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

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Answer 1 · 2015-11-16T16:23:35

See expanation

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

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

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

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

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

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

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

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

$f'(x_0)=2x_0$

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