Derivative of n^x

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We haven’t seen a derivative for n^x and it seemed like this math object might be a good test for our Dark Blue Magic program. Dark Blue Magic is the collection of tools and ideas that students think we would want to have if, for some unexpected reason, we ended up on a deserted island and we had to rediscover all of Algebra and Calculus.

One of the ideas in Dark Blue Magic is that the derivative of a function should be quite similar to the original function. For example, we would expect the derivative of a monomial to be built using monomials. We would expect the derivative of a trigonometric function to be something comprised of trigonometric functions. This idea isn’t a rule; it is a guideline for making guesses.

Our first hypothesis would be that

\dfrac {d} {dx} n^x = an^y

Nothing is known about either ‘a’ or ‘y’ and it is possible that one of them could be unnecessary; if this is the case we will have a=1 or y=1.

We used handheld calculators for getting derivatives of specific examples.

  • For n=3 and x=2
  • the function would be f(x)=3^x
  • We write the following
  • Derivative with respect to x for f(x) = 3^x where x=2[/latex] (this will be used several times below)

We would calculate:

\dfrac {3^{2.000001} - 3^2} {2.000001 - 2} = \dfrac {3^{2.000001} - 3^2} {0.000001}

Some results are shown below

  • Derivative with respect to x for f(x) = 2^x where x=0 –> 0.693147
  • Derivative with respect to x for f(x) = 2^x where x=1 –> 1.386295
  • Derivative with respect to x for f(x) = 2^x where x=2 –> 2.772590
  • Derivative with respect to x for f(x) = 2^x where x=3 –> 5.545179
  • Derivative with respect to x for f(x) = 2^x where x=4 –> 11.090359
  • Derivative with respect to x for f(x) = 2^x where x=5 –> 22.186723

We aren’t getting integer values. Integer values would be nice. Well, we have the next best thing–when we scrutinize these first few values it looks like one divided by another could be an integer. We know about round-off error so if we get 2.9999, we’ll recognize it as being 3.

We notice a place where one of these nonintegers could be twice another of the nonintegers so we try a division. We see another opportunity and we try that division also.

  • 11.090359/5.545179 = 2
  • 22.186723/5.545179 = 4

After staring at what we have and asking “how can things be tied together?”, we notice that 2 is 2^1 and 4 is 2^2 and that the values for the arguments of the derivative have a difference of 1 or 2, respectively. We work this into the formula below (see comments in Appendix A):

  • f(x)=2^x, \dfrac {f'(4)} {f'(3)} = 2 = 2^1 = 2^{4-3}
  • f(x)=2^x, \dfrac {f'(5)} {f'(3)} = 4 = 2^2 = 2^{5-3}

Does this trend continue?

  • 11.090359/1.386295 = 8
  • f(x)=2^x, \dfrac {f'(4)} {f'(1)} = 8 = 2^3 = 2^{4-1}
  • 11.090359/0.693147 = 16
  • f(x)=2^x, \dfrac {f'(4)} {f'(0)} = 16 = 2^4 = 2^{4-0}

Using the variables j and k for the integers, we can write what we know so far from the above examples:

  • \dfrac {f'(j)} {f'(k)} = 2^{j-k}

Our numbers 0.693147, 1.386295, 2.772590, 5.545179, … are mysterious to us. If we had a noninteger like 3.14159… or 1.414214… then we would probably recognize the decimal as leading us to something like \pi or \sqrt{2}.

Mockingbird students will soon be adding a list of nonintegers, like the ones above, that we believe are a part of our Dark Blue Magic.

  • ln 1 = 0
  • ln 2 = 0.69314718056
  • ln 3 = 1.09861228867

This lets another cat out of the bag, at least for the 0.69314718056 noninteger. Could it be that “1.0986122…” will be the result for math just like we’ve done, having 3 instead of 2?

For 2 we had:

Derivative with respect to x for f(x) = 2^x where x=0 –> 0.693147

Let’s do the work for

Derivative with respect to x for f(x) = 3^x where x=0

\dfrac {3^{0.000001} - 3^0} {0.000001 - 0} = \dfrac {3^{0.000001} - 3^2} {0.000001} = 1.09861289221

The trend continues. Let’s try to build something more general given what we now know. Let’s look at our general formula again:

\dfrac {f'(j)} {f'(k)} = 2^{j-k}

Let’s bring up the denominator so we can isolate f'(j) on the left side:

f'(j) = 2^{j-k} f'(k)

We could make it simpler by setting k to zero:

f'(j) = 2^{j-0} f'(0)

f'(j) = 2^{j} f'(0)

Let’s go ahead and replace j with x:

f'(x) = 2^{x} f'(0)

We know that for f(x)=n^x, f'(0)=ln(n). Also, that 2 is actually n–we were doing work with n=2 when we started building this general example:

  • if f(x)=n^x then f'(x) = ln(n) n^{x}

For the work in this monograph, only integers were used. We are happy with the success we have to this point, but we will do some work with nonintegers before we fax a copy of this work out to our friends at another academy (a soon-coming Appendix B will have such an example).

Oh, one very important thing! Assume this actually was the first time that this was done–it would be very important to have something written up showing the whole thing with a date and our name on it! In fact, that’s one of the things our friends at other academies are supposed to do–get mail from us with our name and a date, and keep it so later it can be used as proof that we had developed and divulged the idea!

Appendix A

Ever so often, you can read through a textbook and it looks like the person who figured it out was like Spock on an episode of Star Trek, and he figured it out in five minutes, just in time to save the Enterprise.

Some work was done fast, but in other stories, it took person months or years to decide what the next step should be. Mathematicians with years of experience have superhighways in their brains between commonly linked ideas to help them, and as a student starting out, you have a path through the grass. Keep reading stuff and trying stuff–build your own highways! : – )

Appendix B

  • n=e=2.71828
  • x=pi=3.14159
  • n^x = e^\pi = 23.140582
  • \dfrac {2.71828^{3.14160} - 2.71828^{3.14159}} {3.14160 - 3.14159} = 23.14068

OK- That is close enough! We are ready to write this up in a letter and send it to a friend!

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Published by Robert Sterling

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