derivative (10x^3-6x^2+3)/(4x^2-6x+3)

Lösung

ableitung von

\frac{10 x ^{3} - 6 x ^{2} + 3}{4 x ^{2} - 6 x + 3}

\frac{40 x ^{4} - 120 x ^{3} + 126 x ^{2} - 60 x + 18}{( 4 x ^{2} - 6 x + 3 ) ^{2}}

Schritte zur Lösung

\frac{d}{d x} (\frac{10 x ^{3} - 6 x ^{2} + 3}{4 x ^{2} - 6 x + 3})

Wende die Quotientenregel an:

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

= \frac{\frac{d}{d x} ( 10 x ^{3} - 6 x ^{2} + 3 ) ( 4 x ^{2} - 6 x + 3 ) - \frac{d}{d x} ( 4 x ^{2} - 6 x + 3 ) ( 10 x ^{3} - 6 x ^{2} + 3 )}{( 4 x ^{2} - 6 x + 3 ) ^{2}}

\frac{d}{d x} (10 x^{3} - 6 x^{2} + 3) = 30 x^{2} - 12 x

\frac{d}{d x} (4 x^{2} - 6 x + 3) = 8 x - 6

= \frac{( 30 x ^{2} - 12 x ) ( 4 x ^{2} - 6 x + 3 ) - ( 8 x - 6 ) ( 10 x ^{3} - 6 x ^{2} + 3 )}{( 4 x ^{2} - 6 x + 3 ) ^{2}}

Multipliziere aus

(30 x^{2} - 12 x) (4 x^{2} - 6 x + 3) - (8 x - 6) (10 x^{3} - 6 x^{2} + 3) : 40 x^{4} - 120 x^{3} + 126 x^{2} - 60 x + 18

= \frac{40 x ^{4} - 120 x ^{3} + 126 x ^{2} - 60 x + 18}{( 4 x ^{2} - 6 x + 3 ) ^{2}}

x \to \infty lim (- \frac{1}{3} c e^{- 3 (x)})

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

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

a re a 4 x, x^{2}

\frac{d y}{d x} + y^{5} x + 5 y = 0