Question #14583

Question

930 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2018-01-09T13:06:51

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

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

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

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

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

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