# How do you find f'(x) using the definition of a derivative  f(x) = x^2 + x?

Jul 23, 2018

$2 x + 1$.

#### Explanation:

Recall that, $f ' \left(x\right) = {\lim}_{t \to x} \frac{f \left(t\right) - f \left(x\right)}{t - x} \ldots \ldots \ldots \ldots \left(\ast\right)$.

$f \left(x\right) = {x}^{2} + x \Rightarrow f \left(t\right) = {t}^{2} + t$.

$\therefore f \left(t\right) - f \left(x\right) = \left({t}^{2} - {x}^{2}\right) + \left(t - x\right)$,

$= \left(t - x\right) \left(t + x\right) + \left(t - x\right)$,

$\Rightarrow f \left(t\right) - f \left(x\right) = \left(t - x\right) \left\{\left(t + x\right) + 1\right\}$.

$\therefore \frac{f \left(t\right) - f \left(x\right)}{t - x} = \left\{\left(t + x\right) + 1\right\} , \left(t \ne x\right)$.

$\therefore , \text{ by "(ast), } f ' \left(x\right) = {\lim}_{t \to x} \left\{\left(t + x\right) + 1\right\}$,

$i . e . , f ' \left(x\right) = \left\{\left(x + x\right) + 1\right\} = 2 x + 1$, as desired!

$\textcolor{b l u e}{\text{Enjoy Maths.!}}$

Jul 23, 2018

$f ' \left(x\right) = 2 x + 1$

#### Explanation:

$\text{differentiating from first principles}$

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

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

$= {\lim}_{h \to 0} \frac{\cancel{{x}^{2}} + 2 h x + {h}^{2} \cancel{+ x} + h \cancel{- {x}^{2}} \cancel{- x}}{h}$

$= {\lim}_{h \to 0} \frac{\cancel{h} \left(2 x + h + 1\right)}{\cancel{h}}$

$= 2 x + 1$