# Question #c1a8e

Oct 4, 2014

a. Difference Quotient

Identify f(x) and f(x+h)

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

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

Make substitutions and rationalize the numerator

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

Remember difference of perfect squares

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

Distribute the $3$

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

Simplify

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

Cancel the factor $h$

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

Now you can substitute in 0 for $h$

$\frac{3}{\sqrt{3 \left(x + 0\right)} + \sqrt{3 x}} =$

Simplify and manipulate the square roots

$\frac{3}{\sqrt{3 \left(x\right)} + \sqrt{3 x}} = \frac{3}{\sqrt{3 x} + \sqrt{3 x}} = \frac{3}{2 \sqrt{3 x}} \cdot \frac{\sqrt{3}}{\sqrt{3}} = \frac{3 \sqrt{3}}{6 \sqrt{x}} = \frac{\sqrt{3}}{2 \sqrt{x}} = \frac{\sqrt{3}}{2 \sqrt{x}} \cdot \frac{\sqrt{x}}{\sqrt{x}} = \frac{\sqrt{3 x}}{2 x}$

b. Product Rule

$f \left(x\right) = u \cdot v = \sqrt{3} \cdot \sqrt{x} = \sqrt{3} \cdot {x}^{\frac{1}{2}}$

$u ' = 0$
$v ' = \left(\frac{1}{2}\right) {x}^{\frac{1}{2} - 1} = \left(\frac{1}{2}\right) {x}^{\frac{1}{2} - \frac{2}{2}} = \left(\frac{1}{2}\right) {x}^{- \frac{1}{2}} = \frac{1}{2 \sqrt{x}}$

Apply the product rule by making the appropriate substitutions

$f ' \left(x\right) = u \cdot v ' + v \cdot u ' = \sqrt{3} \cdot \frac{1}{2 \sqrt{x}} + \sqrt{x} \cdot 0 = \frac{\sqrt{3}}{2 \sqrt{x}} = \frac{\sqrt{3}}{2 \sqrt{x}} \cdot \frac{\sqrt{x}}{\sqrt{x}} = \frac{\sqrt{3 x}}{2 x}$

c. Factor out the constant of $\sqrt{3}$

$f \left(x\right) = \sqrt{3} \cdot {x}^{\frac{1}{2}}$

$f ' \left(x\right) = \sqrt{3} \cdot \frac{d}{\mathrm{dx}} \left({x}^{\frac{1}{2}}\right) = \sqrt{3} \cdot \left(\frac{1}{2}\right) {x}^{\frac{1}{2} - 1} = \sqrt{3} \cdot \left(\frac{1}{2}\right) {x}^{\frac{1}{2} - \frac{2}{2}} = \sqrt{3} \cdot \left(\frac{1}{2}\right) {x}^{- \frac{1}{2}} = \sqrt{3} \cdot \left(\frac{1}{2 {x}^{\frac{1}{2}}}\right) = \frac{\sqrt{3}}{2 {x}^{\frac{1}{2}}} = \frac{\sqrt{3}}{2 \sqrt{x}} = \frac{\sqrt{3}}{2 \sqrt{x}} \cdot \frac{\sqrt{x}}{\sqrt{x}} = \frac{\sqrt{3 x}}{2 x}$