derivative of log_{5}((6x^3+4/(2x^3-3)))

解答

\frac{d}{d x} (lo g_{5} (\frac{6 x ^{3} + 4}{2 x ^{3} - 3}))

- \frac{39 x ^{2}}{ln ( 5 ) ( 3 x ^{3} + 2 ) ( 2 x ^{3} - 3 )}

求解步骤

\frac{d}{d x} (lo g_{5} (\frac{6 x ^{3} + 4}{2 x ^{3} - 3}))

使用对数计算法则:

lo g_{a} (b) = \frac{ln ( b )}{ln ( a )}

= \frac{d}{d x} \frac{ln ( \frac{6 x ^{3} + 4}{2 x ^{3} - 3} )}{ln ( 5 )}

将常数提出:

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

= \frac{1}{ln ( 5 )} \frac{d}{d x} (ln (\frac{6 x ^{3} + 4}{2 x ^{3} - 3}))

使用连锁律:

\frac{1}{\frac{6 x ^{3} + 4}{2 x ^{3} - 3}} \frac{d}{d x} (\frac{6 x ^{3} + 4}{2 x ^{3} - 3})

= \frac{1}{\frac{6 x ^{3} + 4}{2 x ^{3} - 3}} \frac{d}{d x} (\frac{6 x ^{3} + 4}{2 x ^{3} - 3})

\frac{d}{d x} (\frac{6 x ^{3} + 4}{2 x ^{3} - 3}) = - \frac{78 x ^{2}}{( 2 x ^{3} - 3 ) ^{2}}

= \frac{1}{ln ( 5 )} \cdot \frac{1}{\frac{6 x ^{3} + 4}{2 x ^{3} - 3}} (- \frac{78 x ^{2}}{( 2 x ^{3} - 3 ) ^{2}})

化简

\frac{1}{ln ( 5 )} \cdot \frac{1}{\frac{6 x ^{3} + 4}{2 x ^{3} - 3}} (- \frac{78 x ^{2}}{( 2 x ^{3} - 3 ) ^{2}}) : - \frac{39 x ^{2}}{ln ( 5 ) ( 3 x ^{3} + 2 ) ( 2 x ^{3} - 3 )}

= - \frac{39 x ^{2}}{ln ( 5 ) ( 3 x ^{3} + 2 ) ( 2 x ^{3} - 3 )}

s l o p e in t erce pt (4, - 3), (12, 0)

\int_{2}^{3} \frac{25}{3 - x} d x

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