derivative of ln(\sqrt[4]{(6x+1/(6-x)}))

Lösung

\frac{d}{d x} (ln (4 \frac{6 x + 1}{6 - x}))

\frac{37}{4 ( - x + 6 ) 4 6 x + 1 ( 6 x + 1 ) ^{\frac{3}{4}}}

Schritte zur Lösung

\frac{d}{d x} (ln (4 \frac{6 x + 1}{6 - x}))

Wende die Kettenregel an:

\frac{1}{4 \frac{6 x + 1}{6 - x}} \frac{d}{d x} (4 \frac{6 x + 1}{6 - x})

= \frac{1}{4 \frac{6 x + 1}{6 - x}} \frac{d}{d x} (4 \frac{6 x + 1}{6 - x})

\frac{d}{d x} (4 \frac{6 x + 1}{6 - x}) = \frac{37}{4 ( 6 x + 1 ) ^{\frac{3}{4}} ( - x + 6 ) ^{\frac{5}{4}}}

= \frac{1}{4 \frac{6 x + 1}{6 - x}} \cdot \frac{37}{4 ( 6 x + 1 ) ^{\frac{3}{4}} ( - x + 6 ) ^{\frac{5}{4}}}

Vereinfache

\frac{1}{4 \frac{6 x + 1}{6 - x}} \cdot \frac{37}{4 ( 6 x + 1 ) ^{\frac{3}{4}} ( - x + 6 ) ^{\frac{5}{4}}} : \frac{37}{4 ( - x + 6 ) 4 6 x + 1 ( 6 x + 1 ) ^{\frac{3}{4}}}

= \frac{37}{4 ( - x + 6 ) 4 6 x + 1 ( 6 x + 1 ) ^{\frac{3}{4}}}

\frac{d}{d x} (\frac{0.001 x}{0.001 x + 1})

d er i v a t i v e y = x^{2} sin (x)

(x^{8})^{^{'}}

\int sec^{2} (x) - 8 d x

t a y l or (x - π)^{3}