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

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

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

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

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

Solution steps

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

Apply the Quotient Rule:

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

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

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

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

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

Simplify

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

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

y = \frac{1}{20 x ^{2}}, y = x, y = \frac{x}{6}

\frac{d}{d x} (a (e^{2 x} + 2 e^{2 x} x))

f (x) = 3 x^{2} + 2 ln (8) (x)

y^{^{'}} = \frac{( x ^{5} )}{y}

\int_{0}^{3 π} 3 x^{2} sin (\frac{1}{6} x) d x

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