derivatives

*courtrigrad*

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

Jameson

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*

A

will yeah

BobG

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

Your derivative isn't correct yet, make sure you check that first!

HallsofIvy

The derivative is correct, assuming that the missing parentheses BobG mentions is put in correctly!

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Jameson	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?

TD Recognitions: Homework Help	Your derivative isn't correct yet, make sure you check that first!

HallsofIvy Recognitions: Gold Member Science Advisor Staff Emeritus	The derivative is correct, assuming that the missing parentheses BobG mentions is put in correctly!