Single Variable Calculus, Chapter 3, 3.5, Section 3.5, Problem 65

Suppose that $f$ and $g$ are the functions whose graphs are shown, let $u(x) = f(g(x)), v(x) = g(f(x))$ and $w(x) = g(g(x))$. Find each derivative, if it exist, explain why.







a.) $u'(1)$


$\begin{equation} \begin{aligned} u'(x) =& f'(g(x)) g'(x) \\ \\ u'(1) =& f'(g(1)) g'(1) \\ \\ u'(1) =& f'(3) g'(1) \\ \\ u'(1) =& \left( \frac{3 - 4}{6 - 1} \right) \left( \frac{0 - 6}{2 - 0} \right) \\ \\ u'(1) =& \left( \frac{-1}{4} \right) (-3) \\ \\ u'(1) =& \frac{3}{4} \end{aligned} \end{equation}$


b.) $v' (1)$


$\begin{equation} \begin{aligned} & v'(x) = g'(f(x)) f'(x) \\ \\ & v'(1) = g'(f(1)) f'(1) \\ \\ & v'(1) = g'(2) f'(1) \\ \\ & v'(1) \text{ does not exist because $g'(2)$ doesn't exist.} \end{aligned} \end{equation}$



c.) $w' (1)$


$\begin{equation} \begin{aligned} w'(x) =& g'(g(x)) g'(x) \\ \\ w'(1) =& g'(g(1)) g'(1) \\ \\ w'(1) =& g'(3) g'(1) \\ \\ w'(1) =& \left( \frac{2 - 0}{5 - 2} \right) \left( \frac{0 - 6}{2 - 0} \right) \\ \\ w'(1) =& \left( \frac{2}{3} \right) (-3) \\ \\ w'(1) =& -2 \end{aligned} \end{equation}$