How do I find the derivative of f(x)=x^3 from first principles?

Oct 11, 2014

First Principles $\to$ Difference Quotient

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

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

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

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

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

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

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

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

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

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

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

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