Tag Archives: derivative

A recent post in the AP Calculus Community expressed some confusion about different ways to compute $\displaystyle \frac{dy}{dx}$ at (0,4) for the function $x=2ln(y-3)$.  I share below the two approaches suggested in the original post, proffer two more, and a slightly more in-depth activity I’ve used in my calculus classes for years.  I conclude with an alternative to derivatives of inverses.

Two Approaches Initially Proposed

1 – Accept the function as posed and differentiate implicitly.

$\displaystyle \frac{d}{dx} \left( x = 2 ln(y-3) \right)$

$\displaystyle 1 = 2*\frac{1}{y-3} * \frac{dy}{dx}$

$\displaystyle \frac{dy}{dx} = \frac{y-3}{2}$

Which gives $\displaystyle \frac{dy}{dx} = \frac{1}{2}$ at (x,y)=(0,4).

2 – Solve for y and differentiate explicitly.

$\displaystyle x = 2ln(y-3) \longrightarrow y = 3 + e^{x/2}$

$\displaystyle \frac{dy}{dx} = e^{x/2} * \frac{1}{2}$

Evaluating this at (x,y)=(0,4) gives $\displaystyle \frac{dy}{dx} = \frac{1}{2}$ .

Two Alternative Approaches

3 – Substitute early.

The question never asked for an algebraic expression of $\frac{dy}{dx}$, only the numerical value of this slope.  Because students tend to make more silly mistakes manipulating algebraic expressions than numeric ones, the additional algebra steps are unnecessary, and potentially error-prone.  Admittedly, the manipulations are pretty straightforward here, in more algebraically complicated cases, early substitutions could significantly simplify work. Using approach #1 and substituting directly into the second line gives

$\displaystyle 1 = 2 * \frac{1}{y-3} * \frac{dy}{dx}$.

At (x,y)=(0,4), this is

$\displaystyle 1 = 2 * \frac{1}{4-3}*\frac{dy}{dx}$

$\displaystyle \frac{dy}{dx} = \frac{1}{2}$

The numeric manipulations on the right side are obviously easier than the earlier algebra.

4 – Solve for $\frac{dx}{dy}$ and reciprocate.

There’s nothing sacred about solving for $\frac{dy}{dx}$ directly.  Why not compute the derivative of the inverse and reciprocate at the end? Differentiating first with respect to y eventually leads to the same solution.

$\displaystyle \frac{d}{dy} \left( x = 2 ln(y-3) \right)$

$\displaystyle \frac{dx}{dy} = 2 * \frac{1}{y-3}$

At (x,y)=(0,4), this is

$\displaystyle \frac{dx}{dy} = \frac{2}{4-3} = 2$, so

$\displaystyle \frac{dy}{dx} = \frac{1}{2}$.

Equivalence = A fundamental mathematical concept

I sometimes wonder if teachers should place much more emphasis on equivalence.  We spend so much time manipulating expressions in mathematics classes at all levels, changing mathematical objects (shapes, expressions, equations, etc.) into a different, but equivalent objects.  Many times, these manipulations are completed under the guise of “simplification.”  (Here is a brilliant Dan Teague post cautioning against taking this idea too far.)

But it is critical for students to recognize that proper application of manipulations creates equivalent expressions, even if when the resulting expressions don’t look the same.   The reason we manipulate mathematical objects is to discover features about the object in one form that may not be immediately obvious in another.

For the function $x = 2 ln(y-3)$, the slope at (0,4) must be the same, no matter how that slope is calculated.  If you get a different looking answer while using correct manipulations, the final answers must be equivalent.

Another Example

A similar question appeared on the AP Calculus email list-server almost a decade ago right at the moment I was introducing implicit differentiation.  A teacher had tried to find $\displaystyle \frac{dy}{dx}$ for

$\displaystyle x^2 = \frac{x+y}{x-y}$

using implicit differentiation on the quotient, manipulating to a product before using implicit differentiation, and finally solving for y in terms of x to use an explicit derivative.

1 – Implicit on a quotient

Take the derivative as given:$$\displaystyle \frac{d}{dx} \left( x^2 = \frac{x+y}{x-y} \right)$ $\displaystyle 2x = \frac{(x-y) \left( 1 + \frac{dy}{dx} \right) - (x+y) \left( 1 - \frac{dy}{dx} \right) }{(x-y)^2}$ $\displaystyle 2x * (x-y)^2 = (x-y) + (x-y)*\frac{dy}{dx} - (x+y) + (x+y)*\frac{dy}{dx}$ $\displaystyle 2x * (x-y)^2 = -2y + 2x * \frac{dy}{dx}$ $\displaystyle \frac{dy}{dx} = \frac{-2x * (x-y)^2 + 2y}{2x}$ 2 – Implicit on a product Multiplying the original equation by its denominator gives $x^2 * (x - y) = x + y$ . Differentiating with respect to x gives $\displaystyle 2x * (x - y) + x^2 * \left( 1 - \frac{dy}{dx} \right) = 1 + \frac{dy}{dx}$ $\displaystyle 2x * (x-y) + x^2 - 1 = x^2 * \frac{dy}{dx} + \frac{dy}{dx}$ $\displaystyle \frac{dy}{dx} = \frac{2x * (x-y) + x^2 - 1}{x^2 + 1}$ 3 – Explicit Solving the equation at the start of method 2 for y gives $\displaystyle y = \frac{x^3 - x}{x^2 + 1}$. Differentiating with respect to x gives $\displaystyle \frac{dy}{dx} = \frac {\left( x^2+1 \right) \left( 3x^2 - 1\right) - \left( x^3 - x \right) (2x+0)}{\left( x^2 + 1 \right) ^2}$ Equivalence Those 3 forms of the derivative look VERY DIFFERENT. Assuming no errors in the algebra, they MUST be equivalent because they are nothing more than the same derivative of different forms of the same function, and a function’s rate of change doesn’t vary just because you alter the look of its algebraic representation. Substituting the y-as-a-function-of-x equation from method 3 into the first two derivative forms converts all three into functions of x. Lots of by-hand algebra or a quick check on a CAS establishes the suspected equivalence. Here’s my TI-Nspire CAS check. Here’s the form of this investigation I gave my students. Final Example I’m not a big fan of memorizing anything without a VERY GOOD reason. My teachers telling me to do so never held much weight for me. I memorized as little as possible and used that information as long as I could until a scenario arose to convince me to memorize more. One thing I managed to avoid almost completely were the annoying derivative formulas for inverse trig functions. For example, find the derivative of $y = arcsin(x)$ at $x = \frac{1}{2}$. Since arc-trig functions annoy me, I always rewrite them. Taking sine of both sides and then differentiating with respect to x gives. $sin(y) = x$ $\displaystyle cos(y) * \frac{dy}{dx} = 1$ I could rewrite this equation to give $\frac{dy}{dx} = \frac{1}{cos(y)}$, a perfectly reasonable form of the derivative, albeit as a less-common expression in terms of y. But I don’t even do that unnecessary algebra. From the original function, $x=\frac{1}{2} \longrightarrow y=\frac{\pi}{6}$, and I substitute that immediately after the differentiation step to give a much cleaner numeric route to my answer. $\displaystyle cos \left( \frac{\pi}{6} \right) * \frac{dy}{dx} = 1$ $\displaystyle \frac{\sqrt{3}}{2} * \frac{dy}{dx} = 1$ $\displaystyle \frac{dy}{dx} = \frac{2}{\sqrt{3}}$ And this is the same result as plugging $x = \frac{1}{2}$ into the memorized version form of the derivative of arcsine. If you like memorizing, go ahead, but my mind remains more nimble and less cluttered. One final equivalent approach would have been differentiating $sin(y) = x$ with respect to y and reciprocating at the end. CONCLUSION There are MANY ways to compute derivatives. For any problem or scenario, use the one that makes sense or is computationally easiest for YOU. If your resulting algebra is correct, you know you have a correct answer, even if it looks different. Be strong! Base-x Numbers and Infinite Series In my previous post, I explored what happened when you converted a polynomial from its variable form into a base-x numerical form. That is, what are the computational implications when polynomial $3x^3-11x^2+2$ is represented by the base-x number $3(-11)02_x$, where the parentheses are used to hold the base-x digit, -11, for the second power of x? So far, I’ve explored only the Natural number equivalents of base-x numbers. In this post, I explore what happens when you allow division to extend base-x numbers into their Rational number counterparts. Level 5–Infinite Series: Numbers can have decimals, so what’s the equivalence for base-x numbers? For starters, I considered trying to get a “decimal” form of $\displaystyle \frac{1}{x+2}$. It was “obvious” to me that $12_x$ won’t divide into $1_x$. There are too few “places”, so some form of decimals are required. Employing division as described in my previous post somewhat like you would to determine the rational number decimals of $\frac{1}{12}$ gives Remember, the places are powers of x, so the decimal portion of $\displaystyle \frac{1}{x+2}$ is $0.1(-2)4(-8)..._x$, and it is equivalent to $\displaystyle 1x^{-1}-2x^{-2}+4x^{-3}-8x^{-4}+...=\frac{1}{x}-\frac{2}{x^2}+\frac{4}{x^3}-\frac{8}{x^4}+...$. This can be seen as a geometric series with first term $\displaystyle \frac{1}{x}$ and ratio $\displaystyle r=\frac{-2}{x}$. It’s infinite sum is therefore $\displaystyle \frac{\frac{1}{x}}{1-\frac{-2}{x}}$ which is equivalent to $\displaystyle \frac{1}{x+2}$, confirming the division computation. Of course, as a geometric series, this is true only so long as $\displaystyle |r|=\left | \frac{-2}{x} \right |<1$, or $2<|x|$. I thought this was pretty cool, and it led to lots of other cool series. For example, if $x=8$,you get $\frac{1}{10}=\frac{1}{8}-\frac{2}{64}+\frac{4}{512}-...$. Likewise, $x=3$ gives $\frac{1}{5}=\frac{1}{3}-\frac{2}{9}+\frac{4}{27}-\frac{8}{81}+...$. I found it quite interesting to have a “polynomial” defined with a rational expression. Boundary Convergence: As shown above, $\displaystyle \frac{1}{x+2}=\frac{1}{x}-\frac{2}{x^2}+\frac{4}{x^3}-\frac{8}{x^4}+...$ only for $|x|>2$. At $x=2$, the series is obviously divergent, $\displaystyle \frac{1}{4} \ne \frac{1}{2}-\frac{2}{4}+\frac{4}{8}-\frac{8}{16}+...$. For $x=-2$, I got $\displaystyle \frac{1}{0} = \frac{1}{-2}-\frac{2}{4}+\frac{4}{-8}-\frac{8}{16}+...=-\frac{1}{2}-\frac{1}{2}-\frac{1}{2}-\frac{1}{2}-...$ which is properly equivalent to $-\infty$ as $x \rightarrow -2$ as defined by the convergence domain and the graphical behavior of $\displaystyle y=\frac{1}{x+2}$ just to the left of $x=-2$. Nice. I did find it curious, though, that $\displaystyle \frac{1}{x}-\frac{2}{x^2}+\frac{4}{x^3}-\frac{8}{x^4}+...$ is a solid approximation for $\displaystyle \frac{1}{x+2}$ to the left of its vertical asymptote, but not for its rotationally symmetric right side. I also thought it philosophically strange (even though I understand mathematically why it must be) that this series could approximate function behavior near a vertical asymptote, but not near the graph’s stable and flat portion near $x=0$. What a curious, asymmetrical approximator. Maclaurin Series: Some quick calculus gives the Maclaurin series for $\displaystyle \frac{1}{x+2}$ : $\displaystyle \frac{1}{2}-\frac{x}{4}+\frac{x^2}{8}-\frac{x^3}{16}+...$, a geometric series with first term $\frac{1}{2}$ and ratio $\frac{-x}{2}$. Interestingly, the ratio emerging from the Maclaurin series is the reciprocal of the ratio from the “rational polynomial” resulting from the base-x division above. As a geometric series, the interval of convergence is $\displaystyle |r|=\left | \frac{-x}{2} \right |<1$, or $|x|<2$. Excluding endpoint results, the Maclaurin interval is the complete Real number complement to the base-x series. For the endpoints, $x=-2$ produces the right-side vertical asymptote divergence to $+ \infty$ that $x=-2$ did for the left side of the vertical asymptote in the base-x series. Again, $x=2$ is divergent. It’s lovely how these two series so completely complement each other to create clean approximations of $\displaystyle \frac{1}{x+2}$ for all $x \ne 2$. Other base-x “rational numbers” Because any polynomial divided by another is absolutely equivalent to a base-x rational number and thereby a base-x decimal number, it will always be possible to create a “rational polynomial” using powers of $\displaystyle \frac{1}{x}$ for non-zero denominators. But, the decimal patterns of rational base-x numbers don’t apply in the same way as for Natural number bases. Where $\displaystyle \frac{1}{12}$ is guaranteed to have a repeating decimal pattern, the decimal form of $\displaystyle \frac{1}{x+2}=\frac{1_x}{12_x}=0.1(-2)4(-8)..._x$ clearly will not repeat. I’ve not explored the full potential of this, but it seems like another interesting field. CONCLUSIONS and QUESTIONS Once number bases are understood, I’d argue that using base-x multiplication might be, and base-x division definitely is, a cleaner way to compute products and quotients, respectively, for polynomials. The base-x division algorithm clearly is accessible to Algebra II students, and even opens the doors to studying series approximations to functions long before calculus. Is there a convenient way to use base-x numbers to represent horizontal translations as cleanly as polynomials? How difficult would it be to work with a base-$(x-h)$ number for a polynomial translated h units horizontally? As a calculus extension, what would happen if you tried employing division of non-polynomials by replacing them with their Taylor series equivalents? I’ve played a little with proving some trig identities using base-x polynomials from the Maclaurin series for sine and cosine. What would happen if you tried to compute repeated fractions in base-x? It’s an open question from my perspective when decimal patterns might terminate or repeat when evaluating base-x rational numbers. I’d love to see someone out there give some of these questions a run! Calculus Derivative Rules Over the past few days I’ve been rethinking my sequencing of introducing derivative rules for the next time I teach calculus. The impetus for this was an approach I encountered in a Coursera MOOC in Calculus I’m taking this summer to see how a professor would run a Taylor Series-centered calculus class. Historically, I’ve introduced my high school calculus classes to the product and quotient rules before turing to the chain rule. I’m now convinced the chain rule should be first because of how beautifully it sets up the other two. Why the chain rule should be first Assuming you know the chain rule, check out these derivations of the product and quotient rules. For each of these, $g_1$ and $g_2$ can be any differentiable functions of x. PRODUCT RULE: Let $P(x)=g_1(x) \cdot g_2(x)$. Applying a logarithm gives, $ln(P)=ln \left( g_1 \cdot g_2 \right) = ln(g_1)+ln(g_2)$. Now differentiate and rearrange. $\displaystyle \frac{P'}{P} = \frac{g_1'}{g_1}+\frac{g_2'}{g_2}$ $\displaystyle P' = P \cdot \left( \frac{g_1'}{g_1}+\frac{g_2'}{g_2} \right)$ $\displaystyle P' = g_1 \cdot g_2 \cdot \left( \frac{g_1'}{g_1}+\frac{g_2'}{g_2} \right)$ $P' = g_1' \cdot g_2+g_1 \cdot g_2'$ QUOTIENT RULE: Let $Q(x)=\displaystyle \frac{g_1(x)}{g_2(x)}$. As before, apply a logarithm, differentiate, and rearrange. $\displaystyle ln(Q)=ln \left( \frac{g_1}{g_2} \right) = ln(g_1)-ln(g_2)$ $\displaystyle \frac{Q'}{Q} = \frac{g_1'}{g_1}-\frac{g_2'}{g_2}$ $\displaystyle Q' = Q \cdot \left( \frac{g_1'}{g_1}-\frac{g_2'}{g_2} \right)$ $\displaystyle Q' = \frac{g_1}{g_2} \cdot \left( \frac{g_1'}{g_1}-\frac{g_2'}{g_2} \right)$ $\displaystyle Q' = \frac{g_1'}{g_2}-\frac{g_1 \cdot g_2'}{\left( g_2 \right)^2} = \frac{g_1'g_2-g_1g_2'}{\left( g_2 \right)^2}$ The exact same procedure creates both rules. (I should have seen this long ago.) Proposed sequencing I’ve always emphasized the Chain Rule as the critical algebra manipulation rule for calculus students, but this approach makes it the only rule required. That completely fits into my overall teaching philosophy: learn a limited set of central ideas and use them as often as possible. With this, I’ll still introduce power, exponential, sine, and cosine derivative rules first, but then I’ll follow with the chain rule. After that, I think everything else required for high school calculus will be a variation on what is already known. That’s a lovely bit of simplification. I need to rethink my course sequencing, but I think it’ll be worth it. Polar Derivatives on TI-Nspire CAS The following question about how to compute derivatives of polar functions was posted on the College Board’s AP Calculus Community bulletin board today. From what I can tell, there are no direct ways to get derivative values for polar functions. There are two ways I imagined to get the polar derivative value, one graphically and the other CAS-powered. The CAS approach is much more accurate, especially in locations where the value of the derivative changes quickly, but I don’t think it’s necessarily more intuitive unless you’re comfortable using CAS commands. For an example, I’ll use $r=2+3sin(\theta )$ and assume you want the derivative at $\theta = \frac{\pi }{6}$. METHOD 1: Graphical Remember that a derivative at a point is the slope of the tangent line to the curve at that point. So, finding an equation of a tangent line to the polar curve at the point of interest should find the desired result. Create a graphing window and enter your polar equation (menu –> 3:Graph Entry –> 4:Polar). Then drop a tangent line on the polar curve (menu –> 8:Geometry –> 1:Points&Lines –> 7:Tangent). You would then click on the polar curve once to select the curve and a second time to place the tangent line. Then press ESC to exit the Tangent Line command. To get the current coordinates of the point and the equation of the tangent line, use the Coordinates & Equation tool (menu –> 1:Actions –> 8:Coordinates and Equations). Click on the point and the line to get the current location’s information. After each click, you’ll need to click again to tell the nSpire where you want the information displayed. To get the tangent line at $\theta =\frac{\pi }{6}$, you could drag the point, but the graph settings seem to produce only Cartesian coordinates. Converting $\theta =\frac{\pi }{6}$ on $r=2+3sin(\theta )$ to Cartesian gives $\left( x,y \right) = \left( r \cdot cos(\theta ), r \cdot sin(\theta ) \right)=\left( \frac{7\sqrt{3}}{4},\frac{7}{4} \right)$ . So the x-coordinate is $\frac{7\sqrt{3}}{4} \approx 3.031$. Drag the point to find the approximate slope, $\frac{dy}{dx} \approx 8.37$. Because the slope of the tangent line changes rapidly at this location on this polar curve, this value of 8.37 will be shown in the next method to be a bit off. Unfortunately, I tried to double-click the x-coordinate to set it to exactly $\frac{7\sqrt{3}}{4}$, but that property is also disabled in polar mode. METHOD 2: CAS Using the Chain Rule, $\displaystyle \frac{dy}{dx} = \frac{dy}{d\theta }\cdot \frac{d\theta }{dx} = \frac{\frac{dy}{d\theta }}{\frac{dx}{d\theta }}$. I can use this and the nSpire’s ability to define user-created functions to create a $\displaystyle \frac{dy}{dx}$ polar differentiator for any polar function $r=a(\theta )$. On a Calculator page, I use the Define function (menu –> 1:Actions –> 1:Define) to make the polar differentiator. All you need to do is enter the expression for a as shown in line 2 below. This can be evaluated exactly or approximately at $\theta=\frac{\pi }{6}$ to show $\displaystyle \frac{dy}{dx} = 5\sqrt{3}=\approx 8.660$. Conclusion: As with all technologies, getting the answers you want often boils down to learning what questions to ask and how to phrase them. Controlling graphs and a free online calculator When graphing functions with multiple local features, I often find myself wanting to explain a portion of the graph’s behavior independent of the rest of the graph. When I started teaching a couple decades ago, the processor on my TI-81 was slow enough that I could actually watch the pixels light up sequentially. I could see HOW the graph was formed. Today, processors obviously are much faster. I love the problem-solving power that has given my students and me, but I’ve sometimes missed being able to see function graphs as they develop. Below, I describe the origins of the graph control idea, how the control works, and then provide examples of polynomials with multiple roots, rational functions with multiple intercepts and/or vertical asymptotes, polar functions, parametric collision modeling, and graphing derivatives of given curves. BACKGROUND: A colleague and I were planning a rational function unit after school last week wanting to be able to create graphs in pieces so that we could discuss the effect of each local feature. In the past, we “rigged” calculator images by graphing the functions parametrically and controlling the input values of t. Clunky and static, but it gave us useful still shots. Nice enough, but we really wanted something dynamic. Because we had the use of sliders on our TI-nSpire software, on Geogebra, and on the Desmos calculator, the solution we sought was closer than we suspected. REALIZATION & WHY IT WORKS: Last week, we discovered that we could use $g(x)=\sqrt \frac{\left | x \right |}{x}$ to create what we wanted. The argument of the root is 1 for $x<0$, making $g(x)=1$. For $x>0$, the root’s argument is -1, making $g(x)=i$, a non-real number. Our insight was that multiplying any function $y=f(x)$ by an appropriate version of g wouldn’t change the output of f if the input to g is positive, but would make the product ungraphable due to complex values if the input to g is negative. If I make a slider for parameter a, then $g_2(x)=\sqrt \frac{\left | a-x \right |}{a-x}$ will have output 1 for all $x. That means for any function $y=f(x)$ with real outputs only, $y=f(x)\cdot g_2(x)$ will have real outputs (and a real graph) for $x only. Aha! Using a slider and $g_2$ would allow me to control the appearance of my graph from left to right. NOTE: While it’s still developing, I’ve become a big fan of the free online Desmos calculator after a recent presentation at the Global Math Department (join our 45-60 minute online meetings every Tuesday at 9PM ET!). I use Desmos for all of the following graphs in this post, but obviously any graphing software with slider capabilities would do. EXAMPLE 1: Graph $y=(x+2)^3x^2(x-1)$, a 6th degree polynomial whose end behavior is up for $\pm \infty$, “wiggles” through the x-axis at -2, then bounces off the origin, and finally passes through the x-axis at 1. Click here to access the Desmos graph that created the image above. You can then manipulate the slider to watch the graph wiggle through, then bounce off, and finally pass through the x-axis. EXAMPLE 2: Graph $y=\frac{(x+1)^2}{(x+2)(x-1)^2}$, a 6th degree polynomial whose end behavior is up for $\pm \infty$, “wiggles” through the x-axis at -2, then bounces off the origin, and finally passes through the x-axis at 1. Click here to access the Desmos graph above and control the creation of the rational function’s graph using a slider. EXAMPLE 3: I believe students understand polar graphing better when they see curves like the limacon $r=2+3cos(\theta )$ moving between its maximum and minimum circles. Controlling the slider also allows users to see the values of $\theta$ at which the limacon crosses the pole. Here is the Desmos graph for the graph below. EXAMPLE 4: Object A leaves (2,3) and travels south at 0.29 units/second. Object B leaves (-2,1) traveling east at 0.45 units/second. The intersection of their paths is (2,1), but which object arrives there first? Here is the live version. OK, I know this is an overly simplistic example, but you’ll get the idea of how the controlling slider works on a parametrically-defined function. The$latex \sqrt{\frac{\left | a-x \right |}{a-x}}\$ term only needs to be on one of parametric equations.  Another benefit of the slider approach is the ease with which users can identify the value of t (or time) when each particle reaches the point of intersection or their axes intercepts.  Obviously those values could be algebraically determined in this problem, but that isn’t always true, and this graphical-numeric approach always gives an alternative to algebraic techniques when investigating parametric functions.

ASIDE 1–Notice the ease of the Desmos notation for parametric graphs.  Enter [r,s] where r is the x-component of the parametric function and s is the y-component.  To graph a point, leave r and s as constants.  Easy.

EXAMPLE 5:  When teaching calculus, I always ask my students to sketch graphs of the derivatives of functions given in graphical forms.  I always create these graphs one part at a time.  As an example, this graph shows $y=x^3+2x^2$ and allows you to get its derivative gradually using a slider.

ASIDE 2–It is also very easy to enter derivatives of functions in the Desmos calculator.  Type “d/dx” before the function name or definition, and the derivative is accomplished.  Desmos is not a CAS, so I’m sure the software is computing derivatives numerically.  No matter.  Derivatives are easy to define and use here.

I’m hoping you find this technology tip as useful as I do.

Exponential Derivatives and Statistics

This post gives a different way I developed years ago to determine the form of the derivative of exponential functions, $y=b^x$.  At the end, I provide a copy of the document I use for this activity in my calculus classes just in case that’s helpful.  But before showing that, I walk you through my set-up and solution of the problem of finding exponential derivatives.

Background:

I use this lesson after my students have explored the definition of the derivative and have computed the algebraic derivatives of polynomial and power functions. They also have access to TI-nSpire CAS calculators.

The definition of the derivative is pretty simple for polynomials, but unfortunately, the definition of the derivative is not so simple to resolve for exponential functions.  I do not pretend to teach an analysis class, so I see my task as providing strong evidence–but not necessarily a watertight mathematical proof–for each derivative rule.  This post definitely is not a proof, but its results have been pretty compelling for my students over the years.

Sketching Derivatives of Exponentials:

At this point, my students also have experience sketching graphs of derivatives from given graphs of functions.  They know there are two basic graphical forms of exponential functions, and conclude that there must be two forms of their derivatives as suggested below.

When they sketch their first derivative of an exponential growth function, many begin to suspect that an exponential growth function might just be its own derivative.  Likewise, the derivative of an exponential decay function might be the opposite of the parent function.  The lack of scales on the graphs obviously keep these from being definitive conclusions, but the hypotheses are great first ideas.  We clearly need to firm things up quite a bit.

Numerically Computing Exponential Derivatives:

Starting with $y=10^x$, the students used their CASs to find numerical derivatives at 5 different x-values.  The x-values really don’t matter, and neither does the fact that there are five of them.  The calculators quickly compute the slopes at the selected x-values.

Each point on $f(x)=10^x$ has a unique tangent line and therefore a unique derivative.  From their sketches above, my students are soundly convinced that all ordered pairs $\left( x,f'(x) \right)$ form an exponential function.  They’re just not sure precisely which one. To get more specific, graph the points and compute an exponential regression.

So, the derivatives of $f(x)=10^x$ are modeled by $f'(x)\approx 2.3026\cdot 10^x$.  Notice that the base of the derivative function is the same as its parent exponential, but the coefficient is different.  So the common student hypothesis is partially correct.

Now, repeat the process for several other exponential functions and be sure to include at least 1 or 2 exponential decay curves.  I’ll show images from two more below, but ultimately will include data from all exponential curves mentioned in my Scribd document at the end of the post.

The following shows that $g(x)=5^x$ has derivative $g'(x)\approx 1.6094\cdot 5^x$.  Notice that the base again remains the same with a different coefficient.

OK, the derivative of $h(x)=\left( \frac{1}{2} \right)^x$ causes a bit of a hiccup.  Why should I make this too easy?  <grin>

As all of its $h'(x)$ values are negative, the semi-log regression at the core of an exponential regression is impossible.  But, I also teach my students regularly that If you don’t like the way a problem appears, CHANGE IT!  Reflecting these data over the x-axis creates a standard exponential decay which can be regressed.

From this, they can conclude that  $h'(x)\approx -0.69315\cdot \left( \frac{1}{2} \right)^x$.

So, every derivative of an exponential function appears to be another exponential function whose base is the same as its parent function with a unique coefficient.  Obviously, the value of the coefficient depends on the base of the corresponding parent function.  Therefore, each derivative’s coefficient is a function of the base of its parent function.  The next two shots show the values of all of the coefficients and a plot of the (base,coefficient) ordered pairs.

OK, if you recognize the patterns of your families of functions, that data pattern ought to look familiar–a logarithmic function.  Applying a logarithmic regression gives

For $y=a+b\cdot ln(x)$, $a\approx -0.0000067\approx 0$ and $b=1$, giving $coefficient(base) \approx ln(base)$.

Therefore, $\frac{d}{dx} \left( b^x \right) = ln(b)\cdot b^x$.

Again, this is not a formal mathematical proof, but the problem-solving approach typically keeps my students engaged until the end, and asking my students to  discover the derivative rule for exponential functions typically results in very few future errors when computing exponential derivatives.

Feedback on the approach is welcome.

Classroom Handout:

Here’s a link to a Scribd document written for my students who use TI-nSpire CASs.  There are a few additional questions at the end.  Hopefully this post and the document make it easy enough for you to adapt this to the technology needs of your classroom.  Enjoy.

An unexpected lesson from technology

This discovery happened a few years ago, but I’ve just started ‘blogging, so I guess it’s time to share this for the “first” time.

I forget whether my calculus class at the time was using the first version of the TI-Nspire CAS or if we were still on the TI-89, but I had planned a very brief introduction to the CAS syntax for computing symbolic derivatives, but my 5-minute introduction in the first week of introducing algebraic rules of derivatives ended up with my students discovering antiderivative rules simply because they had technology tools which allowed them to explore beyond where their teacher had intended them to go.

They had absolutely no problem computing algebraic derivatives of power functions, so the following example was used not to demonstrate the power of CAS, but to give easily confirmed outputs.  I asked them for the derivative of $x^5$, and their CAS gave the top line of the image below.

(In case there are readers who are TI-Nspire CAS users who don’t know the shortcut for computing higher order derivatives, use the left arrow to place the cursor after the dx in the “denominator” of $\frac{d}{dx}$ and press the carot (^) key.  Then type the integer of the derivative you want.)

I wanted my students to compute the 2nd and 3rd derivatives and confirm the power rule which they did with the following screen.

That was the extent of what I wanted at the time–to establish that a CAS could quickly and easily confirm algebraic results whether or not a “teacher” was present.  Students could create as many practice problems as were appropriate for themselves and get their solutions confirmed immediately by a non-judgmental expert.  Of course, one of my students began to explore in ways my “trained” mind had long ago learned not to do.  In my earlier days of CAS, I had forgotten the unboundedness of mathematical exploration.

Shortly after my syntax 5-minutes had passed and I had confirmed everyone could handle it, a young man called me to his desk to show me the following.

He understood what a 1st or 2nd derivative was, but what in the world was a negative 1st derivative?  Rather than answering, I posed it to the class who pondered a few moments before recognizing that “underivatives” (as they called them in that moment) of power functions added one to the current exponent before dividing by the new exponent.  They had discovered and explained (at least algebraically) antiderivatives long before I had intended.  Technology actually inspired and extended my students’ learning!

Then I asked them what the CAS would give if we asked it for a 0th derivative.  It was another great technology-inspired discussion.

I really need to explore more about the connections between mathematics as a language and the parallel language of technology.

Math doesn’t happen the way it’s printed

Learning is messy.  Real math is messy.

If you think about it, there shouldn’t be any wonder why students learning a math idea for the first time get frustrated when the work they produce looks almost nothing like the pristine, sharp, and short solutions provided by textbooks and their classroom teachers.

Mathematics is discovered through pattern recognition or trial-and-error or pushing the boundaries of a situation or equation to explore those great “What if…” questions.  Discovery usually is just plain messy, but by the time mathematical writing gets into publication, it has been edited and refined far beyond its messy origins.  We sweep under the publishing rug all of the mistakes and dead ends that taught us so much, delivering our final results all dressed up in tight, pithy expressions.

My latest reminder of this was an exploration of the product rule in my calculus classes this week.  I’ll give a short-and-sweet “textbook” proof of this at the end of this post, but I’ll start with their first exploration of discovering a rule.

At the start of yesterday’s class, my students understood

• the definition of the derivative,
• if a function is differentiable at $x=a$, then sufficiently zooming in on the function at that point essentially shows the tangent line to the function at that point,
• the derivative rule of power functions, and
• if a function was horizontally translated then its derivative experienced the same horizontal translation.

So how do you get the product rule for some $y=f(x)*g(x)$ from that?

Assuming f and g are differentiable at some point $x=a$, then $f(x) \approx f'(a)*(x-a)+f(a)$ and $g(x) \approx g'(a)*(x-a)+g(a)$ for values of x near $x=a$.  Therefore,

$\frac{d}{dx}(f(x)*g(x)) \approx \frac{d}{dx}([f'(a)*(x-a)+f(a)][g'(a)*(x-a)+g(a)])$
$\approx \frac{d}{dx}(f'(a)g'(a)(x-a)^2+(f'(a)g(a)+f(a)g'(a))(x-a)+f(a)g(a))$
$\approx 2f'(a)g'(a)(x-a)+(f'(a)g(a)+f(a)g'(a))+0$

The derivatives of $(x-a)^2$ and $(x-a)$ are the translated derivatives of $x^2$ and $x$–possibilities with their early knowledge without any need for the chain rule.

Therefore, $\frac{d}{dx}(f(x)*g(x))|_{x=a}=$
$=2f'(a)g'(a)(a-a)+(f'(a)g(a)+f(a)g'(a))$
$=f'(a)g(a)+f(a)g'(a)$, the product rule!

It was not the easiest or cleanest of early investigations for my students yesterday.  AND there was lots of potential fudging around the concept of local linearity. BUT … now that the rule has been named, it can be proved  more formally, something we attempted today.
By definition of the derivative,
$\displaystyle \frac{d}{dx}(f(x)*g(x))=\lim_{h\to 0}\frac{f(x+h)g(x+h)-f(x)g(x)}{h}$
But if an $f'(x)$ term is to emerge from this, then the difference quotient must contain an isolated $\frac{f(x+h)-f(x)}{h}$ term.  The only $f(x+h)$ in the original derivative expression also contains an $g(x+h)$ term, so subtracting $f(x)g(x+h)$ will allow the $g(x+h)$ to factor out.  To balance that subtraction, $f(x)g(x+h)$ also needs to be added back.

From there, the proof becomes$\displaystyle \frac{d}{dx}(f(x)*g(x))=\lim_{h\to 0}\frac{f(x+h)g(x+h)-f(x)g(x)}{h}$
$\displaystyle =\lim_{h\to 0}\frac{f(x+h)g(x+h)-f(x)g(x+h)+f(x)g(x+h)-f(x)g(x)}{h}$
$\displaystyle =\lim_{h\to 0}g(x+h)*\frac{f(x+h)-f(x)}{h}+f(x)*\frac{g(x+h)-g(x)}{h}$
$\displaystyle =f'(x)g(x)+f(x)g'(x)$

It’s pretty, but the addition and subtraction of $f(x)g(x+h)$ is completely black box or black magic if you don’t have a reason to do it.  Hopefully, the reasoning above provides such a reason, but it is a result of deep reasoning.  But that’s the reality of most published mathematics, and THAT is what students see as their production expectation the first time they try.  And THAT is one reason why many students get frustrated with mathematics.  The work you produce when you are learning is rarely (if ever) so pretty.

Students need room to be creative, they need room to experiment, and they need to know the math they produce doesn’t need to look pretty when first created.

CAS Derivatives

Here’s a fun problem from my calculus class today, enhanced by CAS. As a set-up, our last unit focused on interpreting the meaning of derivatives with multiple interpretations of the definition of the derivative as the only algebraic work they’ve done.  In that unit, the students discovered that vertical translations on functions didn’t change their derivatives, and horizontal translations on functions changed the corresponding derivatives by the same horizontal translation.  From their work with derivatives of power functions using a definition of the derivative, they hypothesized and proved $\frac{d}{dx}(x^n)=n*x^{n-1}$ for natural (and a few other) values of n.  Knowing nothing else, I posed this.

Use your CAS, determine the derivatives of $y=ln(x)$, $y=ln(2x)$, $y=ln(3x)$, and $y=ln(4x)$.  Use your results to hypothesize the derivative of $y=ln(n*x)$.  Justify your claim.

The following image from the first part shows that the pattern is easy to spot.

Unfortunately, I posed the problem with only 10 minutes remaining in class, but the students clearly knew $\frac{d}{dx}(ln(n*x))=\frac{1}{x}$, but the looming question wasWHY.  With a couple minutes to spare, one guessed rules of logarithms might apply, but not having used them since their first semester exam, he didn’t recall the property.  Some colleagues may argue that I should have insisted on my students having those rules memorized in advance, but I firmly believe that this particular problem actually gave a reason for my students to relearn their logarithm properties.

I let the awkward moment hang there until another called out with glee,
$ln(n*x)=ln(n)+ln(x)$“, to which a third exclaimed, “and $ln(n)$ is a constant, making the derivative of $ln(n*x)$ the same as the derivative of
$ln(x)$,” clearly using her understanding of the effect of translations on functions and their derivatives that she learned in the last unit.

Two other nice ideas emerged:

1. They thought it convenient that $\frac{d}{dx}(ln(n*x))=\frac{1}{x}$, but now really want to know why.
2. A few observant ones noticed that $ln(x)$ and $\frac{1}{x}$ have different domains.  To these, I pointed out the warning at the bottom of the CAS screen above that most had completely overlooked when getting their initial answers.

These questions still linger for the class, but I argue that the use of CAS in my calculus class this afternoon

• left a need in many to discover why $\frac{d}{dx}(ln(n*x))=\frac{1}{x}$, and
• raised domain issues that ultimately will lead to a deeper understanding for the existence of the absolute value in $\int{\frac{1}{x}dx}=ln(|x|)$.