1. ## Find the derivative!

Find the derivative of by product rule and quotient rule. (Thanks in advance!)

2. Originally Posted by Affinity
Find the derivative of by product rule and quotient rule. (Thanks in advance!)
Define $\displaystyle f(x) = x - 3x \sqrt{x}$ and $\displaystyle g(x) = \sqrt{x}$

Then $\displaystyle f'(x) = 1 - 3 \sqrt{x} - 3x \cdot \frac{1}{2} \frac{1}{\sqrt{x}}$
and
$\displaystyle g'(x) = \frac{1}{2} \frac{1}{\sqrt{x}}$

Quotient Rule:
So $\displaystyle h(x) = \frac{f(x)}{g(x)}$

$\displaystyle h'(x) = \frac{f'(x)g(x) - f(x)g'(x)}{(g(x))^2}$

$\displaystyle h'(x) = \frac{ \left ( 1 - 3 \sqrt{x} - 3x \cdot \frac{1}{2} \frac{1}{\sqrt{x}} \right ) \sqrt{x} - (x - 3x \sqrt{x}) \frac{1}{2} \frac{1}{\sqrt{x}} }{x}$

$\displaystyle h'(x) = ...= \frac{\sqrt{x} - 6x}{2x}$
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Define $\displaystyle f(x) = x - 3x \sqrt{x}$ and $\displaystyle g(x) = \frac{1}{\sqrt{x}}$

Then $\displaystyle f'(x) = 1 - 3 \sqrt{x} - 3x \cdot \frac{1}{2} \frac{1}{\sqrt{x}}$
and
$\displaystyle g'(x) = -\frac{1}{2} \frac{1}{\sqrt{x^3}} = -\frac{1}{2x\sqrt{x}}$

Product rule:
So $\displaystyle h(x) = f(x)g(x)$

$\displaystyle h'(x) = f'(x)g(x) + f(x)g(x)$

$\displaystyle h'(x) = \left (1 - 3 \sqrt{x} - 3x \cdot \frac{1}{2} \frac{1}{\sqrt{x}} \right ) \frac{1}{\sqrt{x}} + (x - 3x \sqrt{x}) \left ( -\frac{1}{2x\sqrt{x}} \right )$

$\displaystyle h'(x) = ... = \frac{1}{2 \sqrt{x}} - 3 = \frac{\sqrt{x} - 6}{2x}$

-Dan

3. Hello, Affinity!

Differentiate: .$\displaystyle f(x) \:=\:\frac{x - 3x\sqrt{x}}{\sqrt{x}}$
by the product rule and quotient rule.

Product Rule
We have: .$\displaystyle f(x)\;=\;x^{-\frac{1}{2}}\left(x - 3x^{\frac{3}{2}}\right)$

Then: .$\displaystyle f'(x)\;=\;x^{-\frac{1}{2}}\left(1 - \frac{9}{2}x^{\frac{1}{2}}\right) + \left(-\frac{1}{2}x^{-\frac{3}{2}}\right)\left(x - 3x^{\frac{3}{2}}\right)$

. . . . . $\displaystyle f'(x) \;= \;x^{-\frac{1}{2}} - \frac{9}{2} - \frac{1}{2}x^{-\frac{1}{2}} + \frac{3}{2}$

. . . . . $\displaystyle \boxed{f'(x) \;= \;\frac{1}{2}x^{-\frac{1}{2}} - 3}$

Quotient Rule
We have: .$\displaystyle f(x) \;= \;\frac{x - 3x^{\frac{3}{2}}}{x^{\frac{1}{2}}}$

Then: .$\displaystyle f'(x) \;= \;\frac{x^{\frac{1}{2}}\left(1 - \frac{9}{2}x^{\frac{1}{2}}\right) - \left(\frac{1}{2}x^{-\frac{1}{2}}\right)\left(x - 3x^{\frac{3}{2}}\right)}{\left(x^{\frac{1}{2}}\rig ht)^2}$

. . . . . $\displaystyle f'(x) \;= \;\frac{x^{\frac{1}{2}} - \frac{9}{2}x - \frac{1}{2}x^{\frac{1}{2}} + \frac{3}{2}x}{x}$

. . . . . $\displaystyle f'(x) \;= \;\frac{\frac{1}{2}x^{\frac{1}{2}} - 3x}{x}$

. . . . . $\displaystyle \boxed{f'(x) \;= \;\frac{x^{\frac{1}{2}} - 6x}{2x}}$