Find the derivative!

Affinity · November 12th 2006, 12:28 PM

Find the derivative of

by product rule and quotient rule. (Thanks in advance!)

**topsquark** · November 12th 2006, 12:48 PM

Originally Posted by Affinity

Find the derivative of

by product rule and quotient rule. (Thanks in advance!)

Define $f(x) = x - 3x \sqrt{x}$ and $g(x) = \sqrt{x}$

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

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

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

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

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

Define $f(x) = x - 3x \sqrt{x}$ and $g(x) = \frac{1}{\sqrt{x}}$

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

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

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

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

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

-Dan

**Soroban** · November 12th 2006, 04:38 PM

Hello, Affinity!

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

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

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

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

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

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

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

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

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

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

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