# Using the limit definition for derivatives, how do you find the derivative of (a) f(x)=x^2-6x; (b) f(x)=4sqrt(x)?

Feb 12, 2016

1) $f ' \left(x\right) = 2 x - 6$

2) $f ' \left(x\right) = \frac{2}{\sqrt{x}}$

#### Explanation:

1) $f \left(x\right) = {x}^{2} - 6 x$

According to the limit definition, the derivative of $f \left(x\right)$ is

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

To compute $f \left(x + h\right)$, plug $x + h$ for every occurance of $x$ in $f \left(x\right)$:

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

Thus, you can compute your derivative as follows:

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

$= {\lim}_{h \to 0} \frac{{x}^{2} + 2 x h + {h}^{2} - 6 x - 6 h - {x}^{2} + 6 x}{h}$

$= {\lim}_{h \to 0} \frac{\textcolor{b l u e}{\cancel{{x}^{2}}} + 2 x h + {h}^{2} - \textcolor{red}{\cancel{6 x}} - 6 h - \textcolor{b l u e}{\cancel{{x}^{2}}} + \textcolor{red}{\cancel{6 x}}}{h}$

$= {\lim}_{h \to 0} \frac{2 x h + {h}^{2} - 6 h}{h}$

... factor $h$ in the numerator...

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

... cancel $h$...

$= {\lim}_{h \to 0} \left(2 x + h - 6\right)$

... apply the limit, so plug $h = 0$...

$= 2 x - 6$

So, we have found that

$f ' \left(x\right) = 2 x - 6$.

================================

2) $f \left(x\right) = 4 \sqrt{x}$

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

$= {\lim}_{h \to 0} \frac{4 \sqrt{x + h} - 4 \sqrt{x}}{h}$

$= {\lim}_{h \to 0} \frac{4 \left(\sqrt{x + h} - \sqrt{x}\right)}{h}$

... multiply the numerator and the denominator with $\left(\sqrt{x + h} + \sqrt{x}\right)$...

$= {\lim}_{h \to 0} \frac{4 \left(\sqrt{x + h} - \sqrt{x}\right) \textcolor{b l u e}{\left(\sqrt{x + h} + \sqrt{x}\right)}}{h \textcolor{b l u e}{\left(\sqrt{x + h} + \sqrt{x}\right)}}$

... use the formula $\left(a + b\right) \left(a - b\right) = {a}^{2} - {b}^{2}$ to expand the numerator...

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

$= {\lim}_{h \to 0} \frac{4 \left(x + h - x\right)}{h \left(\sqrt{x + h} + \sqrt{x}\right)}$

$= {\lim}_{h \to 0} \frac{4 h}{h \left(\sqrt{x + h} + \sqrt{x}\right)}$

... cancel $h$....

$= {\lim}_{h \to 0} \frac{4}{\sqrt{x + h} + \sqrt{x}}$

... apply the limit, so plug $h = 0$...

$= \frac{4}{\sqrt{x + 0} + \sqrt{x}}$

$= \frac{4}{2 \sqrt{x}}$

$= \frac{2}{\sqrt{x}}$

So you have found your derivative and it's

$f ' \left(x\right) = \frac{2}{\sqrt{x}}$

Feb 12, 2016

If $y = {x}^{2} - 6 x$ then $\frac{\mathrm{dy}}{\mathrm{dx}} = 2 x - 6$

If $f \left(x\right) = 4 \sqrt{x}$ then $f ' \left(x\right) = \frac{2}{\sqrt{x}}$

#### Explanation:

Question 1: $\textcolor{b l a c k}{y = {x}^{2} - 6 x}$
(using limit definition for derivative)
$\frac{\mathrm{dy}}{\mathrm{dx}} = {\lim}_{h \rightarrow 0} \frac{\left({\left(x + h\right)}^{2} - 6 \left(x + h\right)\right) - \left({x}^{2} - 6 x\right)}{h}$

$\textcolor{w h i t e}{\text{XXX}} = {\lim}_{h \rightarrow o} \frac{\cancel{{x}^{2}} + 2 h x + {h}^{2} \cancel{- 6 x} - 6 h - \cancel{- {x}^{2}} \cancel{+ 6 x}}{h}$

$\textcolor{w h i t e}{\text{XXX}} = {\lim}_{h \rightarrow 0} \frac{2 \cancel{h} x - 6 \cancel{h}}{\cancel{h}}$

$\textcolor{w h i t e}{\text{XXX}} = 2 x - 6$

Question 2: $\textcolor{b l a c k}{f \left(x\right) = 4 \sqrt{x}}$
(using limit definition for derivative)
f'(x)=lim_(hrarr0)=(4sqrt(x+h)-4sqrt(x)/h

$\textcolor{w h i t e}{\text{XXX}} = 4 {\lim}_{h \rightarrow 0} \left(\frac{\sqrt{x + h} - \sqrt{x}}{h}\right) \cdot \left(\frac{\sqrt{x + h} + \sqrt{x}}{\sqrt{x + h} + \sqrt{x}}\right)$

$\textcolor{w h i t e}{\text{XXX}} = 4 {\lim}_{h \rightarrow 0} \frac{x + h - x}{h \sqrt{x + h} + \sqrt{x}}$

$\textcolor{w h i t e}{\text{XXX}} = 4 {\lim}_{h \rightarrow 0} \frac{\cancel{h}}{\cancel{h} \sqrt{x + h} + \sqrt{x}}$

$\textcolor{w h i t e}{\text{XXX}} = 4 \cdot \frac{1}{2 \sqrt{x}}$

$\textcolor{w h i t e}{\text{XXX}} = \frac{2}{\sqrt{x}}$