anonymous
  • anonymous
f(x)' x√x
Mathematics
jamiebookeater
  • jamiebookeater
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anonymous
  • anonymous
hmm?
Diyadiya
  • Diyadiya
\[\Large \sqrt{x}=x^ \frac{1}{2}\]Do you know this?
anonymous
  • anonymous
ya .. i know derivite of that is 1/2√x

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Diyadiya
  • Diyadiya
The easier way is to do by this metho\[\Large x \sqrt{x}= x.x^ \frac{1}{2}\]
anonymous
  • anonymous
ok.. then?
Diyadiya
  • Diyadiya
\[\Large x^m \times x^n = x^{m+n}\]
Diyadiya
  • Diyadiya
\[\Large x \times x^ \frac{1}{2}= ?\]
anonymous
  • anonymous
x^3/2
anonymous
  • anonymous
0_o.. didn;t think of that..but aren't u supposed to use roduct rule?
anonymous
  • anonymous
product rule
anonymous
  • anonymous
no product rule is not applicable here 3/2 x^1/2 answer
anonymous
  • anonymous
now that makes sense 0_o
Diyadiya
  • Diyadiya
Yes you can use product rule ,You'll get the same answer
anonymous
  • anonymous
why not tough?
anonymous
  • anonymous
:o ok
anonymous
  • anonymous
thx
anonymous
  • anonymous
\[3/2x^{1/2}\]
Diyadiya
  • Diyadiya
\[\frac{d}{dx}x \sqrt{x}=x \frac{d}{dx}\sqrt{x}+\sqrt{x}\frac{d}{dx}x\]\[=x .\frac{1}{2\sqrt{x}}+\sqrt{x}=\frac{x}{2\sqrt{x}}+\sqrt{x} =\frac{\sqrt{x}}{2}+\sqrt{x}\]\[=\frac{\sqrt{x}}{2}+ \frac{\sqrt{x}}{1}=\frac{\sqrt{x}}{2}+\frac{\sqrt{x} \times 2 }{1 \times 2} =\frac{\sqrt{x}}{2}+\frac{2\sqrt{x}}{2}=\frac{3\sqrt{x}}{2}\]
Diyadiya
  • Diyadiya
Using the Product Rule^^
anonymous
  • anonymous
wow diya you are really good at maths :)
Diyadiya
  • Diyadiya
\[\Large \frac{3}{2}x^ \frac{1}{2}= \frac{3}{2}\sqrt{x}= \frac{3\sqrt{x}}{2}\]Same answer! :)
Diyadiya
  • Diyadiya
Lol Thanks Annas!
anonymous
  • anonymous
welcome !
anonymous
  • anonymous
lol ye.. ur really good
anonymous
  • anonymous
hmm hey, if u dont mind can u help me with another question =/?
Diyadiya
  • Diyadiya
Post another question :)
anonymous
  • anonymous
ok

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