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

Lösung

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

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

Vereinfache

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

\int_{0}^{5} f (t) d t

d er i v a t i v e - π sin (π x)

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

x \to 6 + lim (f (x))

y^{^{'}} = (x y^{^{'}} + y)