Single Variable Calculus, Chapter 3, Review Exercises, Section Review Exercises, Problem 12

a.) Suppose that $f(x) = \sqrt{3-5x}$, use the definition of a derivative to find $f'(x)$ using the definition of a derivative.
Using the definition of a derivative
$\displaystyle f'(x) = \lim\limits_{h \to 0} \frac{f(x+h) - f(x) }{h}$

$
\begin{equation}
\begin{aligned}
f'(x) &= \lim\limits_{h \to 0} \frac{\sqrt{3-5(x+h)}-\sqrt{3-5x}}{h}\\
\\
f'(x) &= \lim\limits_{h \to 0} \frac{\sqrt{3-5x-5h}-\sqrt{3-5x}}{h} \cdot \frac{\sqrt{3-5x-5h}+\sqrt{3-5x}}{\sqrt{3-5x-5h}+\sqrt{3-5x}}\\
\\
f'(x) &= \lim\limits_{h \to 0} \frac{3-5x-5h(3-5x)}{h\left(\sqrt{3-5x-5h}+\sqrt{3-5x} \right)}\\
\\
f'(x) &= \lim\limits_{h \to 0} \frac{\cancel{3} - \cancel{5x} - 5h - \cancel{3} + \cancel{5x}}{h\left(\sqrt{3-5x-5h}+\sqrt{3-5x} \right)}\\
\\
f'(x) &= \lim\limits_{h \to 0} \frac{-5\cancel{h}}{\cancel{h}\left(\sqrt{3-5x-5h}+\sqrt{3-5x} \right)}\\
\\
f'(x) &= \lim\limits_{h \to 0} \left( \frac{-5}{\sqrt{3-5x-5h}+\sqrt{3-5x}} \right) = \frac{-5}{\sqrt{3-5x-5(0)}+\sqrt{3-5x}} = \frac{-5}{\sqrt{3-5x}+\sqrt{3-5x}}\\
\\
f'(x) &= \frac{-5}{2 \sqrt{3-5x}}
\end{aligned}
\end{equation}
$

b.) Determine the domains of $f$ and $f'$.
The function $f$ involves a root function that can not have a negative value. So,

$
\begin{equation}
\begin{aligned}
3-5x & \leq 0 \\
3 & \leq 5x\\
x & \leq \frac{3}{5}
\end{aligned}
\end{equation}
$

Therefore, the domain of $f$ is $\displaystyle \left( -\inf, \frac{3}{5} \right]$. The function $f'$ is a rational function which has a root function on the denominator, so the value should bot be equal to zero and does not have a negative value. Therefore, the domain of $f'$ is $\displaystyle \left( -\infty, \frac{3}{5} \right)$.