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

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