dig in products/quotients

.gitignore

@@ -4,3 +4,5 @@
 *.log
 *.out
 *.toc
+*.synctex.gz
+*.jax

textbook/productAndQuotientRules/digInProductRuleAndQuotientRule.tex

@@ -2,10 +2,20 @@
 
 \input{../preamble.tex}
 
+\outcome{Identify products of functions.}
+\outcome{Use the product rule to calculate derivatives.}
+\outcome{Identify quotients of functions.}
+\outcome{Use the quotient rule to calculate derivatives.}
+\outcome{Combine derivative rules to take derivatives of more complicated functions.}
+\outcome{Explain the signs of the terms in the numerator of the quotient rule.}
+\outcome{Multiply tangent lines to justify the product rule.}
+\outcome{Use the product and quotient rule to calculate derivatives from a table of values.}
+
 \title[Dig-In:]{The Product Rule and Quotient Rule}
 
 \begin{document}
 \begin{abstract}
+Here we compute derivatives of products and quotients of functions
 \end{abstract}
 \maketitle
 
@@ -20,7 +30,7 @@ \section{The Product Rule}
 where $f(x)=x^2+1$ and $g(x)=x^3-3x$. An obvious guess for the
 derivative of $f(x)g(x)$ is the product of the derivatives:
 \begin{align*}
-f'(x)g'(x) &= (2x)(3x^2-3)\\
+f'(x)g'(x) &= (\answer[given]{2x})(\answer[given]{3x^2-3})\\
 &= 6x^3-6x.
 \end{align*}
 Is this guess correct? We can check by rewriting $f(x)$
@@ -33,7 +43,7 @@ \section{The Product Rule}
 \end{align*} 
 Hence
 \[
-\ddx f(x) g(x) = 5x^4-6x^2-3, 
+\ddx f(x) g(x) = \ddx(x^5 - 2x^3 - 3x) = \answer[given]{5x^4-6x^2-3}, 
 \]
 so we see that 
 \[
@@ -112,7 +122,7 @@ \section{The Product Rule}
 \begin{proof}
 From the limit definition of the derivative, write
 \[
-\ddx (f(x)g(x)) = \lim_{h \to0} \frac{f(x+h)g(x+h) - f(x)g(x)}{h}
+\ddx (f(x)g(x)) = \lim_{h \to0} \frac{f(\answer[given]{x+h})g(\answer[given]{x+h}) - f(\answer[given]{x})g(\answer[given]{x})}{\answer[given]{h}}
 \]
 Now we use the exact same trick we used in the proof of
 Theorem~\ref{theorem:limit-product}, we add $0 = -f(x+h)g(x) + f(x+h)g(x)$:
@@ -138,22 +148,25 @@ \section{The Product Rule}
 \ddx f(x)g(x).
 \]
 
+\begin{explanation}
 Write
 \begin{align*}
 \ddx f(x)g(x) &= f(x)g'(x) + f'(x)g(x)\\
-&=(x^2+1)(3x^2-3) + 2x(x^3-3x).
+&=(x^2+1)(\answer[given]{3x^2-3}) + (\answer[given]{2x})(x^3-3x).
 \end{align*}
 We could stop here---but we should show that expanding this out recovers our previous result. Write
 \begin{align*}
 (x^2+1)(3x^2-3) + 2x(x^3-3x) &= 3x^4-3x^2 +3x^2 -3 + 2x^4-6x^2\\
-&=5x^4-6x^2-3,
+&=\answer[given]{5x^4-6x^2-3},
 \end{align*}
 which is precisely what we obtained before.
+\end{explanation}
 \end{example}
 
 
 
 
+
 \section{The Quotient Rule}
 
 \index{quotient rule} 
@@ -182,7 +195,7 @@ \section{The Quotient Rule}
 then we could use the product rule to complete our proof.  Write
 \begin{align*}
 \ddx\frac{1}{g(x)}&=\lim_{h\to0} \frac{\frac{1}{g(x+h)}-\frac{1}{g(x)}}{h} \\
-&=\lim_{h\to0} \frac{\frac{g(x)-g(x+h)}{g(x+h)g(x)}}{h} \\
+&=\lim_{h\to0} \frac{\frac{g(\answer[given]{x})-g(\answer[given]{x+h})}{g(x+h)g(x)}}{h} \\
 &=\lim_{h\to0} \frac{g(x)-g(x+h)}{g(x+h)g(x)h} \\
 &=\lim_{h\to0} -\frac{g(x+h)-g(x)}{h} \frac{1}{g(x+h)g(x)} \\
 &=-\frac{g'(x)}{g(x)^2}.
@@ -203,11 +216,13 @@ \section{The Quotient Rule}
 \ddx \frac{x^2+1}{x^3-3x}.
 \]
 
+\begin{explanation}
 Write
 \begin{align*}
-\ddx \frac{x^2+1}{x^3-3x} &= \frac{2x(x^3-3x)-(x^2+1)(3x^2-3)}{(x^3-3x)^2}\\
-&=\frac{-x^4-6x^2+3}{(x^3-3x)^2}.
+\ddx \frac{x^2+1}{x^3-3x} &= \frac{2x(\answer[given]{x^3-3x})-(\answer[given]{x^2+1})(3x^2-3)}{(\answer[given]{x^3-3x})^2}\\
+&=\frac{-x^4-6x^2+3}{(\answer[given]{x^3-3x})^2}.
 \end{align*}
+\end{explanation}
 \end{example}
 
 It is often possible to calculate derivatives in more than one way, as
@@ -222,20 +237,23 @@ \section{The Quotient Rule}
 \] 
 in two ways. First using the quotient rule and then using the product
 rule.
-
+\begin{explanation}
 First, we'll compute the derivative using the quotient rule. Write
 \[
-\ddx \frac{625-x^2}{\sqrt{x}} = \frac{\left(-2x\right)\left(\sqrt{x}\right) - (625-x^2)\left(\frac{1}{2}x^{-1/2}\right)}{x}.
+\ddx \frac{625-x^2}{\sqrt{x}} = \frac{\left(-2x\right)\left(\answer[given]{\sqrt{x}}\right) - (\answer[given]{625-x^2})\left(\frac{1}{2}x^{-1/2}\right)}{\answer[given]{x}}.
 \]
+\end{explanation}
+\begin{explanation}
 Second, we'll compute the derivative using the product rule:
 \begin{align*}
 \ddx \frac{625-x^2}{\sqrt{x}} &= \ddx \left(625-x^2\right)x^{-1/2}\\
-&=\left(625-x^2\right)\left(\frac{-x^{-3/2}}{2}\right)+ (-2x)\left(x^{-1/2}\right).
+&=\left(625-x^2\right)\left(\answer[given]{\frac{-x^{-3/2}}{2}}\right)+ (\answer[given]{-2x})\left(x^{-1/2}\right).
 \end{align*}
 With a bit of algebra, both of these simplify to
 \[
 -\frac{3x^2+625}{2x^{3/2}}.
 \]
+\end{explanation}
 \end{example}