Question

Using the limit definition, how do you find the derivative of `f(x) = 4 -2x -x^2`?

Answer 1

`d/dx (4-2x-x^2) = -2-2x`

Explanation:

By definition:

`d/dx (4-2x-x^2) = lim_(h->0) ( ( 4-2(x+h) -(x+h)^2 ) - (4-2x-x^2))/h`

`d/dx (4-2x-x^2) = lim_(h->0) ( ( color(blue)(4) color(red)(-2x)-2h color(green)(-x^2)-2hx -h^2 color(blue)(- 4) color(red)(+2x) color(green)(+x^2)))/h`

`d/dx (4-2x-x^2) = lim_(h->0) ( -2h-2hx-h^2)/h`

`d/dx (4-2x-x^2) = lim_(h->0) ( -2-2x-h)`

`d/dx (4-2x-x^2) = -2-2x`