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How can you simplify 1/(3sqrt(1-x^2/9))? | HIX Tutor

note: This question comes originally from <a href="http://socratic.org/questions/what-is-the-derivative-of-y-arcsin-x-3">this page</a>.
To simplify this expression I brought the <mathjax>#3#</mathjax> "inside" the radical.
<mathjax>#3#</mathjax> is equivalent to <mathjax>#sqrt(9)#</mathjax>. Thus,
1.) <mathjax>#1/(3sqrt(1-x^2/9))#</mathjax>
is equivalent to
2.) <mathjax>#1/(sqrt(9)*sqrt(1-x^2/9))#</mathjax>.
A common law of radicals is the law of "combining," which basically looks like:
<mathjax>#sqrt(A)*sqrt(B) = sqrt(A*B)#</mathjax>
where <mathjax>#A#</mathjax> and <mathjax>#B#</mathjax> can be anything.
Using this law we can simplify our expression a little further.
2.) <mathjax>#1/(sqrt(9)*sqrt(1-x^2/9))#</mathjax>
will become:
3.) <mathjax>#1/(sqrt(9(1-x^2/9)))#</mathjax>
Now, all that's left is to distribute the <mathjax>#9#</mathjax>. This gives us:
4.) <mathjax>#1/(sqrt(9 - x^2)#</mathjax>
which looks a lot less ugly than what we started with in 1 .

note: This question comes originally from this page.
To simplify this expression I brought the <mathjax>#3#</mathjax> "inside" the radical.
<mathjax>#3#</mathjax> is equivalent to <mathjax>#sqrt(9)#</mathjax>. Thus,
1.) <mathjax>#1/(3sqrt(1-x^2/9))#
is equivalent to
2.) <mathjax>#1/(sqrt(9)*sqrt(1-x^2/9))#</mathjax>.
A common law of radicals is the law of "combining," which basically looks like:
<mathjax>#sqrt(A)*sqrt(B) = sqrt(A*B)#
where <mathjax>#A#</mathjax> and <mathjax>#B#</mathjax> can be anything.
Using this law we can simplify our expression a little further.
2.) <mathjax>#1/(sqrt(9)*sqrt(1-x^2/9))#
will become:
3.) <mathjax>#1/(sqrt(9(1-x^2/9)))#
Now, all that's left is to distribute the <mathjax>#9#</mathjax>. This gives us:
4.) <mathjax>#1/(sqrt(9 - x^2)#
which looks a lot less ugly than what we started with in 1 .

How can you simplify #1/(3sqrt(1-x^2/9))#?