derivative f(x)=((sqrt(x)-2))/((2x^2+3))

Lösung

ableitung von

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

\frac{- 6 x ^{2} + 16 x x + 3}{2 x ( 2 x ^{2} + 3 ) ^{2}}

Schritte zur Lösung

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

Wende die Quotientenregel an:

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

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

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

\frac{d}{d x} (2 x^{2} + 3) = 4 x

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

Vereinfache

\frac{\frac{1}{2 x} ( 2 x ^{2} + 3 ) - 4 x ( x - 2 )}{( 2 x ^{2} + 3 ) ^{2}} : \frac{- 6 x ^{2} + 16 x x + 3}{2 x ( 2 x ^{2} + 3 ) ^{2}}

= \frac{- 6 x ^{2} + 16 x x + 3}{2 x ( 2 x ^{2} + 3 ) ^{2}}

\int_{2}^{3} \frac{1}{x ^{2} - 4 x} d x

d er i v a t i v e 3 x^{4} - 5 x + 2 3 x

\frac{d}{d x} (ln (1 - 6 x))

t an g e n t - x^{2} + 3 x + 4, at x = 1

\frac{d y}{d x} = \frac{- e ^{x}}{8}