derivative of (4x-sqrt(1+x)/(x^2-1))

Lösung

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

\frac{- 8 x ^{2} + 3 x x + 1 + x + 1 - 8}{2 ( x ^{2} - 1 ) ^{2}}

Schritte zur Lösung

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

Wende die Quotientenregel an:

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

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

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

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

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

Vereinfache

\frac{( 4 - \frac{1}{2 1 + x} ) ( x ^{2} - 1 ) - 2 x ( 4 x - 1 + x )}{( x ^{2} - 1 ) ^{2}} : \frac{- 8 x ^{2} + 3 x x + 1 + x + 1 - 8}{2 ( x ^{2} - 1 ) ^{2}}

= \frac{- 8 x ^{2} + 3 x x + 1 + x + 1 - 8}{2 ( x ^{2} - 1 ) ^{2}}

n = 1 \sum \infty \frac{n}{n ^{3} + 5}

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

s im pl i f y (\frac{sin ( x )}{1 + cos ( x )})^{2}

d er i v a t i v e f (x) = (x^{2} \cdot 3^{x})^{7}

t y^{^{'}} + 2 y = 4 t^{2}, y (1) = 2