Use H-definition to compute the dervative F(x)-(3x+1)^2
Unless you're referring to the limit process this is all I could drudge up anywhere near "h-definition."
Here's to just in case.
$\displaystyle \frac{f(x + h) - f(x)}{h}$
$\displaystyle \frac{(3(x + h) + 1)^2 - (3x + 1)^2}{h}$
$\displaystyle \frac{(9(x + h)^2 + 6(x + h) + 1) - (9x^2 + 6x + 1)}{h}$
$\displaystyle \frac{(9x^2 + 18xh + 9h^2 + 6x + 6h + 1) - (9x^2 + 6x + 1}{h}$
$\displaystyle \frac{18xh + 9h^2 + 6h}{h}$
$\displaystyle 18x + 9h + 6$
Now taking the limit as h tends toward 0, 9h gets thrown out and we're left with
$\displaystyle 18x + 6$
And surely enough, if you use chain rule:
$\displaystyle 2(3x + 1)3 = 18x + 6$
if you mean the limit definion then
$\displaystyle \lim_{h \to 0}\frac{f(x+h)-f(x)}{h}$
$\displaystyle \lim_{h \to 0}\frac{[3(x+h)+1]^2-(3x+1)^2}{h}$
$\displaystyle \lim_{h \to 0}\frac{9(x+h)^2+6(x+h)+1-(3x+1)^2}{h}$
$\displaystyle \lim_{h \to 0}\frac{9x^2+18xh+9h^2+6x+6h+1-9x^2-6x-1}{h}$
$\displaystyle \lim_{h \to 0}\frac{18xh+9h^2+6h}{h}$
$\displaystyle \lim_{h \to 0}18x+9h+6=18x+6$