dig in chain rule

textbook/chainRule/digInChainRule.tex

@@ -2,10 +2,20 @@
 
 \input{../preamble.tex}
 
+	\outcome{Recognize a composition of functions.}
+	\outcome{Take derivatives of compositions of functions using the chain rule.}
+	\outcome{Take derivatives that require the use of multiple derivative rules.}
+	\outcome{Use the chain rule to calculate derivatives from a table of values.}
+	\outcome{Understand rate of change when quantities are dependent upon each other.}
+	\outcome{Use order of operations in situations requiring multiple derivative rules.}
+    \outcome{Justify the chain rule via the composition of linear approximations.}
+    \outcome{Apply chain rule to relate quantities expressed with different units.}
+
 \title[Dig-In:]{The chain rule}
 
 \begin{document}
 \begin{abstract}
+Here we compute derivatives of compositions of functions
 \end{abstract}
 \maketitle
 
@@ -24,7 +34,7 @@
 
 While there are several different ways to differentiate this function,
 if we let $f(x) = x^5$ and $g(x) = 1+2x$, then we can express $h(x) =
-f(g(x))$. The question is, can we compute the derivative of a
+\answer[given]{f}(\answer[given]{g}(x))$. The question is, can we compute the derivative of a
 composition of functions using the derivatives of the constituents
 $f(x)$ and $g(x)$? To do so, we need the \textit{chain rule}.
 
@@ -186,18 +196,19 @@
 \ddx (1+2x)^5
 \]
 
-
+\begin{explanation}
 Set $f(x) = x^5$ and $g(x) = 1+2x$, now
 \[
-f'(x) = 5x^4 \qquad\text{and}\qquad g'(x) = 2.
+f'(x) = \answer[given]{5x^4} \qquad\text{and}\qquad g'(x) = \answer[given]{2}.
 \]
 Hence
 \begin{align*}
 \ddx (1+2x)^5 &= \ddx f(g(x))\\ 
 &=f'(g(x))g'(x) \\
-&= 5(1+2x)^4\cdot 2\\
+&= 5(\answer[given]{1+2x})^4\cdot \answer[given]{2}\\
 &= 10(1+2x)^4.
 \end{align*}
+\end{explanation}
 \end{example}
 
 
@@ -209,20 +220,21 @@
 \ddx \sqrt{1+\sqrt{x}}
 \]
 
-
+\begin{explanation}
 Set 
 $f(x)=\sqrt{x}$ and $g(x)=1+x$. Hence,
 \[
-\sqrt{1+\sqrt{x}}=f(g(f(x))) \qquad\text{and}\qquad\ddx f(g(f(x))) = f'(g(f(x)))g'(f(x))f'(x).
+\sqrt{1+\sqrt{x}}=f(g(\answer[given]{f}(x))) \qquad\text{and}\qquad\ddx f(g(f(x))) = f'(g(f(x)))g'(f(x))f'(x).
 \]
 Since 
 \[
-f'(x) = \frac{1}{2\sqrt{x}} \qquad\text{and}\qquad g'(x) = 1
+f'(x) = \answer[given]{\frac{1}{2\sqrt{x}}} \qquad\text{and}\qquad g'(x) = \answer[given]{1}
 \]
 We have that
 \[
-\ddx \sqrt{1+\sqrt{x}} = \frac{1}{2\sqrt{1+\sqrt{x}}}\cdot 1\cdot  \frac{1}{2\sqrt{x}}.
+\ddx \sqrt{1+\sqrt{x}} = \frac{1}{2\sqrt{1+\sqrt{x}}}\cdot 1\cdot  \answer[given]{\frac{1}{2\sqrt{x}}}.
 \]
+\end{explanation}
 \end{example}
 
 Using the chain rule, the power rule, and the product rule it is
@@ -234,19 +246,20 @@
 \ddx \frac{x^3}{x^2+1}
 \]
 
-
+\begin{explanation}
 Rewriting this as 
 \[
 \ddx x^3(x^2+1)^{-1}, 
 \]
-set $f(x) = x^{-1}$ and $g(x) = x^2+1$. Now
+set $f(x) = \answer[given]{x^{-1}}$ and $g(x) = \answer[given]{x^2+1}$ so that $f(g(x)) = (x^2 + 1)^{-1}$. Now
 \[
-x^3(x^2+1)^{-1} = x^3 f(g(x)) \qquad\text{and}\qquad \ddx x^3 f(g(x)) = 3x^2f(g(x))+ x^3 f'(g(x))g'(x).
+x^3(x^2+1)^{-1} = x^3 f(g(x)) \qquad\text{and}\qquad \ddx x^3 f(g(x)) = \answer[given]{3x^2} \cdot f(g(x))+ \answer[given]{x^3} \cdot f'(g(x))g'(x).
 \]
-Since $f'(x) = \frac{-1}{x^2}$ and $g'(x) = 2x$, write
+Since $f'(x) = \answer[given]{\frac{-1}{x^2}}$ and $g'(x) = \answer[given]{2x}$, write
 \[
 \ddx \frac{x^3}{x^2+1} = \frac{3x^2}{x^2+1}-\frac{2x^4}{(x^2+1)^2}.
 \]
+\end{explanation}
 \end{example}
 
 \end{document}