derivative f(x)=(ln(15-23x^2))/(8x^5)

Lösung

ableitung von

f (x) = \frac{ln ( 15 - 23 x ^{2} )}{8 x ^{5}}

- \frac{46 x ^{2} + 5 ln ( - 23 x ^{2} + 15 ) ( - 23 x ^{2} + 15 )}{8 x ^{6} ( - 23 x ^{2} + 15 )}

Schritte zur Lösung

\frac{d}{d x} (\frac{ln ( 15 - 23 x ^{2} )}{8 x ^{5}})

Entferne die Konstante:

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

= \frac{1}{8} \frac{d}{d x} (\frac{ln ( 15 - 23 x ^{2} )}{x ^{5}})

Wende die Quotientenregel an:

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

= \frac{1}{8} \cdot \frac{\frac{d}{d x} ( ln ( 15 - 23 x ^{2} ) ) x ^{5} - \frac{d}{d x} ( x ^{5} ) ln ( 15 - 23 x ^{2} )}{( x ^{5} ) ^{2}}

\frac{d}{d x} (ln (15 - 23 x^{2})) = - \frac{46 x}{15 - 23 x ^{2}}

\frac{d}{d x} (x^{5}) = 5 x^{4}

= \frac{1}{8} \cdot \frac{( - \frac{46 x}{15 - 23 x ^{2}} ) x ^{5} - 5 x ^{4} ln ( 15 - 23 x ^{2} )}{( x ^{5} ) ^{2}}

Vereinfache

\frac{1}{8} \cdot \frac{( - \frac{46 x}{15 - 23 x ^{2}} ) x ^{5} - 5 x ^{4} ln ( 15 - 23 x ^{2} )}{( x ^{5} ) ^{2}} : - \frac{46 x ^{2} + 5 ln ( - 23 x ^{2} + 15 ) ( - 23 x ^{2} + 15 )}{8 x ^{6} ( - 23 x ^{2} + 15 )}

= - \frac{46 x ^{2} + 5 ln ( - 23 x ^{2} + 15 ) ( - 23 x ^{2} + 15 )}{8 x ^{6} ( - 23 x ^{2} + 15 )}

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