# How does this work? (Derivatives)

• February 17th 2008, 11:57 AM
CenturionMonkey
How does this work? (Derivatives)
I don't understand how the derivative of
$R(x)=\frac{1}{10000}(60000x-x^2)$

is

$R'(x)=\frac{1}{5000}(30000-x)$

I tried it with both the quotient rule and the product rule and I didn't get the same answer. Because isn't the derivative of a constant always 0?

• February 17th 2008, 12:19 PM
Cursed
Think of $R(x)=\frac{1}{10000}(60000x-x^2)$ as a constant times a variable.

Apply the product rule:
$R(x)=\frac{1}{10000}(60000-2x) + 0(60000x-x^2)$

Combine terms:
$R(x)=\frac{60000-2x}{10000}$

Simplify:
$R(x)=\frac{30000-x}{5000}$ $= \frac{1}{5000}(30000-x)$
• February 17th 2008, 12:54 PM
Soroban
Hello, CenturionMonkey!

You don't need any fancy rules.
. . That fraction is a constant.

Quote:

$R(x)\:=\:\frac{1}{10,000}(60,000x-x^2)$

We have: . $R'(x) \;=\;\frac{1}{10,000}(60,000 - 2x) \;=\;\frac{2(30,000-x)}{10,000} \;=\;\frac{1}{5,000}(30,000-x)$

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If you insist on doing it the Long Way . . .

Product Rule

$R(x) \;=\;\overbrace{\left(\frac{1}{10.000}\right)}^{f( x)}\cdot\overbrace{(60,000x-x^2)}^{g(x)}$

$R'(x)\;=\;\overbrace{\left(\frac{1}{10,000}\right) }^{f(x)}\cdot\overbrace{(60000 - 2x)}^{g'(x)} + \overbrace{0}^{f'(x)}\cdot\overbrace{(60,000x - x^2)}^{g(x)} \;= \;\frac{1}{10,000}\cdot(60,000-2x)$

. . . . $= \;\frac{1}{10,000}\cdot2(30,000 - x) \;=\;\frac{1}{5,000}(30,000-x)$

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Quotient Rule

$R(x) \;=\;\frac{\overbrace{60,000x - x^2}^{f(x)}}{\underbrace{10,000}_{g(x)}}$

$R'(x)\;=\;\frac{\overbrace{(10,000)}^{g(x)}\cdot\o verbrace{(60,000-2x)}^{f'(x)} - \overbrace{(60,000x-x^2)}^{f(x)}\cdot\overbrace{0}^{g'(x)}}{\underbrac e{10,000^2}_{[g(x)]^2}}$

. . . . $= \;\frac{10,000(60,000 - 2x)}{10,000^2} \;=\;\frac{20,000(30,000-x)}{10,000^2} \;=\;\frac{30,000 - x}{5,000}$

• February 17th 2008, 01:36 PM
CenturionMonkey
Oh i see it now, thanks a lot guys. Great help