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

Lösung

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

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

Schritte zur Lösung

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

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

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

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

Vereinfache

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

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

in v erse l a pl a ce \frac{16}{( s - 1 ) ( s + 1 )}

in v erse l a pl a ce \frac{2}{s ^{2} ( s ^{2} + 25 )}

\frac{d}{d x} (5 x^{7} sin (x))

\frac{d y}{d t} = 0.3 y + 26 e^{0.3 t}

d er i v a t i v e \frac{8 x ^{2} + 4 x + 8}{x}