## Calculus: Early Transcendentals 8th Edition

$f'(x)=\dfrac{2cx}{(x^{2}+c)^{2}}$
$f(x)=\dfrac{x}{x+\dfrac{c}{x}}$ (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 $...=\dfrac{2x^{3}+2cx-2x^{3}}{(x^{2}+c)^{2}}=\dfrac{2cx}{(x^{2}+c)^{2}}$