Answer
(a) $f'(x) = \frac{10-3x}{2\sqrt{5-x}}$
(b) At the point $(1,2)$:
$y = \frac{7}{4}x+\frac{1}{4}$
At the point $(4,4)$:
$y = -x+8$
(c) We can see a sketch of the graph of $f$ and the tangent lines below.
(d) We can see a sketch of the graphs of $f$ and $f'$ below.
Work Step by Step
(a) $f(x) = x~\sqrt{5-x}$
We can find $f'(x)$:
$f'(x) = \sqrt{5-x}+(x)(\frac{-1}{2~\sqrt{5-x}})$
$f'(x) = \frac{2(5-x)}{2\sqrt{5-x}}-\frac{x}{2~\sqrt{5-x}}$
$f'(x) = \frac{10-3x}{2\sqrt{5-x}}$
(b) We can find the slope of the tangent line at the point $(1,2)$:
$m = f'(1) = \frac{10-3(1)}{2\sqrt{5-(1)}} = \frac{7}{4}$
We can find the equation of the tangent line at the point $(1,2)$:
$y-2 = \frac{7}{4}(x-1)$
$y = \frac{7}{4}x+\frac{1}{4}$
We can find the slope of the tangent line at the point $(4,4)$:
$m = f'(4) = \frac{10-3(4)}{2\sqrt{5-(4)}} = -1$
We can find the equation of the tangent line at the point $(4,4)$:
$y-4 = (-1)(x-4)$
$y = -x+8$
(c) We can see a sketch of the graph and the tangent lines below.
(d) We can see a sketch of the graphs of $f$ and $f'$ below.
The answer to part (a) seems reasonable since $f'(x)$ has negative values when the slope of $f(x)$ has a negative slope, $f'(x)$ is 0 when the slope of $f(x)$ is 0, and $f'(x)$ has positive values when the slope of $f(x)$ has a positive slope.
