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

Lösung

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

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

Schritte zur Lösung

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

Wende die Quotientenregel an:

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

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

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

\frac{d}{d x} (1 + 3 x^{2}) = 6 x

= \frac{( 2 x - 1 ) ( 1 + 3 x ^{2} ) - 6 x ( x ^{2} - x )}{( 1 + 3 x ^{2} ) ^{2}}

Multipliziere aus

(2 x - 1) (1 + 3 x^{2}) - 6 x (x^{2} - x) : 3 x^{2} + 2 x - 1

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

k = 0 \sum \infty k! 2^{k} x^{k}

t an g e n t f (x) = - 3 cos (x), at x = \frac{3}{4} π

\int_{0}^{1} (\frac{4 t ^{3}}{1 + t ^{4}}) d t

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

s l o p e (0, - 3), (2, 0)