Ask your own question, for FREE!
Mathematics
OpenStudy (anonymous):

f(x)' x√x

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?

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

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

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 :)

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 =/?

OpenStudy (diyadiya):

Post another question :)

OpenStudy (anonymous):

ok

Can't find your answer? Make a FREE account and ask your own questions, OR help others and earn volunteer hours!

Join our real-time social learning platform and learn together with your friends!
Can't find your answer? Make a FREE account and ask your own questions, OR help others and earn volunteer hours!

Join our real-time social learning platform and learn together with your friends!