derivative of (8x^2-2x+1/(x^2-5x))

Lösung

\frac{d}{d x} (\frac{8 x ^{2} - 2 x + 1}{x ^{2} - 5 x})

\frac{- 38 x ^{2} - 2 x + 5}{( x ^{2} - 5 x ) ^{2}}

Schritte zur Lösung

\frac{d}{d x} (\frac{8 x ^{2} - 2 x + 1}{x ^{2} - 5 x})

Wende die Quotientenregel an:

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

= \frac{\frac{d}{d x} ( 8 x ^{2} - 2 x + 1 ) ( x ^{2} - 5 x ) - \frac{d}{d x} ( x ^{2} - 5 x ) ( 8 x ^{2} - 2 x + 1 )}{( x ^{2} - 5 x ) ^{2}}

\frac{d}{d x} (8 x^{2} - 2 x + 1) = 16 x - 2

\frac{d}{d x} (x^{2} - 5 x) = 2 x - 5

= \frac{( 16 x - 2 ) ( x ^{2} - 5 x ) - ( 2 x - 5 ) ( 8 x ^{2} - 2 x + 1 )}{( x ^{2} - 5 x ) ^{2}}

Multipliziere aus

(16 x - 2) (x^{2} - 5 x) - (2 x - 5) (8 x^{2} - 2 x + 1) : - 38 x^{2} - 2 x + 5

= \frac{- 38 x ^{2} - 2 x + 5}{( x ^{2} - 5 x ) ^{2}}

\int csc (y) d y

t an g e n t y = \frac{8 x}{( x ^{2} + 1 )}

e x p an d (2 x - 3)^{4}

\frac{d}{d x} (e^{x} ln (x^{5}))

x^{^{'}} + x = e^{t}