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anonymous
 4 years ago
f(x)' x√x
anonymous
 4 years ago
f(x)' x√x

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

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0ya .. i know derivite of that is 1/2√x

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

Diyadiya
 4 years ago
Best ResponseYou've already chosen the best response.3\[\Large x^m \times x^n = x^{m+n}\]

Diyadiya
 4 years ago
Best ResponseYou've already chosen the best response.3\[\Large x \times x^ \frac{1}{2}= ?\]

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.00_o.. didn;t think of that..but aren't u supposed to use roduct rule?

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0no product rule is not applicable here 3/2 x^1/2 answer

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0now that makes sense 0_o

Diyadiya
 4 years ago
Best ResponseYou've already chosen the best response.3Yes you can use product rule ,You'll get the same answer

Diyadiya
 4 years ago
Best ResponseYou've already chosen the best response.3\[\frac{d}{dx}x \sqrt{x}=x \frac{d}{dx}\sqrt{x}+\sqrt{x}\frac{d}{dx}x\]\[=x .\frac{1}{2\sqrt{x}}+\sqrt{x}=\frac{x}{2\sqrt{x}}+\sqrt{x} =\frac{\sqrt{x}}{2}+\sqrt{x}\]\[=\frac{\sqrt{x}}{2}+ \frac{\sqrt{x}}{1}=\frac{\sqrt{x}}{2}+\frac{\sqrt{x} \times 2 }{1 \times 2} =\frac{\sqrt{x}}{2}+\frac{2\sqrt{x}}{2}=\frac{3\sqrt{x}}{2}\]

Diyadiya
 4 years ago
Best ResponseYou've already chosen the best response.3Using the Product Rule^^

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0wow diya you are really good at maths :)

Diyadiya
 4 years ago
Best 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! :)

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0lol ye.. ur really good

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0hmm hey, if u dont mind can u help me with another question =/?

Diyadiya
 4 years ago
Best ResponseYou've already chosen the best response.3Post another question :)
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