From 79b10f886eb18b2e168a3230483b54b7ee376b61 Mon Sep 17 00:00:00 2001 From: Hans Parshall Date: Mon, 27 Jul 2015 15:54:59 -0400 Subject: [PATCH] dig in products/quotients --- .gitignore | 2 + .../digInProductRuleAndQuotientRule.tex | 40 ++++++++++++++----- 2 files changed, 31 insertions(+), 11 deletions(-) diff --git a/.gitignore b/.gitignore index 97bb9a618..d54f30e9e 100644 --- a/.gitignore +++ b/.gitignore @@ -4,3 +4,5 @@ *.log *.out *.toc +*.synctex.gz +*.jax \ No newline at end of file diff --git a/textbook/productAndQuotientRules/digInProductRuleAndQuotientRule.tex b/textbook/productAndQuotientRules/digInProductRuleAndQuotientRule.tex index 0f2a87bd3..37ad1d8f4 100644 --- a/textbook/productAndQuotientRules/digInProductRuleAndQuotientRule.tex +++ b/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}