Single Variable Calculus, Chapter 3, Review Exercises, Section Review Exercises, Problem 10

Determine $f'(x)$ of the function $\displaystyle f(x) = \frac{4-x}{3+x}$ using the definition of a derivative.
Using the definition of derivative



$
\begin{equation}
\begin{aligned}
f'(x) &= \lim_{h \to 0} \frac{f(x+h) - f(x) }{h}\\
\\
f'(x) &= \lim_{h \to 0} \frac{\left[\frac{4-(x+h)}{3+x+h}\right] - \left( \frac{4-x}{3+x}\right) }{h}\\
\\
f'(x) &= \lim_{h \to 0} \frac{\frac{4-x-h}{3+x+h}-\frac{4-x}{3+x} }{h}\\
\\
f'(x) &= \lim_{h \to 0} \frac{}{} \frac{\frac{(4-x-h)(3+x)-(3+x+h)(4-x)}{(3+x+h)(3+x)} }{h}\\
\\
f'(x) &= \lim_{h \to 0} \frac{\cancel{12} - \cancel{3x} - 3h + \cancel{4x} - \cancel{x^2} - \cancel{xh} - \cancel{12} + \cancel{3x} - \cancel{4x} + \cancel{x^2} - 4h + \cancel{xh} }{h(3+x+h)(3+x)}\\
\\
f'(x) &= \lim_{h \to 0} \frac{-7\cancel{h}}{\cancel{h} (3+x+h)(3+x)}\\
\\
f'(x) &= \lim_{h \to 0} \left[\frac{-7}{(3+x+h)(3+x)}\right] = \frac{-7}{(3+x+0)(3+x)} = \frac{-7}{(3+x)(3+x)}\\
\\
f'(x) &= \frac{-7}{(3+x)^2}
\end{aligned}
\end{equation}
$