# Tag Archives: derivative

## 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 $\int{\frac{1}{x}dx}=ln(|x|)$.