derivative of (ln(-2x^3+5x^2)/(-x))

Lösung

\frac{d}{d x} (\frac{ln ( - 2 x ^{3} + 5 x ^{2} )}{- x})

- \frac{2 ( - 3 x + 5 ) - ln ( - 2 x ^{3} + 5 x ^{2} ) ( - 2 x + 5 )}{x ^{2} ( - 2 x + 5 )}

Schritte zur Lösung

\frac{d}{d x} (\frac{ln ( - 2 x ^{3} + 5 x ^{2} )}{- x})

Vereinfache

\frac{ln ( - 2 x ^{3} + 5 x ^{2} )}{- x} : - \frac{ln ( - 2 x ^{3} + 5 x ^{2} )}{x}

= \frac{d}{d x} (- \frac{ln ( - 2 x ^{3} + 5 x ^{2} )}{x})

Entferne die Konstante:

(a \cdot f)^{'} = a \cdot f^{'}

= - \frac{d}{d x} (\frac{ln ( - 2 x ^{3} + 5 x ^{2} )}{x})

Wende die Quotientenregel an:

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

= - \frac{\frac{d}{d x} ( ln ( - 2 x ^{3} + 5 x ^{2} ) ) x - \frac{d x}{d x} ln ( - 2 x ^{3} + 5 x ^{2} )}{x ^{2}}

\frac{d}{d x} (ln (- 2 x^{3} + 5 x^{2})) = \frac{2 ( - 3 x + 5 )}{x ( - 2 x + 5 )}

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

= - \frac{\frac{2 ( - 3 x + 5 )}{x ( - 2 x + 5 )} x - 1 \cdot ln ( - 2 x ^{3} + 5 x ^{2} )}{x ^{2}}

Vereinfache

- \frac{\frac{2 ( - 3 x + 5 )}{x ( - 2 x + 5 )} x - 1 \cdot ln ( - 2 x ^{3} + 5 x ^{2} )}{x ^{2}} : - \frac{2 ( - 3 x + 5 ) - ln ( - 2 x ^{3} + 5 x ^{2} ) ( - 2 x + 5 )}{x ^{2} ( - 2 x + 5 )}

= - \frac{2 ( - 3 x + 5 ) - ln ( - 2 x ^{3} + 5 x ^{2} ) ( - 2 x + 5 )}{x ^{2} ( - 2 x + 5 )}

f (x) = \frac{1}{1 + x}

x \to 1 lim (x^{3} - x + 2)

2 y^{^{''}} + y^{^{'}} - 4 y = 0

\int \frac{4}{3} π r^{3} d r

\frac{d}{d x} ((x + 1)^{3} - x^{3})