Question

Question #1299f

Answer 1

`k'(x)=((x+2)(2(x-1)-(x+2)))/(x-1)^2`

Explanation:

The quotient rule says:
`d/dx(f(x)/g(x))=(f'(x)g(x)-f(x)g'(x))/[g(x)]^2`

In our case, `f(x)=(x+2)^2` and `g(x)=x-1`.

`f'(x)=2(x+2)` (by the chain rule)

`g'(x)=1`

This gives:
`k'(x)=(2(x+2)(x-1)-(x+2)^2*1)/(x-1)^2=`

`=((x+2)(2(x-1)-(x+2)))/(x-1)^2`