derivative of (2x-1/(2x+1))

Lösung

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

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

Schritte zur Lösung

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

Wende die Quotientenregel an:

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

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

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

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

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

Multipliziere aus

2 (2 x + 1) - 2 (2 x - 1) : 4

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

t an g e n t f (x) = 9 x^{2} + 94, at x = 6

\frac{\partial}{\partial x} (\frac{3}{x} - 1 - y)

a re a 4 x + 8, x^{2} + 19 x + 62, [- 9, - 6]

\frac{d}{d x} ((8 x - 2 x^{2}) cos (y))

d er i v a t i v e y = (x^{3} + 8) e^{x}