Question

Find from first principles the derivative of f:x maps to (x+2)all squared

Answer 1

Given:

$f(x)=(x+2)^2$

Required:

To find the first principles

Explanation:

First principle,

$\lim_{h\to0}\frac{f(x+h)-f(x)}{h}$$=\lim_{h\to0}\frac{(x+h+2)^2-(x+2)^2}{h}$$=\lim_{h\to0}\frac{x^2+(h+2)^2+2x(h+2)-x^2-4-4x}{h}$$=\lim_{h\to0}\frac{h^2+4+4h+2xh+4x-4-4x}{h}$$\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}$

Final Answer:

$2x+4$