Question

How do you find the second derivative of `f(x) = x/(1-ln(x-1))` ?

Answer 1

`f'(x)={2x-1-(x-1)ln(x-1)}/[(x-1){1-ln(x-1)}^2].`

Explanation:

Let us write,

`f(x)=g(x)/(h(x))," where, "g(x)=x, and, h(x)=1-ln(x-1).`

Using the Quotient Rule for Diffn., we have,

`f'(x)={h(x)g'(x)-g(x)h'(x)}/((h(x))^2.`

Now, `h(x)=1-ln(x-1) rArr h'(x)=0-1/(x-1)*d/dx(x-1)`

`:. h'(x)=-1/(x-1).`

Also, `g(x)=x rArr g'(x)=1.`

Utilising these, we get,

`f'(x)={(1-ln(x-1))*1-x(-1/(x-1))}/(1-ln(x-1))^2, i.e.,`

`f'(x)={1-ln(x-1)+x/(x-1)}/{1-ln(x-1)}^2, or,`

`f'(x)={2x-1-(x-1)ln(x-1)}/[(x-1){1-ln(x-1)}^2].`