The derivative of [tex]f(x)[/tex] is defined by the limit,
[tex]f'(x)=\displaystyle\lim_{h\to0}\frac{f(x+h)-f(x)}h[/tex]
In this case, [tex]f(x)=8x-x^2-4[/tex], so the derivative would be
[tex]f'(x)=\displaystyle\lim_{h\to0}\frac{(8(x+h)-(x+h)^2-4)-(8x-x^2-4)}h[/tex]
Simplifying the numerator gives
[tex]\dfrac{(8x+8h-x^2-2xh-h^2-4)-(8x-x^2-4)}h=\dfrac{8h-2xh-h^2}h[/tex]
The numerator and denominator share a common factor of [tex]h[/tex], which we can cancel because [tex]h\to0[/tex] means that we're considering [tex]h\neq0[/tex] so that [tex]\dfrac hh=1[/tex]:
[tex]\dfrac{8h-2xh-h^2}h=8-2x-h[/tex]
Then as [tex]h\to0[/tex], we're left with the derivative
[tex]f'(x)=8-2x[/tex]
