use the definition of derivative to find d/dx(1/(3-x))

Answer 1

at x+h
y = 1/(3-x-h)
at x
y = 1/(3-x)

subtract
1/(3-x-h) -1/(3-x)

[(3-x) -(3-x-h)]/[(3-x-h)(3-x)]

[ h]/[ x^2-6x+xh-3h+9]

divide by h
1/[x^2-6x+9 +xh-3h]

let h --->0
1/(x-3)^2

Answer 2

the dividing by h is confusing me. wouldn't you be left with (x^2/h)-(6x/h)+x-3+(9/h)?