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

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

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

The first derivative of f(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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derivative of

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

\frac{d x}{d x} = 1

\frac{d}{d 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}}

\int y 9 - y d y

x \to 3 - lim (\frac{5}{x ^{2} - 9})

x \to 3 lim (e^{\frac{7}{x - 3}})

x \to 1 lim (\frac{7}{( x - 1 ) ^{2}})

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