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anonymous
 4 years ago
derivative of y=x(9x)^1/2?
anonymous
 4 years ago
derivative of y=x(9x)^1/2?

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anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0product rule for this one

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0this is what I have: x[1/x(9x)^1/2]+(9x)^1/2(1)

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0if so how to simplify

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0\[y'=1\times \sqrt{9x}+x\times \frac{1}{\sqrt{9x}}\times 1\] and then clean up with some algebra

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0really not much to clean up, just \[y'=\sqrt{9x}\frac{x}{\sqrt{9x}}\]

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0or if you want to put it over one denominator, you can

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0\[\frac{9xx}{\sqrt{9x}}\] \[\frac{92x}{\sqrt{9x}}\]

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0if you want to "simplify" things usually best not to use exponents, but rather radicals

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0so√(9x) *x equals 92x right

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0so how did u getthe numerator

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0\[y'=\sqrt{9x}\frac{x}{\sqrt{9x}}\]

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0you want to put i to over the same denominator, so you multiply top and bottom of the first term by \[\sqrt{9x}\]

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0\[y'=\sqrt{9x}\frac{x}{\sqrt{9x}}\] \[y'=\frac{\sqrt{9x}}{\sqrt{9x}}\times \sqrt{9x}\frac{x}{\sqrt{9x}}\]

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0how come wolfram has: 3(x6)/2(9x)^1/2 ?
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