derivatives


by courtrigrad
Tags: derivatives
courtrigrad
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#1
Oct5-05, 07:23 PM
P: 1,239
Suppose [tex] f(x) = \frac{(x-3)^{4}}{x^{2}+2x} [/tex]. Find the derivative an determine the values for which it is equal to 0. So [tex] f'(x) = \frac{x^{2}+2x(4(x-3)^{3}) - (x-3)^{4}(2x+2)}{(x^{2}+2x)^{2}} [/tex]. But now how would I go about finding the values for which the derivative equals 0? [tex] f'(x) = \frac{x^{2}+2x(4(x-3)^{3}) (x-3)^{4}(2x+2)}{(x^{2}+2x)^{2}} = 0 [/tex]. Is it possible to factor?

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Jameson
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#2
Oct5-05, 07:53 PM
P: 789
For any fraction, let's call it [itex]\frac{A}{B}[/itex], which part will make the whole thing equal to 0? A or B?
courtrigrad
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#3
Oct5-05, 08:08 PM
P: 1,239
A

will yeah

BobG
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#4
Oct5-05, 09:08 PM
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derivatives


Quote Quote by plugpoint
Suppose [tex] f(x) = \frac{(x-3)^{4}}{x^{2}+2x} [/tex]. Find the derivative an determine the values for which it is equal to 0. So [tex] f'(x) = \frac{(x^{2}+2x)(4(x-3)^{3}) - (x-3)^{4}(2x+2)}{(x^{2}+2x)^{2}} [/tex]. But now how would I go about finding the values for which the derivative equals 0? [tex] f'(x) = \frac{x^{2}+2x(4(x-3)^{3}) (x-3)^{4}(2x+2)}{(x^{2}+2x)^{2}} = 0 [/tex]. Is it possible to factor?

Thanks
You missed a set of parentheses.
TD
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#5
Oct6-05, 03:32 AM
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P: 1,024
Your derivative isn't correct yet, make sure you check that first!
HallsofIvy
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#6
Oct6-05, 08:16 AM
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PF Gold
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The derivative is correct, assuming that the missing parentheses BobG mentions is put in correctly!


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