Answer
a) The first image is the first graph of $f(x) = \sqrt{6-x}$
b) The second image is the graph of $f'(x) = \sqrt{6-x} = \frac{-1}{2\sqrt{6-x}}$
c) $f'(x) = \sqrt{6-x} = \frac{-1}{2\sqrt{6-x}}$
The domain of $f(x)$ = $(-\infty, 6]$
The domain of $f'(x)$ = $(-\infty, 6)$
Work Step by Step
a) The first image is the graph of $ f(x) = \sqrt{6-x}$. We obtain this by using the transformations: reflect $\sqrt{x}$ around $y$-axis and move right 6 units.
b) $f(x) = \sqrt{6-x}$
$f'(x) = \frac{d}{dx} (6-x)^{\frac{1}{2}}$
$f'(x) = \frac{1}{2} (6-x)^{\frac{1}{2} - 1} \times (-1)$
$f'(x) = \frac{1}{2} (6-x)^{-\frac{1}{2}} \times (-1)$
$f'(x) = \frac{1}{2\sqrt{6-x}} \times (-1)$
$f'(x) = \frac{-1}{2\sqrt{6-x}}$
(See second image.)
(c) $f'(x) = \lim\limits_{h \to 0} \frac{f(x+h) - f(x)}{h}$
$f'(x) = \lim\limits_{h \to 0} \frac{f(\sqrt{6-(x+h)}) - f(\sqrt{6-x})}{h}$
$f'(x) = \lim\limits_{h \to 0} \frac{\sqrt{6-x-h} - \sqrt{6-x}}{h}$
$f'(x) = \lim\limits_{h \to 0} \frac{\sqrt{6-x-h} - \sqrt{6-x}}{h} \times \frac{\sqrt{6-x-h} + \sqrt{6-x}}{\sqrt{6-x-h} + \sqrt{6-x}}$
$f'(x) = \lim\limits_{h \to 0} \frac{(\sqrt{6-x-h})^2 - (\sqrt{6-x})^2}{h(\sqrt{6-x-h} + \sqrt{6-x})}$
$f'(x) = \lim\limits_{h \to 0} \frac{(6-x-h) - (6-x)}{h(\sqrt{6-x-h} + \sqrt{6-x})}$
$f'(x) = \lim\limits_{h \to 0} \frac{6-x-h - 6+x}{h(\sqrt{6-x-h} + \sqrt{6-x})}$
$f'(x) = \lim\limits_{h \to 0} \frac{-h}{h(\sqrt{6-x-h} + \sqrt{6-x})}$
$f'(x) = \lim\limits_{h \to 0} \frac{-1}{\sqrt{6-x-h} + \sqrt{6-x}}$
Replace $h$ for $0$
$f'(x) = \frac{-1}{\sqrt{6-x-0} + \sqrt{6-x}}$
$f'(x) = \frac{-1}{\sqrt{6-x} + \sqrt{6-x}}$
$f'(x) = \frac{-1}{2\sqrt{6-x}}$
Domain of $f(x)$ is $(-\infty, 6]$
Domain of $f'(x)$ is $(-\infty, 6)$
