F(x) = 7x^2 + 4, find and simplify using the derivative function f(a+h) -f(a)/ (h) ?

Aug 23, 2017

$\frac{\mathrm{dF}}{\mathrm{dx}} = 14 x$

Explanation:

I'm not 100% sure what you are asking but I'm going to take a punt that you want to find the derivative of $F \left(x\right)$ using the limit definition? You have left the limit out of your 'derivative function' but this is what I think you meant.

$\frac{\mathrm{df}}{\mathrm{da}} = {\lim}_{h \to 0} \frac{f \left(a + h\right) - f \left(a\right)}{h}$

With $F \left(x\right) = 7 {x}^{2} + 4$ we have

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

Expanding out the squared bracket some terms helpfully cancel out:

$= {\lim}_{h \to 0} \frac{\textcolor{red}{7 {x}^{2}} + 14 x h + 7 {h}^{2} + \textcolor{b l u e}{4} \textcolor{red}{- 7 {x}^{2}} \textcolor{b l u e}{- 4}}{h}$

$= {\lim}_{h \rightarrow 0} 14 x + 7 h = 14 x$

$\therefore \frac{\mathrm{dF}}{\mathrm{dx}} = 14 x$

Pleasingly this is what we would expect had we differentiated normally using the power rule.

Aug 23, 2017

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

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

$f ' \left(x\right) = {\lim}_{h \to 0} \frac{7 {x}^{2} + 14 x h + {h}^{2} + 4 - 7 {x}^{2} - 4}{h}$

$f ' \left(x\right) = {\lim}_{h \to 0} \frac{14 x h + {h}^{2}}{h}$

$f ' \left(x\right) = {\lim}_{h \to 0} \frac{h \left(14 x + h\right)}{h}$

$f ' \left(x\right) = {\lim}_{h \to 0} 14 x + h$

$f ' \left(x\right) = 14 x$

This matches what we would obtain using the power rule, where $f ' \left(x\right) = 7 \left(2\right) {x}^{2 - 1} = 14 {x}^{1} = 14 x$.

Hopefully this helps!