derivative of (arccos(x)/((x+1)^2))

Lösung

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

\frac{- x ^{2} - 2 x - 1 - 2 x arccos ( x ) 1 - x ^{2} - 2 arccos ( x ) 1 - x ^{2}}{( x + 1 ) ^{4} 1 - x ^{2}}

Schritte zur Lösung

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

Wende die Quotientenregel an:

(\frac{f}{g})^{'} = \frac{f ^{'} \cdot g - g ^{'} \cdot f}{g ^{2}}

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

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

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

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

Vereinfache

\frac{( - \frac{1}{1 - x ^{2}} ) ( x + 1 ) ^{2} - 2 ( x + 1 ) arccos ( x )}{( ( x + 1 ) ^{2} ) ^{2}} : \frac{- x ^{2} - 2 x - 1 - 2 x arccos ( x ) 1 - x ^{2} - 2 arccos ( x ) 1 - x ^{2}}{( x + 1 ) ^{4} 1 - x ^{2}}

= \frac{- x ^{2} - 2 x - 1 - 2 x arccos ( x ) 1 - x ^{2} - 2 arccos ( x ) 1 - x ^{2}}{( x + 1 ) ^{4} 1 - x ^{2}}

\frac{d}{d x} (e^{g (x)} sin (x))

\int_{- \infty}^{12} x e^{\frac{x}{4}} d x

im pl i c i t \frac{d y}{d x}, 3 x^{2} + 2 y^{2} = 30

\frac{\partial}{\partial x} (x^{3} + e^{y})

\frac{d}{d x} (ln (\frac{x}{6}))