derivative of y=x(9-x)^1/2?

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derivative of y=x(9-x)^1/2?

Mathematics
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At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.

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\[y=x\sqrt{(9-x)}\]?
yes
product rule for this one

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Other answers:

this is what I have: x[-1/x(9-x)^1/2]+(9-x)^1/2(1)
is that right?
if so how to simplify
\[y'=1\times \sqrt{9-x}+x\times \frac{1}{\sqrt{9-x}}\times -1\] and then clean up with some algebra
really not much to clean up, just \[y'=\sqrt{9-x}-\frac{x}{\sqrt{9-x}}\]
simplifiedmore?
or if you want to put it over one denominator, you can
\[\frac{9-x-x}{\sqrt{9-x}}\] \[\frac{9-2x}{\sqrt{9-x}}\]
if you want to "simplify" things usually best not to use exponents, but rather radicals
so√(9-x) *x equals 9-2x right
on no
so how did u getthe numerator
\[y'=\sqrt{9-x}-\frac{x}{\sqrt{9-x}}\]
is that part ok?
oh so u subtract
you want to put i to over the same denominator, so you multiply top and bottom of the first term by \[\sqrt{9-x}\]
\[y'=\sqrt{9-x}-\frac{x}{\sqrt{9-x}}\] \[y'=\frac{\sqrt{9-x}}{\sqrt{9-x}}\times \sqrt{9-x}-\frac{x}{\sqrt{9-x}}\]
how come wolfram has: -3(x-6)/2(9-x)^1/2 ?

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