How do you use the definition of a derivative to find the derivative of f(x)=x^3-2x^2+5x-6?

Aug 14, 2018

$f ' \left(x\right) = 3 {x}^{2} - 4 x + 5$

Explanation:

Here ,

$f \left(x\right) = {x}^{3} - 2 {x}^{2} + 5 x - 6 \implies f \left(t\right) = {t}^{3} - 2 {t}^{2} + 5 t - 6$

We know that ,

$f ' \left(x\right) = {\lim}_{t \to x} \frac{f \left(t\right) - f \left(x\right)}{t - x} \to \text{definition}$

Substitute values of $f \left(t\right) \mathmr{and} f \left(x\right)$

$f ' \left(x\right) = {\lim}_{t \to x} \frac{\left({t}^{3} - 2 {t}^{2} + 5 t - 6\right) - \left({x}^{3} - 2 {x}^{2} + 5 x - 6\right)}{t - x}$

$f ' \left(x\right) = {\lim}_{t \to x} \frac{\left({t}^{3} - {x}^{3}\right) - 2 \left({t}^{2} - {x}^{2}\right) + 5 \left(t - x\right)}{t - x}$

$f ' \left(x\right) = {\lim}_{t \to x} \frac{\textcolor{red}{\cancel{\left(t - x\right)}} \left\{\left({t}^{2} + t x + {x}^{2}\right) - 2 \left(t + x\right) + 5 \left(1\right)\right\}}{\textcolor{red}{\cancel{t - x}}}$

$f ' \left(x\right) = {\lim}_{t \to x} \left\{\left({t}^{2} + t x + {x}^{2}\right) - 2 \left(t + x\right) + 5 \left(1\right)\right\}$

$f ' \left(x\right) = \left({x}^{2} + x \cdot x + {x}^{2}\right) - 2 \left(x + x\right) + 5 \left(1\right)$

$f ' \left(x\right) = \left(3 {x}^{2}\right) - 2 \left(2 x\right) + 5 \left(1\right)$

$\therefore f ' \left(x\right) = 3 {x}^{2} - 4 x + 5$

Aug 14, 2018

$f ' \left(x\right) = 3 {x}^{2} - 4 x + 5$

Explanation:

$\text{The definition of the derivative of "f(x)" is : }$

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

$\text{So here we have }$

${\lim}_{h \to 0} \frac{{\left(x + h\right)}^{3} - 2 {\left(x + h\right)}^{2} + 5 \left(x + h\right) - 6 - {x}^{3} + 2 {x}^{2} - 5 x + 6}{h}$

$= \frac{\left({x}^{3} + 3 h {x}^{2} + 3 {h}^{2} x + {h}^{3}\right) - 2 \left({x}^{2} + 2 h x + {h}^{2}\right) + 5 \left(x + h\right) - {x}^{3} + 2 {x}^{2} - 5 x}{h}$

$= {\lim}_{h \to 0} \frac{3 h {x}^{2} + 3 {h}^{2} x + {h}^{3} - 4 h x - 2 {h}^{2} + 5 h}{h}$

$= {\lim}_{h \to 0} 3 {x}^{2} + 3 h x + {h}^{2} - 4 x - 2 h + 5$

$= {\lim}_{h \to 0} 3 {x}^{2} - 4 x + 5 + {h}^{2} + 3 h x - 2 h$

$= 3 {x}^{2} - 4 x + 5 + {\lim}_{h \to 0} {h}^{2} + 3 h x - 2 h$

$= 3 {x}^{2} - 4 x + 5 + 0$

$= 3 {x}^{2} - 4 x + 5$