Single Variable Calculus, Chapter 3, 3.3, Section 3.3, Problem 41

Differentiate $\displaystyle f(x) = \frac{x }{x + \displaystyle \frac{c}{x}} $



$
\begin{equation}
\begin{aligned}

f(x) =& \frac{x }{x + \displaystyle \frac{c}{x}}
&& \text{Get the LCD on the denominator}
\\
\\
f(x) =& \frac{x}{\displaystyle \frac{x^2 + c}{x}}
&& \text{Simplify the equation}
\\
\\
f(x) =& \frac{x^2}{x^2 + c}
&& \text{}
\\
\\
f'(x) =& \frac{(x^2 + c) \displaystyle \frac{d}{dx} (x^2) - \left[ (x^2) \frac{d}{dx} (x^2 + c) \right]}{(x^2 + c)^2}
&& \text{Apply Quotient Rule}
\\
\\
f'(x) =& \frac{(x^2 + c) (2x) - [(x^2)(2x)]}{(x^2 + c)^2}
&& \text{Expand the equation}
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\\
f'(x) =& \frac{\cancel{2x^3} + 2cx - \cancel{2x^3}}{(x^2 + c)^2}
&& \text{Combine like terms}
\\
\\
f'(x) =& \frac{2cx}{(x^2 + c)^2}
&& \text{}

\end{aligned}
\end{equation}
$