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

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



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