derivative of (3x-1/(2x+1))

Lösung

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

\frac{5}{( 2 x + 1 ) ^{2}}

Schritte zur Lösung

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

\frac{d}{d x} (3 x - 1) = 3

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

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

Multipliziere aus

3 (2 x + 1) - 2 (3 x - 1) : 5

= \frac{5}{( 2 x + 1 ) ^{2}}

y^{^{'}} + 4 x^{- 1} y = x^{2}, y (1) = - 2

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

t a y l or \frac{1}{x + 2}, 0

k = 2 \sum \infty \frac{1}{k} - \frac{1}{4 ^{k}}

(1 + ln (x) + \frac{y}{x}) d x - (1 - ln (x)) d y = 0