Answer
$f'(x)=-\frac{4c^2x}{(x^2+c^2)^2}$
Work Step by Step
Using the quotient rule: $fâ(x)=(\frac{u(x)}{v(x)})'=\frac{u'(x)v(x)-v'(x)u(x)}{(v(x))^2}$
$u(x)=(c^2-x^2); u'(x)=-2x$
$v(x)=(c^2+x^2); v'(x)=2x$
$f'(x)=\frac{(-2x)(c^2+x^2)-(2x)(c^2-x^2)}{(c^2+x^2)^2}$
$=\frac{(-2c^2x-2x^3)+(-2c^2x+2x^3)}{(x^2+c^2)^2}$
$=-\frac{4c^2x}{(x^2+c^2)^2}$