derivative of x/(1-ln(x-9))

Lösung

\frac{d}{d x} (\frac{x}{1 - ln ( x - 9 )})

\frac{2 x - ln ( x - 9 ) ( x - 9 ) - 9}{( x - 9 ) ( 1 - ln ( x - 9 ) ) ^{2}}

Schritte zur Lösung

\frac{d}{d x} (\frac{x}{1 - ln ( x - 9 )})

Wende die Quotientenregel an:

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

= \frac{\frac{d x}{d x} ( 1 - ln ( x - 9 ) ) - \frac{d}{d x} ( 1 - ln ( x - 9 ) ) x}{( 1 - ln ( x - 9 ) ) ^{2}}

\frac{d x}{d x} = 1

\frac{d}{d x} (1 - ln (x - 9)) = - \frac{1}{x - 9}

= \frac{1 \cdot ( 1 - ln ( x - 9 ) ) - ( - \frac{1}{x - 9} ) x}{( 1 - ln ( x - 9 ) ) ^{2}}

Vereinfache

\frac{1 \cdot ( 1 - ln ( x - 9 ) ) - ( - \frac{1}{x - 9} ) x}{( 1 - ln ( x - 9 ) ) ^{2}} : \frac{2 x - ln ( x - 9 ) ( x - 9 ) - 9}{( x - 9 ) ( 1 - ln ( x - 9 ) ) ^{2}}

= \frac{2 x - ln ( x - 9 ) ( x - 9 ) - 9}{( x - 9 ) ( 1 - ln ( x - 9 ) ) ^{2}}

x y^{^{'}} - y = \frac{5}{4} x ln (x)

t an g e n t y = 8 x arccos (x - 1), at x = 1

d er i v a t i v e (t^{2} + 5)^{2} - 7 (t^{2} + 5)

2 x \frac{d y}{d x} - 8 y = x^{- 8}

a re a \frac{2}{x}, \frac{1}{2} x, 1, 2