# First Principles Example 2: x³

## Key Questions

• First Principles $\to$ Difference Quotient

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

$f \left(x\right) = {x}^{3} - 2 {x}^{2} + \frac{x}{4} + 6$

$f \left(x + h\right) = {\left(x + h\right)}^{3} - 2 {\left(x + h\right)}^{2} + \frac{x + h}{4} + 6$

$f ' \left(x\right) = {\lim}_{h \to 0} \frac{{\left(x + h\right)}^{3} - 2 {\left(x + h\right)}^{2} + \frac{x + h}{4} + 6 - \left({x}^{3} - 2 {x}^{2} + \frac{x}{4} + 6\right)}{h}$

$f ' \left(x\right) = {\lim}_{h \to 0} \frac{{\left(x + h\right)}^{3} - 2 {\left(x + h\right)}^{2} + \frac{x + h}{4} + 6 - {x}^{3} + 2 {x}^{2} - \frac{x}{4} - 6}{h}$

$f ' \left(x\right) = {\lim}_{h \to 0} \frac{{\left(x + h\right)}^{3} - 2 {\left(x + h\right)}^{2} + \frac{x + h}{4} - {x}^{3} + 2 {x}^{2} - \frac{x}{4}}{h}$

$f ' \left(x\right) = {\lim}_{h \to 0} \frac{{\left(x + h\right)}^{3} - 2 \left({x}^{2} + 2 x h + {h}^{2}\right) + \frac{x + h}{4} - {x}^{3} + 2 {x}^{2} - \frac{x}{4}}{h}$

$f ' \left(x\right) = {\lim}_{h \to 0} \frac{{\left(x + h\right)}^{3} - 2 {x}^{2} - 4 x h - 2 {h}^{2} + \frac{x + h}{4} - {x}^{3} + 2 {x}^{2} - \frac{x}{4}}{h}$

$f ' \left(x\right) = {\lim}_{h \to 0} \frac{{\left(x + h\right)}^{3} - 4 x h - 2 {h}^{2} + \frac{x + h}{4} - {x}^{3} - \frac{x}{4}}{h}$

$f ' \left(x\right) = {\lim}_{h \to 0} \frac{\left(x + h\right) \left({x}^{2} + 2 x h + {h}^{2}\right) - 4 x h - 2 {h}^{2} + \frac{x}{4} + \frac{h}{4} - {x}^{3} - \frac{x}{4}}{h}$

$f ' \left(x\right) = {\lim}_{h \to 0} \frac{\left(x + h\right) \left({x}^{2} + 2 x h + {h}^{2}\right) - 4 x h - 2 {h}^{2} + \frac{h}{4} - {x}^{3}}{h}$

$f ' \left(x\right) = {\lim}_{h \to 0} \frac{{x}^{3} + 2 {x}^{2} h + x {h}^{2} + h {x}^{2} + 2 x {h}^{2} + {h}^{3} - 4 x h - 2 {h}^{2} + \frac{h}{4} - {x}^{3}}{h}$

$f ' \left(x\right) = {\lim}_{h \to 0} \frac{2 {x}^{2} h + x {h}^{2} + h {x}^{2} + 2 x {h}^{2} + {h}^{3} - 4 x h - 2 {h}^{2} + \frac{h}{4}}{h}$

$f ' \left(x\right) = {\lim}_{h \to 0} \frac{h \cdot \left(2 {x}^{2} + x h + {x}^{2} + 2 x h + {h}^{2} - 4 x - 2 h + \frac{1}{4}\right)}{h}$

$f ' \left(x\right) = {\lim}_{h \to 0} 2 {x}^{2} + x h + {x}^{2} + 2 x h + {h}^{2} - 4 x - 2 h + \frac{1}{4}$

$f ' \left(x\right) = 2 {x}^{2} + x \left(0\right) + {x}^{2} + 2 x \left(0\right) + {\left(0\right)}^{2} - 4 x - 2 \left(0\right) + \frac{1}{4}$

$f ' \left(x\right) = 2 {x}^{2} + {x}^{2} - 4 x + \frac{1}{4}$

$f ' \left(x\right) = 3 {x}^{2} - 4 x + \frac{1}{4}$

• First Principles $\to$ Difference Quotient

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

$f \left(x\right) = {x}^{3}$

$f \left(x + h\right) = {\left(x + h\right)}^{3}$

$f ' \left(x\right) = {\lim}_{h \to 0} \frac{{\left(x + h\right)}^{3} - {x}^{3}}{h}$

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

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

$f ' \left(x\right) = {\lim}_{h \to 0} \frac{2 {x}^{2} h + x {h}^{2} + {x}^{2} h + 2 x {h}^{2} + {h}^{3}}{h}$

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

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

$f ' \left(x\right) = {\lim}_{h \to 0} 3 {x}^{2} + x h + 2 x h + {h}^{2}$

$f ' \left(x\right) = 3 {x}^{2} + x \left(0\right) + 2 x \left(0\right) + {\left(0\right)}^{2}$

$f ' \left(x\right) = 3 {x}^{2}$