Answer
$f'(x)=\dfrac{2cx}{(x^{2}+c)^{2}}$
Work Step by Step
$f(x)=\dfrac{x}{x+\dfrac{c}{x}}$ (Here $c$ is a constant)
Let's use algebra to make some changes to the function:
$f(x)=\dfrac{x}{x+\dfrac{c}{x}}=\dfrac{x}{\dfrac{x^{2}+c}{x}}=\dfrac{x^{2}}{x^{2}+c}$
Differentiate using the quotient rule:
$f'(x)=\dfrac{(x^{2}+c)(x^{2})'-(x^{2})(x^{2}+c)'}{(x^{2}+c)^{2}}=\dfrac{(x^{2}+c)(2x)-(x^{2})(2x)}{(x^{2}+c)^{2}}$
Simplify
$...=\dfrac{2x^{3}+2cx-2x^{3}}{(x^{2}+c)^{2}}=\dfrac{2cx}{(x^{2}+c)^{2}}$
