derivatives

[courtrigrad]

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

Thanks

[Jameson]

For any fraction, let's call it \frac{A}{B}, which part will make the whole thing equal to 0? A or B?

[courtrigrad]

A

will yeah

[BobG]

Suppose f(x) = \frac{(x-3)^{4}}{x^{2}+2x} . Find the derivative an determine the values for which it is equal to 0. So f'(x) = \frac{(x^{2}+2x)(4(x-3)^{3}) - (x-3)^{4}(2x+2)}{(x^{2}+2x)^{2}} . But now how would I go about finding the values for which the derivative equals 0? f'(x) = \frac{x^{2}+2x(4(x-3)^{3}) (x-3)^{4}(2x+2)}{(x^{2}+2x)^{2}} = 0 . 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!