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