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

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

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

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

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derivative of

ln (2 x - 1) - ln (x - 1)

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

Solution steps

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

Apply the Sum/Difference Rule:

(f \pm g)^{'} = f^{'} \pm g^{'}

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

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

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

= \frac{2}{2 x - 1} - \frac{1}{x - 1}

Simplify

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

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

f (x) = sin (2 x)

y = 3 x + 5

y = \frac{x ^{2}}{x - 1}, (0, 0)

f (x) = (x^{2} - 3 x) ln (x^{2} - 3 x)

z = - 3 - 3 i

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