$\bf \stackrel{\textit{de} \textit{finition of a derivative as a limit}}{\lim\limits_{h\to 0}~\cfrac{f(x+h)-f(x)}{h}}\\\\
-------------------------------\\\\
f(x)=2x^2+1\implies \lim\limits_{h\to 0}~\cfrac{[2(x+h)^2+1]~~-~~[2x^2+1]}{h}
\\\\\\
\lim\limits_{h\to 0}~\cfrac{[2(x^2+2xh+h^2)+1]~~-~~[2x^2+1]}{h}
\\\\\\
\lim\limits_{h\to 0}~\cfrac{[2x^2+4xh+2h^2+1]~~-~~[2x^2+1]}{h}$

$\bf \lim\limits_{h\to 0}~\cfrac{\underline{2x^2}+4xh+2h^2\underline{+1}~~~~\underline{-2x^2-1}}{h}\implies \lim\limits_{h\to 0}~\cfrac{4xh+2h^2}{h}
\\\\\\
\lim\limits_{h\to 0}~\cfrac{2\underline{h}(2x+h)}{\underline{h}}\implies \lim\limits_{h\to 0}~2(2x+h)\implies \lim\limits_{h\to 0}~2(2x+0)\implies 4x$

3 0

4 years ago

How do you do it? What do you take away and do after?

Nuetrik [128]

hi sorry I can't help with that

5 0

3 years ago

Read 2 more answers