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

The derivative of (ln(2x-1)/(x^2-1)) is (2(x^2-1)-2xln(2x-1)(2x-1))/((2x-1)(x^2-1)^2)

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

The first derivative of (ln(2x-1)/(x^2-1)) is (2(x^2-1)-2xln(2x-1)(2x-1))/((2x-1)(x^2-1)^2)

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

\frac{2 ( x ^{2} - 1 ) - 2 x ln ( 2 x - 1 ) ( 2 x - 1 )}{( 2 x - 1 ) ( x ^{2} - 1 ) ^{2}}

Solution steps

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

Apply the Quotient Rule:

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

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

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

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

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

Simplify

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

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

x \to 0 lim (x^{2} + 5 x + 6)

x \to 1 + lim (1 - x)

\int_{0}^{1} \frac{x ^{9}}{x ^{10} - 1} d x

\frac{d}{d x} (sin (x) - cos (x))

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

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