Question

How do you differentiate `y= ln (x/(x-1))`?

Answer 1

`(-1)/(x(x-1)`

Explanation:

Using `color(red)(|bar(ul(color(white)(a/a)color(black)(d/dx(lnf(x))=1/(f(x)))color(white)(a/a)|)))`

combined with the `color(blue)" chain rule"`

`color(red)(|bar(ul(color(white)(a/a)color(black)(d/dx[f(g(x))]=f'(g(x)).g'(x))color(white)(a/a)|)))color(green)" A"`
`"------------------------------------------------------------"`

here `f(g(x))=ln(x/(x-1))rArrf'(g(x))=1/(x/(x-1))=(x-1)/x`

and `g(x)=x/(x-1)rArrg'(x) = "see below"`
`"-----------------------------------------------------------"`
To differentiate g(x) we require to use the `color(blue)"quotient rule"`

If f(x)`=(g(x))/(h(x))" then " f'(x)=(h(x).g'(x)-g(x).h'(x))/(h(x)^2)color(green)"B"`
`"-----------------------------------------------------------------"`

here g(x) = x `rArrg'(x)=1`

and h(x) = x-1 `rArrh'(x)=1`
`"----------------------------------------------------------------"`
Substitute these values into `color(green)" B"`

`rArrf'(x)=((x-1).1-x.1)/(((x-1)^2))=(x-1-x)/(x-1)^2=(-1)/(x-1)^2`
`"------------------------------------------------------------"`
Now this is the value of g'(x) that we set out to obtain from above

Finally , 'plug' these all back into `color(green)"A"`

`rArrdy/dx=(x-1)/x xx(-1)/(x-1)^2`

`=cancel((x-1))/x xx(-1)/(cancel((x-1)) (x-1))=(-1)/(x(x-1))`