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DiyadiyaBest ResponseYou've already chosen the best response.3
\[\Large \sqrt{x}=x^ \frac{1}{2}\]Do you know this?
 one year ago

abz_techBest ResponseYou've already chosen the best response.0
ya .. i know derivite of that is 1/2√x
 one year ago

DiyadiyaBest ResponseYou've already chosen the best response.3
The easier way is to do by this metho\[\Large x \sqrt{x}= x.x^ \frac{1}{2}\]
 one year ago

DiyadiyaBest ResponseYou've already chosen the best response.3
\[\Large x^m \times x^n = x^{m+n}\]
 one year ago

DiyadiyaBest ResponseYou've already chosen the best response.3
\[\Large x \times x^ \frac{1}{2}= ?\]
 one year ago

abz_techBest ResponseYou've already chosen the best response.0
0_o.. didn;t think of that..but aren't u supposed to use roduct rule?
 one year ago

annasBest ResponseYou've already chosen the best response.2
no product rule is not applicable here 3/2 x^1/2 answer
 one year ago

abz_techBest ResponseYou've already chosen the best response.0
now that makes sense 0_o
 one year ago

DiyadiyaBest ResponseYou've already chosen the best response.3
Yes you can use product rule ,You'll get the same answer
 one year ago

DiyadiyaBest ResponseYou've already chosen the best response.3
\[\frac{d}{dx}x \sqrt{x}=x \frac{d}{dx}\sqrt{x}+\sqrt{x}\frac{d}{dx}x\]\[=x .\frac{1}{2\sqrt{x}}+\sqrt{x}=\frac{x}{2\sqrt{x}}+\sqrt{x} =\frac{\sqrt{x}}{2}+\sqrt{x}\]\[=\frac{\sqrt{x}}{2}+ \frac{\sqrt{x}}{1}=\frac{\sqrt{x}}{2}+\frac{\sqrt{x} \times 2 }{1 \times 2} =\frac{\sqrt{x}}{2}+\frac{2\sqrt{x}}{2}=\frac{3\sqrt{x}}{2}\]
 one year ago

DiyadiyaBest ResponseYou've already chosen the best response.3
Using the Product Rule^^
 one year ago

annasBest ResponseYou've already chosen the best response.2
wow diya you are really good at maths :)
 one year ago

DiyadiyaBest ResponseYou've already chosen the best response.3
\[\Large \frac{3}{2}x^ \frac{1}{2}= \frac{3}{2}\sqrt{x}= \frac{3\sqrt{x}}{2}\]Same answer! :)
 one year ago

abz_techBest ResponseYou've already chosen the best response.0
lol ye.. ur really good
 one year ago

abz_techBest ResponseYou've already chosen the best response.0
hmm hey, if u dont mind can u help me with another question =/?
 one year ago

DiyadiyaBest ResponseYou've already chosen the best response.3
Post another question :)
 one year ago
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