Find the derivative an determine the values

[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!