Question 1117134
{{{y}}} is a function of a function of {{{x}}} .
The function {{{g(x)=(2x-1)/(x^2+1)}}} operates on {{{x}}} ,
and then a mysterious {{{f(g)}}} function, such that {{{"f ' ( x )"=sin(x^2)}}}  operates on the result.
You calculate the derivative of a function like that using the “chain rule.”
 
{{{dg/dx}}}{{{"="}}}{{{(2(x^2+1)-2x(2x-1))/(x^2+1)^2}}}{{{"="}}}{{{(2x^2+2-4x^2+2x)/(x^2+1)^2}}}{{{"="}}}{{{(-2x^2+2x+2)/(x^2+1)^2}}}{{{"="}}}{{{-2(x^2-x-1)/(x^2+1)^2}}}
 
It was hard to wrap my head around this problem,
but if I did not get myself hopeless confused,
{{{dy/dx}}}{{{"="}}}{{{(dy/d("( 2x-1 ) /"(x^2+1)))(d("( 2x-1 ) /"(x^2+1))/dx)}}}{{{"="}}}{{{"f ' "}}}{{{((2x-1)/(x^2+1))(-2(x^2-x-1)/(x^2+1)^2)}}}{{{"="}}}{{{(sin((2x-1)^2/(x^2+1)^2))}}}{{{(-2(x^2-x-1)/(x^2+1)^2)}}}{{{"="}}}{{{(-2(x^2-x-1)/(x^2+1)^2)sin((2x-1)^2/(x^2+1)^2)}}} .