Answer:
8x
Step-by-step explanation:
Using differentiation from first principles
f'(x) = [tex]lim_{h>0}[/tex] [tex]\frac{f(x+h)-f(x)}{h}[/tex]
= [tex]lim_{h>0}[/tex] [tex]\frac{4(x+h)^2-2-(4x^2-2)}{h}[/tex]
= [tex]lim_{h>0}[/tex] [tex]\frac{4x^2+8hx+4h^2-4x^2+2}{h}[/tex]
= [tex]lim_{h>0}[/tex] [tex]\frac{8hx+4h^2}{h}[/tex]
= [tex]lim_{h>0}[/tex] [tex]\frac{4h(2x+h)}{h}[/tex] ← cancel the h on numerator/denominator
= 4(2x)
= 8x
