derivative of x/(3x-4ln(x))

Lösung

\frac{d}{d x} (\frac{x}{3 x - 4 ln ( x )})

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

Schritte zur Lösung

\frac{d}{d x} (\frac{x}{3 x - 4 ln ( x )})

Wende die Quotientenregel an:

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

= \frac{\frac{d x}{d x} ( 3 x - 4 ln ( x ) ) - \frac{d}{d x} ( 3 x - 4 ln ( x ) ) x}{( 3 x - 4 ln ( x ) ) ^{2}}

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

\frac{d}{d x} (3 x - 4 ln (x)) = 3 - \frac{4}{x}

= \frac{1 \cdot ( 3 x - 4 ln ( x ) ) - ( 3 - \frac{4}{x} ) x}{( 3 x - 4 ln ( x ) ) ^{2}}

Vereinfache

\frac{1 \cdot ( 3 x - 4 ln ( x ) ) - ( 3 - \frac{4}{x} ) x}{( 3 x - 4 ln ( x ) ) ^{2}} : \frac{- 4 ln ( x ) + 4}{( 3 x - 4 ln ( x ) ) ^{2}}

= \frac{- 4 ln ( x ) + 4}{( 3 x - 4 ln ( x ) ) ^{2}}

\frac{\partial}{\partial x} ((- 3 x y) 0)

\frac{d}{d t} (V (t))

\frac{\partial}{\partial x} (2 x cos (y))

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

y^{^{'}} = y^{2} - 6