Question

Question #14583

Answer 1

`y'=dy/dx=1/(2(x-2))-0.5`

Explanation:

`y=0.5ln(x-2)-0.5x-1`

First, differentiate `0.5ln(x-2)`.

`d/dx(0.5ln(x-2))=0.5d/dx(ln(x-2))`

Using the chain rule ,

let `u=x-2`

`0.5d/dx(ln(x-2))=0.5d/dx(ln(u))*d/dx(x-2)`

`0.5d/dx(ln(x-2))=0.5*1/u*1`

`0.5d/dx(ln(x-2))=0.5/u`

Substitute `u=x-2` back in

`0.5d/dx(ln(x-2))=0.5/(x-2)=1/(2(x-2))`

`y'=1/(2(x-2))-(0.5x)'-(1)'`

`y'=1/(2(x-2))-0.5`