Question

Using the limit definition, how do you differentiate `f(x) = x^2+3`?

Answer 1

See expanation

Explanation:

`f'(x_0)=lim_{h->0}(f(x_0+h)-f(x_0))/h`

If we substitute `x_0^2+3` we get:

`f'(x_0)=lim_{h->0}((x_0+h)^2+3-(x_0^2+3))/h`

`f'(x_0)=lim_{h->0}((x_0^2+2x_0h+h^2)+3-x_0^2-3)/h`

`f'(x_0)=lim_{h->0}(x_0^2+2x_0h+h^2+3-x_0^2-3)/h`

`f'(x_0)=lim_{h->0}(2x_0h+h^2)/h`

`f'(x_0)=lim_{h->0}(h(2x_0+h))/h`

`f'(x_0)=lim_{h->0}(2x_0+h)`

`f'(x_0)=2x_0`