Differentiating exponents: two wrongs make a right

I was talking to a student about his calculus last week. He was trying to differentiate xx. (Actually he was trying to differentiate x ln(x) and had decided the best place to start was to raise e to the power of it, thus producing xx.) At first he tried this:

I asked him what he thought about that and he said he wasn’t sure it was right because of the x in the power. I asked him how you normally dealt with an x in the power. He thought about it and then produced this:

I confirmed that this is indeed what you would do if there was an x in the power but the base was constant. But I also confirmed that because there was an x in both the power and the base, it was going to be a bit trickier. He asked how you did deal with that, and I started out by writing xx = (eln(x))x. (On reflection, if I was doing SQWIGLES properly, I should have helped him find this definition in his notes.) Together we continued this working as follows. Interestingly, we ended up having to differentiate x ln(x) as part of this, but by this time he realised it would be much simpler to use the product rule!

Do you notice anything interesting about that result? Well, I did: it’s the sum of the other two wrong answers! If we treat the base as a constant we get a (wrong) result, and if we treat the power as a constant we get a (wrong) result, but if we add those two results together we get the correct result! This is WILD!

After we did the summing up what we’ve learned bit, I absolutely had to try this out generally to see what would happen (you can read this working in a pdf here):

Oh my goodness! It worked again! I am reeling with the wildness of this! Currently I don’t know if there’s a more conceptual/visual approach to this that will help me feel like I understand why it is true, but for now I’m just going to relish the wildness of these two wrong results adding up to the correct one.

PS: Isn’t it cool the stuff you learn when you listen to students? If I’d just told him what to do I’d never have noticed this.

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3 Responses

  1. David says:

    This can be explained using the multivariable chain rule. Let f(u,v) = u^v, u=x, v=x. Then df/dx = ∂f/∂u du/dx + ∂f/∂v dv/dx = ∂f/∂u + ∂f/∂v. The product rule is also a special case of the multivariable chain rule.

    • David Butler says:

      Ah you beat me to it! I was just going to do an update to say exactly this!
      In my head it was something like f=u^v with u=f(x) and v=g(x).
      So df/dx = df/du*du/dx + df/dv*dv/dx
      = (derivative of f treating v as constant) + (derivative of f treating u as constant)

      Of course, this student won’t see multivariable calculus in their course for another several months yet…

  2. […] seems like ages ago — but it was only yesterday — that I wrote about differentiating functions with the variable in both the base and the power. Back there, I had learned that the derivative of a function like f(x)g(x) is the sum of the […]

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