Given:
[tex]f(x)=(x+2)^2[/tex]Required:
To find the first principles
Explanation:
First principle,
[tex]\lim_{h\to0}\frac{f(x+h)-f(x)}{h}[/tex][tex]=\lim_{h\to0}\frac{(x+h+2)^2-(x+2)^2}{h}[/tex][tex]=\lim_{h\to0}\frac{x^2+(h+2)^2+2x(h+2)-x^2-4-4x}{h}[/tex][tex]=\lim_{h\to0}\frac{h^2+4+4h+2xh+4x-4-4x}{h}[/tex][tex]\begin{gathered} =\lim_{h\to0}\frac{h^2+4h+2xh}{h} \\ \\ =\lim_{h\to0}\frac{h(h+4+2x)}{h} \\ \\ =\lim_{h\to0}(h+4+2x) \\ =2x+4 \end{gathered}[/tex]Final Answer:
[tex]2x+4[/tex]