How do I find the derivative of x^2 + 7x -4 using first principles?

Oct 11, 2014

First Principles $\to$ Difference Quotient

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

$f \left(x\right) = {x}^{2} + 7 x - 4$

$f \left(x + h\right) = {\left(x + h\right)}^{2} + 7 \left(x + h\right) - 4$

$f ' \left(x\right) = {\lim}_{h \to 0} \frac{{\left(x + h\right)}^{2} + 7 \left(x + h\right) - 4 - \left({x}^{2} + 7 x - 4\right)}{h}$

$f ' \left(x\right) = {\lim}_{h \to 0} \frac{{\left(x + h\right)}^{2} + 7 \left(x + h\right) - 4 - {x}^{2} - 7 x + 4}{h}$

$f ' \left(x\right) = {\lim}_{h \to 0} \frac{{\left(x + h\right)}^{2} + 7 x + 7 h - 4 - {x}^{2} - 7 x + 4}{h}$

$f ' \left(x\right) = {\lim}_{h \to 0} \frac{{x}^{2} + 2 x h + {h}^{2} + 7 x + 7 h - 4 - {x}^{2} - 7 x + 4}{h}$

$f ' \left(x\right) = {\lim}_{h \to 0} \frac{2 x h + {h}^{2} + 7 h}{h}$

$f ' \left(x\right) = {\lim}_{h \to 0} \frac{h \left(2 x + h + 7\right)}{h}$

$f ' \left(x\right) = {\lim}_{h \to 0} \left(2 x + h + 7\right)$

$f ' \left(x\right) = 2 x + \left(0\right) + 7$

$f ' \left(x\right) = 2 x + 7$