What is the derivative of ln((2x+1/(2x-1))) ?

The derivative of ln((2x+1/(2x-1))) is -4/((2x+1)(2x-1))

What is the first derivative of ln((2x+1/(2x-1))) ?

The first derivative of ln((2x+1/(2x-1))) is -4/((2x+1)(2x-1))

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\frac{d}{d x} (ln (\frac{2 x + 1}{2 x - 1}))

- \frac{4}{( 2 x + 1 ) ( 2 x - 1 )}

Solution steps

\frac{d}{d x} (ln (\frac{2 x + 1}{2 x - 1}))

Apply the chain rule:

\frac{1}{\frac{2 x + 1}{2 x - 1}} \frac{d}{d x} (\frac{2 x + 1}{2 x - 1})

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

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

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

Simplify

\frac{1}{\frac{2 x + 1}{2 x - 1}} (- \frac{4}{( 2 x - 1 ) ^{2}}) : - \frac{4}{( 2 x + 1 ) ( 2 x - 1 )}

= - \frac{4}{( 2 x + 1 ) ( 2 x - 1 )}

x \to a lim (1 - 1 - x^{2})

\frac{d y}{d x} = e^{(6 y + 3 x)}

\int (x^{4} - 2 x) d x

\frac{d}{d x} (x arcsec (x^{3}))

x y d x + (x + 1) d y = 0

What is the derivative of ln((2x+1/(2x-1))) ?
The derivative of ln((2x+1/(2x-1))) is -4/((2x+1)(2x-1))
What is the first derivative of ln((2x+1/(2x-1))) ?
The first derivative of ln((2x+1/(2x-1))) is -4/((2x+1)(2x-1))