Question

Question #55c21

Answer 1

Look below

Explanation:

First, apply the quotient rule of `F(x)=ln(x)/(x-1)`

`(\frac{d}{dx}[ln(x)]times(x-1)-ln(x)times\frac{d}{dx}[x-1])/((x-1)^2)`

`\frac{d}{dx}[ln(x)] = 1/x`

`\frac{d}{dx}[x-1] = \frac{d}{dx}[x]+\frac{d}{dx}[-1]`

`\frac{\frac{1}{x}(x-1)-ln(x)times(1+0)}{(x-1)^2`

the one goes into the `ln(x)`

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