What is the derivative of ln((3x+1/(3x-1))) ?

The derivative of ln((3x+1/(3x-1))) is -6/((3x+1)(3x-1))

What is the first derivative of ln((3x+1/(3x-1))) ?

The first derivative of ln((3x+1/(3x-1))) is -6/((3x+1)(3x-1))

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\frac{d}{d x} (ln (\frac{3 x + 1}{3 x - 1}))

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

Solution steps

\frac{d}{d x} (ln (\frac{3 x + 1}{3 x - 1}))

Apply the chain rule:

\frac{1}{\frac{3 x + 1}{3 x - 1}} \frac{d}{d x} (\frac{3 x + 1}{3 x - 1})

= \frac{1}{\frac{3 x + 1}{3 x - 1}} \frac{d}{d x} (\frac{3 x + 1}{3 x - 1})

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

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

Simplify

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

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

8 x + x

\int \frac{x ^{6} - x}{x ^{3}} d x

\int \frac{x + 2}{x ^{2} + 7 x - 8} d x

2.87 y^{^{''}} + 10.5 y^{^{'}} - 1.83 y = 0

y^{^{'}} = y^{2} (1 + e^{x})

What is the derivative of ln((3x+1/(3x-1))) ?
The derivative of ln((3x+1/(3x-1))) is -6/((3x+1)(3x-1))
What is the first derivative of ln((3x+1/(3x-1))) ?
The first derivative of ln((3x+1/(3x-1))) is -6/((3x+1)(3x-1))