{3} + 1} d x

x \cdot \frac{d y}{d x} - 2 \cdot y = 3 \cdot x^{2} + 2 \cdot x

\frac{\partial}{\partial y} (arctan (y^{3} x))

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

d er i v a t i v e y = sin (arccos (t))