# How you you find the derivative f(x)=x^2 using First Principles?

Apr 1, 2018

$f ' \left(x\right) = 2 x$

#### Explanation:

$f ' \left(x\right) = {\lim}_{h \to 0} \frac{f \left(x + h\right) - f \left(x\right)}{h}$

$f ' \left(x\right) = {\lim}_{h \to 0} \frac{{\left(x + h\right)}^{2} - {x}^{2}}{h}$

$f ' \left(x\right) = {\lim}_{h \to 0} \frac{{\cancel{x}}^{2} + 2 x h + {h}^{2} \cancel{-} \left({x}^{2}\right)}{h}$

$f ' \left(x\right) = {\lim}_{h \to 0} \frac{\cancel{h} \left(2 x + h\right)}{\cancel{h}}$

$f ' \left(x\right) = {\lim}_{h \to 0} \left(2 x + h\right)$

$f ' \left(x\right) = 2 x$

Apr 1, 2018

$f ' \left(x\right) = 2 x$

#### Explanation:

$f ' \left(x\right) = {\lim}_{h \to 0} \frac{f \left(x + h\right) - f \left(x\right)}{h}$

$\Rightarrow f ' \left(x\right) = {\lim}_{h \to 0} \frac{{\left(x + h\right)}^{2} - {x}^{2}}{h}$

$\textcolor{w h i t e}{\Rightarrow f ' \left(x\right)} = {\lim}_{h \to 0} \frac{\cancel{{x}^{2}} + 2 h x + {h}^{2} \cancel{- {x}^{2}}}{h}$

$\textcolor{w h i t e}{\Rightarrow f ' \left(x\right)} = {\lim}_{h \to 0} \frac{\cancel{h} \left(2 x + h\right)}{\cancel{h}}$

$\textcolor{w h i t e}{\Rightarrow f ' \left(x\right)} = 2 x$