## Calculus 8th Edition

$f'(x) = \dfrac{dy}{dx}[\dfrac{x^2}{x^2+c}] = \dfrac{2cx }{(x^2+c)^2}$
Given $f(x) = \dfrac{x}{x+\dfrac{c}{x}}$ and knowing tha $c$ is a constant First, we simplify the expression as shown: 1.) Put each term in $x+\dfrac{c}{x}$ over the common denominator x: $x+\dfrac{c}{x} = \dfrac{x^2}{x} + \dfrac{c}{x} = \dfrac{x^2 + c}{x} \longrightarrow \dfrac{x}{\dfrac{x^2}{x} + \dfrac{c}{x}} = \dfrac{x}{\dfrac{x^2 + c}{x}}$ 2.) Multiply the numerator of $\dfrac{x}{\dfrac{x^2 + c}{x}}$ by the reciprocal of the denominator: $\dfrac{x}{d\frac{x^2 + c}{x}} = \dfrac{x x}{x^2+c} = \dfrac{x^2}{x^2+c}$ After the expression has been simplified, we proceed to differentiate: $f'(x) = \dfrac{dy}{dx}[\dfrac{x^2}{x^2+c}]$ 1.) By the quotient rule we have: $\dfrac{dy}{dx}[\dfrac{x^2}{x^2+c}] = \dfrac{(x^2+c)\times \dfrac{dy}{dx}[x^2] - \dfrac{dy}{dx}[x^2+c] \times (x^2) }{(x^2+c)^2}$ 2.) By the power rule we have: $\dfrac{(x^2+c)\times \dfrac{dy}{dx}[x^2] - \dfrac{dy}{dx}[x^2+c] \times (x^2) }{(x^2+c)^2} = \dfrac{(x^2+c)\times (2x) - (2x+0) \times (x^2) }{(x^2+c)^2}$ 3.)Performing multiplications we have: $\dfrac{(x^2+c)\times (2x) - (2x+0) \times (x^2) }{(x^2+c)^2} = \dfrac{(2x^3 + 2cx) - (2x^3) }{(x^2+c)^2}$ 4.) Simplifying the numerator we have: $\dfrac{2x^3 + 2cx - 2x^3 }{(x^2+c)^2} = \dfrac{2cx }{(x^2+c)^2}$