# Tag Archives: product rule

## Implicit Derivative Question

Yesterday, this question posted within the AP Calculus Community about implicit derivatives.

Below, I argue why the derivatives MUST be the same, show how four different variations can all be shown to give the same derivative, and provide a final conclusion.

INITIAL INTUITION

The Desmos graph of the given relation, is $y^2 = \frac{x-1}{x+1}$, is shown below.  Logically, it seems that even when the terms of the relation are algebraically rearranged, the graph should be invariant.   The other two forms mentioned in the Community post are on lines 2 and 3.  Lines 4, 5, and 6 show three other variations.  Here is the link to my Desmos graph allowing you to change between the forms to visually confirm the graphical invariance intuition.

If calculus “works”, it also shouldn’t matter how one calculates a derivative.   While the forms of the derivative certainly could LOOK different,  because any point on the invariant graph has the same tangent line no matter what the form of its equation, and the derivative of a relation at a point is the slope of that invariant tangent line, then the derivative also MUST be invariant.

CALCULATING “DIFFERENT” DERIVATIVES

To show the derivatives are fundamentally all the same (as suspected by the initial post), I calculate the derivatives of the equations on lines 1 and 3 given in the initial post as well as my variations on lines 4 and 6.

LINE 1:

Using the Chain Rule on the left and the Quotient Rule on the right gives

LINE 3:

This version is more complicated, requiring the Product Rule in addition to the earlier Chain and Quotient Rules.  In the penultimate line, I used the original equation to substitute for $y^2$ to transform the derivative into the same form as line 1.

LINE 4:

This time, differentiation requires only the Chain and Product Rules.

After the usual substitution for $y^2$, I multiplied both sides by $(x+1)$ to clear the denominator and solved for $y'$, returning the same result.

LINE 6:

This time, the relation is solved for x, resulting in a much more complicated Quotient+Chain Rule calculation, but substituting for $y^2$ and changing the form leads once again to the same answer.

Hopefully this is convincing evidence that all derivative forms can be shown to be equivalent.   If you’re still learning implicit differentiation, I encourage you to show the derivatives from the lines 2 and 5 variations are also equivalent.

CONCLUSION

So which approach is “best”?  In my opinion, it all depends on your personal comfort with algebraic manipulations.  Some prefer to just take a derivative from the given form of $y^2 = \frac{x-1}{x+1}$.  I avoid the more complicated quotient rule whenever I can, so the variation from line 4 would have been my approach.

The cool part is that it doesn’t matter what approach you use, so long as your algebraic manipulations are sound.  You don’t have to accept the form in which a problem is given; CHANGE IT to a form that works for you!

## 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.

## 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.