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

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

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

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

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

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

Solution steps

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

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

(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}}

\int (x^{3} + e^{3 x}) d x

\frac{\partial}{\partial x} (4 x^{3} - 5 x y + y^{2})

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

\frac{d}{d x} (x^{7} e^{4 x})

\int x^{5} sin (x^{6}) d x

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