New (to me) phenomenon

[JeffM]

Yesterday I was working on answering a question about logarithmic differentiation and came across something I had not noticed before. I did not know if I had made a mistake or if it was a coincidence, but it stopped me from posting an answer.

[MATH]y = a^n, \text { where n is an integer and}[/MATH]
a is a very messy function in x, but one easily disaggretated by logarithms. In the actual problem a was a product of powers of four functions, but let's keep things simple here with

[MATH]a = bc.[/MATH] ................................................................edited

To compute the derivative by hand, there are two obvious ways to go. Power and product rules.

[MATH]y = a^n \implies y' = na^{(n-1)}a'.[/MATH]
But that a' will be ugly because, by hypothesis, a' is supposed to be awful to compute

Logarithmic differentiation.

[MATH]y = a^n \implies ln(y) = n * ln(a) = n\{ln(b) + \ln(c)\} \implies\\ \dfrac{1}{y} * y' = n * \left ( \dfrac{1}{b} * b' + \dfrac{1}{c} * c \right ) \implies y' = ny * \left ( \dfrac{b'}{b} + \dfrac{c'}{c} \right ) \implies \\ y' = na^n * \left ( \dfrac{b'}{b} + \dfrac{c'}{c} \right ).[/MATH]All systems froze. How can a be to the power of n rather than n - 1?

In the context of a very messy problem, that had me really scratching my head. Had I made an error? If so, where? Of course, the solution is obvious in a context this simple.

[MATH]y' = na^n * \left ( \dfrac{b'}{b} + \dfrac{c'}{c} \right ) \implies \\ y' = na^n * \dfrac{b'c + bc'}{bc} = na^n * \dfrac{bc' + b'c}{a} = na^{(n-1)}(bc' +b'c) \implies \\ y' = na^{(n-1)}a'.[/MATH]This may be something everyone else has been conscious of since their teens, but it was brand new to me. Again, it makes me wonder whether how things are taught is sane.

 

[Steven G]

I am not sure what you are asking. You arrived at a^n(something else). Now that something else had an a in the denominator. As you noted that resulted in the answer being a^(n-1)*(something). Of course when you multiply both sides by y you will have a factor on the rhs of a^n. There obviously is something in your post that I am missing. Please try again. Thanks!

 

[HallsofIvy]

Yesterday I was working on answering a question about logarithmic differentiation and came across something I had not noticed before. I did not know if I had made a mistake or if it was a coincidence, but it stopped me from posting an answer.

[MATH]y = a^n, \text { where n is an integer and}[/MATH]
a is a very messy function in x, but one easily disaggretated suitable by logarithms. In the actual problem a was a product of powers of four functions, but let's keep things simple here with

[MATH]a = b.[/MATH]
To compute the derivative by hand, there are two obvious ways to go. Power and product rules.

[MATH]y = a^n \implies y' = na^{(n-1)}a'.[/MATH]
But that a' will be ugly because, by hypothesis, a' is supposed to be awful to compute

Logarithmic differentiation.

[MATH]y = a^n \implies ln(y) = n * ln(a) = n\{ln(b) + \ln(c)\} \implies\\[/math]

Well, first, you didn't say that a= bc. Did you mean to write that and accidently write, above "a= b"?

[math]\dfrac{1}{y} * y' = n * \left ( \dfrac{1}{b} * b' + \dfrac{1}{c} * c \right ) \implies y' = ny * \left ( \dfrac{b'}{b} + \dfrac{c'}{c} \right ) \implies \\ y' = na^n * \left ( \dfrac{b'}{b} + \dfrac{c'}{c} \right ).[/MATH]All systems froze. How can a be to the power of n rather than n - 1?

It is [math]a^n[/math] because, if you were to add the fractions [math]\dfrac{b'}{b}+ \frac{c'}{c}[/math] you would have a denominator of bc= a.

In the context of a very messy problem, that had me really scratching my head. Had I made an error? If so, where? Of course, the solution is obvious in a context this simple.

[MATH]y' = na^n * \left ( \dfrac{b'}{b} + \dfrac{c'}{c} \right ) \implies \\ y' = na^n * \dfrac{b'c + bc'}{bc} = na^n * \dfrac{bc' + b'c}{a} = na^{(n-1)}(bc' +b'c) \implies \\ y' = na^{(n-1)}a'.[/MATH]This may be something everyone else has been conscious of since their teens, but it was brand new to me. Again, it makes me wonder whether how things are taught is sane.

 

[JeffM]

Well, first, you didn't say that a= bc. Did you mean to write that and accidently write, above "a= b"?



It is [math]a^n[/math] because, if you were to add the fractions [math]\dfrac{b'}{b}+ \frac{c'}{c}[/math] you would have a denominator of bc= a.

Yes, it should have said bc.

 

[JeffM]

Halls and Jomo

Sorry about a = b when what was meant was a = bc. The time limit on editing means I can't fix it now.

I am not asking for an explanation. I found and wrote out the explanation myself once I simplified the context. It is not any deep result.

But I was surprised to find an unexpected power pop up in a more complex context. It sent me of into all sorts of unnecessary work. And yet it turns out to be a natural consequence of logarithmic differentiation. Had I known that, it would have been confirmation that things were proceeding naturally rather than a distraction. I suspect it would distract students as well if they are not told about it because the cancellation is not going to be obvious in exactly the situations where logarithmic differentiation is useful.

Maybe you both have explained this to your students for years. Or maybe you do not see it as a possible distraction to students. I brought it up because I think it may be helpful to students. My opinion only.

 

[Deleted member 4993]

Halls and Jomo

Sorry about a = b when what was meant was a = bc. The time limit on editing means I can't fix it now.

I am not asking for an explanation. I found and wrote out the explanation myself once I simplified the context. It is not any deep result.

But I was surprised to find an unexpected power pop up in a more complex context. It sent me of into all sorts of unnecessary work. And yet it turns out to be a natural consequence of logarithmic differentiation. Had I known that, it would have been confirmation that things were proceeding naturally rather than a distraction. I suspect it would distract students as well if they are not told about it because the cancellation is not going to be obvious in exactly the situations where logarithmic differentiation is useful.

Maybe you both have explained this to your students for years. Or maybe you do not see it as a possible distraction to students. I brought it up because I think it may be helpful to students. My opinion only.

The time limit on editing means I can't fix it now.

I donned my cape and "S" emblazoned underwear - and it is all fixed now. UP....UP... and away.....

 

[HallsofIvy]

All I can say is that I would expect a student who is taking Calculus, and so has known algebra for some time, to recognize that [math]a^n\left(\frac{b'}{b}+ \frac{c'}{c}\right)= a^n\frac{b'c+ bc'}{bc}=a^n\frac{b'c+ bc'}{a}= a^{n-1}(b'c+ bc')[/math]

 

[JeffM]

All I can say is that I would expect a student who is taking Calculus, and so has known algebra for some time, to recognize that [math]a^n\left(\frac{b'}{b}+ \frac{c'}{c}\right)= a^n\frac{b'c+ bc'}{bc}=a^n\frac{b'c+ bc'}{a}= a^{n-1}(b'c+ bc')[/math]

Halls

You persist in failing to see the point I was trying to make. I made up this example to be ridiculously simple and obvious. The actual example was

[MATH]y' = 4 * \left ( \dfrac{e^x\{x^2 + cos(x)\}}{arctan(x) \{ln(3x^4)\}^5} \right )^4 \left \{ 1 + \left (\dfrac{d}{dx} ln\{x^2 + cos(x)\} \right ) - \left ( \dfrac{d}{dx} arctan(x) \right ) - 5 * \left ( \dfrac{d}{dx} ln\{4ln(3x)\} \right ) \right \}.[/MATH]
I do not see that it is obvious from inspection that the exponent on the fraction will end up being reduced by 1 after further computations In fact, I suspect the parenthetical with the derivatives might never be consolidated into a single fraction at all but rather left as a sum, in which case the exponent will remain at 4 and not be reduced.

 

[JeffM]

I shall try one last time. I am interested in where and why students have difficulties. I recognize that those diffculties are frequently purely psychological and therefore of no interest whatsoever to pure mathematicians. Quite a few of us, however, are interested in getting people to learn mathematics, and psychological barriers are important in that regard.

[MATH]y = \{f(x)\}^r, \text { where } r \in \mathbb R.[/MATH]
A student who knows the chain and power rules can see immediately that y' should look like

[MATH]y' = r * \{f(x)\}^{(r-1)} * \text {something}.[/MATH]
That format is what they have been taught to expect. Now we are assuming that f(x) is a messy function, but subject to logarithmic disaggregation. So, if we are trying to explain differential calculus, we introduce the topic of logarithmic differentiation.

[MATH]y = \{f(x)\}^r \implies ln(y) = ln(\{fx)\}^r) = r * ln\{f(x)\}.[/MATH]
The student who grasped logs sees this instantly.

And obviously if z = ln(y), z is a function of y, but y is a function of x so z is also a function of x. Let's calculate z's derivative with respect to x according to the chain rule.

[MATH]z' = \dfrac{dz}{dy} * \dfrac{dy}{dx} = \dfrac{1}{y} * y'.[/MATH]
The student has no trouble here but may be wondering where all this is going.

[MATH]z = ln(y) \text { and } ln(y) = r * ln\{f(x)\} \implies z = r * ln(y) \implies\\ z' = \text {the derivative of } r * ln\{f(x)\} \implies\\ \dfrac{1}{y} * y' = \text {the derivative of } r * ln\{f(x)\} \implies \\ y' = ry * \text {the derivative of } r * ln\{f(x)\} \implies \\ y' = r\{f(x)\}^r * \text {the derivative of } ln\{f(x)\}.[/MATH]And the teacher then concludes with "And that derivative of ln{fx)} is easy to compute" That is a logical place for the teacher to conclude because we teach logarithmic differentiation to simplify differentiation of a class of functions. But that exponent of r is a potential bomb of cognitive dissonance that can explode at any time. The student does not expect it; it seems to contradict what was previously learned.

Yes, the apparent discrepancy can be dispelled without too much trouble.

[MATH]w = ln\{f(x)\} \implies w' = \dfrac{1}{f(x)} * f'(x) \implies\\ y' = r\{f(x)\}^r * \dfrac{1}{f(x)} * f'(x) = r\{f(x)\}^{(r-1)}f'(x).[/MATH]Of course, the whole point of the exercise is to avoid computing f'(x) so there is no logical reason to even mention how to resolve the apparent discrepancy. If someone notices it and is bothered by it, they are illogical people who do not belong in math class and should study history like me.