derivative of (5x^2-9x+13/(2x+4))

Lösung

\frac{d}{d x} (\frac{5 x ^{2} - 9 x + 13}{2 x + 4})

\frac{5 x ^{2} + 20 x - 31}{2 ( x + 2 ) ^{2}}

Schritte zur Lösung

\frac{d}{d x} (\frac{5 x ^{2} - 9 x + 13}{2 x + 4})

Wende die Quotientenregel an:

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

= \frac{\frac{d}{d x} ( 5 x ^{2} - 9 x + 13 ) ( 2 x + 4 ) - \frac{d}{d x} ( 2 x + 4 ) ( 5 x ^{2} - 9 x + 13 )}{( 2 x + 4 ) ^{2}}

\frac{d}{d x} (5 x^{2} - 9 x + 13) = 10 x - 9

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

= \frac{( 10 x - 9 ) ( 2 x + 4 ) - 2 ( 5 x ^{2} - 9 x + 13 )}{( 2 x + 4 ) ^{2}}

Vereinfache

\frac{( 10 x - 9 ) ( 2 x + 4 ) - 2 ( 5 x ^{2} - 9 x + 13 )}{( 2 x + 4 ) ^{2}} : \frac{5 x ^{2} + 20 x - 31}{2 ( x + 2 ) ^{2}}

= \frac{5 x ^{2} + 20 x - 31}{2 ( x + 2 ) ^{2}}

x \to 6 lim (\frac{6 + x}{x})

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

d er i v a t i v e f (x) = tan^{2} (x^{3})

\int ln (21 t) d t

t an g e n t f (x) = 2 x^{3} + 3 x^{2} - 12 x + 4, at x = 0