What is (\partial)/(\partial x)(ln((2x-1)/(x-1))) ?

The answer to (\partial)/(\partial x)(ln((2x-1)/(x-1))) is -1/((2x-1)(x-1))

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(\partial)/(\partial x)(ln((2x-1)/(x-1)))

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

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

Solution steps

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

Apply the chain rule:

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

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

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

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

Simplify

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

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

\frac{\partial}{\partial y} ((2 x + y - 4) e^{2 y})

\int sin (t) cos (t) d t

x \to 0.1 lim (\frac{8 sin ( x )}{x})

2 x^{4} 25 - x^{2}

y^{^{''''}} + 1 = 0

What is (\partial)/(\partial x)(ln((2x-1)/(x-1))) ?
The answer to (\partial)/(\partial x)(ln((2x-1)/(x-1))) is -1/((2x-1)(x-1))