Question

Question #9a98d

Answer 1

`f'(x)=(x-3)/|x-3|`

Explanation:

`f(x)=|x-3|`

Since `|x|=sqrt(x^2)`, we can rewrite as `f(x)=sqrt((x-3)^2)`.

Use the chain rule `dy/dx=dy/(du)*(du)/(dv)*(dv)/dx` where `y=sqrt((x-3)^2),u=(x-3)^2,v=x-3`.

This gives `dy/(du)=1/(2sqrt((x-3)^2)),(du)/(dv)=2(x-3),(dv)/(dx)=1`.

Thus, `f'(x)=1/(2sqrt((x-3)^2))*2(x-3)*1=(x-3)/sqrt((x-3)^2)=(x-3)/|x-3|`.

So, when `x-3>0` (i.e. `x>3`), `f'(x)=1`. When `x-3<0` (i.e. `x<3`), `f'(x)=-1`. Else, when `x=3`, `f'(x)` is undefined.