derivative ln((6x^2+5)/(x^{4/5)})

Lösung

ableitung von

ln (\frac{6 x ^{2} + 5}{x ^{\frac{4}{5}}})

\frac{36 x ^{2} - 20}{5 x ( 6 x ^{2} + 5 )}

Schritte zur Lösung

\frac{d}{d x} (ln (\frac{6 x ^{2} + 5}{x ^{\frac{4}{5}}}))

Wende die Kettenregel an:

\frac{1}{\frac{6 x ^{2} + 5}{x ^{\frac{4}{5}}}} \frac{d}{d x} (\frac{6 x ^{2} + 5}{x ^{\frac{4}{5}}})

= \frac{1}{\frac{6 x ^{2} + 5}{x ^{\frac{4}{5}}}} \frac{d}{d x} (\frac{6 x ^{2} + 5}{x ^{\frac{4}{5}}})

\frac{d}{d x} (\frac{6 x ^{2} + 5}{x ^{\frac{4}{5}}}) = \frac{36 x ^{2} - 20}{5 x ^{\frac{9}{5}}}

= \frac{1}{\frac{6 x ^{2} + 5}{x ^{\frac{4}{5}}}} \cdot \frac{36 x ^{2} - 20}{5 x ^{\frac{9}{5}}}

Vereinfache

\frac{1}{\frac{6 x ^{2} + 5}{x ^{\frac{4}{5}}}} \cdot \frac{36 x ^{2} - 20}{5 x ^{\frac{9}{5}}} : \frac{36 x ^{2} - 20}{5 x ( 6 x ^{2} + 5 )}

= \frac{36 x ^{2} - 20}{5 x ( 6 x ^{2} + 5 )}

in v erse l a pl a ce \frac{s + 4}{s ^{2} + 2 s - 3}

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y^{^{''}} - \frac{2}{x} y^{^{'}} - \frac{10}{x ^{2}} y = 0

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