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

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

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

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

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

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

Solution steps

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

Apply the Quotient Rule:

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

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

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

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

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

Simplify

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

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

x \to 0 lim ((x + \frac{1}{x}) \cdot e^{- \frac{1}{x ^{2}}})

y = 4 x^{2} - 3

x \to 0 lim (\frac{( 9 x ^{2} - 2 x )}{x})

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

\int_{0}^{1} (x - x^{k}) d x

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