derivative of-(ln(1+6x)/x)

Lösung

\frac{d}{d x} (- \frac{ln ( 1 + 6 x )}{x})

- \frac{6 x - ln ( 1 + 6 x ) ( 1 + 6 x )}{x ^{2} ( 1 + 6 x )}

Schritte zur Lösung

\frac{d}{d x} (- \frac{ln ( 1 + 6 x )}{x})

Entferne die Konstante:

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

= - \frac{d}{d x} (\frac{ln ( 1 + 6 x )}{x})

Wende die Quotientenregel an:

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

= - \frac{\frac{d}{d x} ( ln ( 1 + 6 x ) ) x - \frac{d x}{d x} ln ( 1 + 6 x )}{x ^{2}}

\frac{d}{d x} (ln (1 + 6 x)) = \frac{6}{1 + 6 x}

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

= - \frac{\frac{6}{1 + 6 x} x - 1 \cdot ln ( 1 + 6 x )}{x ^{2}}

Vereinfache

- \frac{\frac{6}{1 + 6 x} x - 1 \cdot ln ( 1 + 6 x )}{x ^{2}} : - \frac{6 x - ln ( 1 + 6 x ) ( 1 + 6 x )}{x ^{2} ( 1 + 6 x )}

= - \frac{6 x - ln ( 1 + 6 x ) ( 1 + 6 x )}{x ^{2} ( 1 + 6 x )}

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

x \frac{d y}{d x} = 6 y

t an g e n t f (x) = 2 x^{2} + x - 3, at x = - 1

y^{^{'}} = (t + y)^{2}

a re a 2 x, 3 x^{2}, 0, 1