Answer
(a) $u'(1) = \frac{3}{4}$
(b) $v'(1)$ does not exist
(c) $w'(1) = -2$
Work Step by Step
(a) $u(x) = f(g(x))$
$u'(x) = f'(g(x))\cdot g'(x)$
$u'(1) = f'(g(1))\cdot g'(1)$
$u'(1) = f'(3)\cdot g'(1)$
$u'(1) = (-\frac{1}{4})\cdot (-3)$
$u'(1) = \frac{3}{4}$
(b) $v(x) = g(f(x))$
$v'(x) = g'(f(x))\cdot f'(x)$
$v'(1) = g'(f(1))\cdot f'(1)$
$v'(1) = g'(2)\cdot f'(1)$
Since the graph of $g$ has a sharp point at $x=2,$ then $g'(2)$ does not exist.
$v'(1)$ does not exist
(c) $w(x) = g(g(x))$
$w'(x) = g'(g(x))\cdot g'(x)$
$w'(1) = g'(g(1))\cdot g'(1)$
$w'(1) = g'(3)\cdot g'(1)$
$w'(1) = (\frac{2}{3})\cdot (-3)$
$w'(1) = -2$
