SOLUTION: i need the derivative of
{{{ -2*x/(x^2+5) }}}
i got
{{{ 2/(x^2+5) }}}
using the y=u/v ; y'= (vu'-uv')/v^2 rule.. but there has to be a relative max and a relative mi
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Question 248394: i need the derivative of
i got
using the y=u/v ; y'= (vu'-uv')/v^2 rule.. but there has to be a relative max and a relative min based on the graph of the original function, so there must be some value for which the derivative = 0. But if the numerator is a constant in the derivative, there is no possible x value for which the derivative = 0, so i feel like i must have gotten something wrong...
Answer by jim_thompson5910(35256) (Show Source): You can put this solution on YOUR website!
Let 
and 
. So 
and 
.
=\frac{vu^{\prime}-uv^{\prime}}{v^2})
... Start with the quotient rule
=\frac{(x^2+5)(-2)-(-2x)(2x)}{(x^2+5)^2})
... Plug in 
=\frac{-2x^2-10+4x^2}{(x^2+5)^2})
... Distribute
=\frac{2x^2-10}{(x^2+5)^2})
... Combine like terms.
So the derivative of is
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