Mathematics

OpenStudy (anonymous):
f(x)' x√x

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OpenStudy (anonymous):
hmm?

OpenStudy (diyadiya):
\[\Large \sqrt{x}=x^ \frac{1}{2}\]Do you know this?

OpenStudy (anonymous):
ya
.. i know derivite of that is 1/2√x

OpenStudy (diyadiya):
The easier way is to do by this metho\[\Large x \sqrt{x}= x.x^ \frac{1}{2}\]

OpenStudy (anonymous):
ok.. then?

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OpenStudy (diyadiya):
\[\Large x^m \times x^n = x^{m+n}\]

OpenStudy (diyadiya):
\[\Large x \times x^ \frac{1}{2}= ?\]

OpenStudy (anonymous):
x^3/2

OpenStudy (anonymous):
0_o.. didn;t think of that..but aren't u supposed to use roduct rule?

OpenStudy (anonymous):
product rule

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OpenStudy (anonymous):
no product rule is not applicable here
3/2 x^1/2 answer

OpenStudy (anonymous):
now that makes sense 0_o

OpenStudy (diyadiya):
Yes you can use product rule ,You'll get the same answer

OpenStudy (anonymous):
why not tough?

OpenStudy (anonymous):
:o ok

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OpenStudy (anonymous):
thx

OpenStudy (anonymous):
\[3/2x^{1/2}\]

OpenStudy (diyadiya):
\[\frac{d}{dx}x \sqrt{x}=x \frac{d}{dx}\sqrt{x}+\sqrt{x}\frac{d}{dx}x\]\[=x .\frac{1}{2\sqrt{x}}+\sqrt{x}=\frac{x}{2\sqrt{x}}+\sqrt{x} =\frac{\sqrt{x}}{2}+\sqrt{x}\]\[=\frac{\sqrt{x}}{2}+ \frac{\sqrt{x}}{1}=\frac{\sqrt{x}}{2}+\frac{\sqrt{x} \times 2 }{1 \times 2} =\frac{\sqrt{x}}{2}+\frac{2\sqrt{x}}{2}=\frac{3\sqrt{x}}{2}\]

OpenStudy (diyadiya):
Using the Product Rule^^

OpenStudy (anonymous):
wow diya you are really good at maths :)

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OpenStudy (diyadiya):
\[\Large \frac{3}{2}x^ \frac{1}{2}= \frac{3}{2}\sqrt{x}= \frac{3\sqrt{x}}{2}\]Same answer! :)

OpenStudy (diyadiya):
Lol Thanks Annas!

OpenStudy (anonymous):
welcome !

OpenStudy (anonymous):
lol ye.. ur really good

OpenStudy (anonymous):
hmm hey, if u dont mind can u help me with another question =/?

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OpenStudy (diyadiya):
Post another question :)

OpenStudy (anonymous):
ok