I assume you know a few properties of the derivative, namely the product and quotient rules.
Product rule:
h(x) = f(x) g(x)
h'(x) = f'(x) g(x) + f(x) g'(x)
Quotient rule:
k(x) = f(x) / g(x)
k'(x) = ( f'(x) g(x) - f(x) g'(x) ) / g(x) ²
For j(x), simply swap f and g.
(9) Use the plot to approximate f (1) = 2 and g (1) = 3.
Both f and g are made up of line segments. For 0 < x < 2, it looks like f passes through the points (0, 0) and (2, 4), which means the line segment over this interval has slope 2, so f'(x) = 2 for all x between 0 and 2. For 2 < x < 4, f passes through (2, 4) and (4, 0), so the slope is f'(x) = -2 for all x between 2 and 4.
Meanwhile, g passes through the points (0, 4) and (4, 0), so the slope is -1.
This tells us that, by the product rule,
h' (1) = f' (1) g (1) + f (1) g' (1) = 2•3 + 2•(-1) = 4
Next, the plot tells us f (2) = 4 and g (2) = 2. When x = 2, however, f has a sudden sharp turn, so f' (2) does not exist, and thus
h' (2) does not exist
According to the plot, f (3) = 2 and g (3) = 1, so
h' (3) = (-2)•1 + 2•(-1) = -4
(10) Use the quotient rule and the values underlined above:
k' (1) = ( f' (1) g (1) - f (1) g' (1) ) / g (1)² = (2•3 - 2•(-1)) / 3² = 8/9
k' (2) does not exist
k' (3) = ((-2)•1 - 2•(-1)) / 3² = 0
(11) Same procedure as in (10):
j' (1) = ( g' (1) f (1) - g (1) f' (1) ) / f (1)² = ((-1)•2 - 3•2) / 2² = -2
j' (2) does not exist
j' (3) = ((-1)•2 - 1•(-2)) / 2² = 0