Question 990668
<pre>

{{{y}}}{{{""=""}}}{{{x^2/sqrt(x^2-1)}}}

{{{y}}}{{{""=""}}}{{{x^2/(x^2-1)^(1/2)}}}

We could differentiate it as a quotient but it's a 
little easier to first turn it into a product:

{{{y}}}{{{""=""}}}{{{x^2(x^2-1)^(-1/2)}}}

{{{(dy)/(dx)}}}{{{""=""}}}{{{x^2(-1/2)(x^2-1)^(-3/2)(2x)}}}{{{""+""}}}{{{(x^2-1)^(-1/2)*(2x) }}}

{{{(dy)/(dx)}}}{{{""=""}}}{{{x^2(-1/cross(2))(x^2-1)^(-3/2)(cross(2)x)}}}{{{""+""}}}{{{(x^2-1)^(-1/2)*(2x) }}}

{{{(dy)/(dx)}}}{{{""=""}}}{{{-x^3(x^2-1)^(-3/2)}}}{{{""+""}}}{{{(x^2-1)^(-1/2)*(2x) }}}

A clever trick we often play here is to factor out the smallest exponent, the
{{{-3/2}}} power.

When we factor the {{{-3/2}}} power out of the {{{-1/2}}} power what's left inside is 
the difference of the exponents {{{(-1/2)-(-3/2)=-1/2+3/2=2/2=1}}}
It's just the first power!

We factor {{{x(x^2-1)^(-3/2)}}} out:

{{{(dy)/(dx)}}}{{{""=""}}}{{{x(x^2-1)^(-3/2) (-x^2+2(x^2-1)^1) }}}

{{{(dy)/(dx)}}}{{{""=""}}}{{{x(x^2-1)^(-3/2) (-x^2+2(x^2-1)^"") }}}

{{{(dy)/(dx)}}}{{{""=""}}}{{{x(x^2-1)^(-3/2) (-x^2+2x^2-2) }}}

{{{(dy)/(dx)}}}{{{""=""}}}{{{x(x^2-1)^(-3/2) (x^2-2) }}}

We write the negative exponent in the bottom as a positive exponent:

{{{(dy)/(dx)}}}{{{""=""}}}{{{x(x^2-2)/(x^2-1)^(3/2) }}}

Edwin</pre>