General powers

In the first section we looked at the formula AB as an algebraic object, when A and B are fixed numbers. Then we imagined what would happen if we tried to make A or B into a variable, thus obtaining powers as functions and exponentials. What happens if we have a "general power", that is, a power where a variable x is present both in the base A and the exponent B?

The simplest example is xx, but general powers can be as complicated as one can imagine, a slightly less simple example might be [1/sin(x)]2x + cos(x).

Unlike the previous two cases, general powers are hard to grasp based on the properties of the usual power. In fact, if we tried this approach, looking at general powers through the lens of algebraic powers, we would get very weird things. One does not have to go far, when given a function, we always start by determining its domain. What happens if we ask for which values of x does the expression xx make sense? We get an answer that is sort of useless, as we can see here.

This shows that we have to adopt a wholly new approach to general powers. The previous section holds the key, we have a formula there that transforms general powers into products.

Let f,g be functions. We define the general power f g by

f (x)g(x) = eln[f (x)]⋅g(x) = eg(x)⋅ln[f (x)].

For instance, the meaning of the function xx is, by the definition, ex⋅ln(x).

Note that in the expression on the right there are no general powers, just a simple composition of logarithm, multiplication and exponential, all of which are nice operations. Everything that we do with functions that are general powers - domain, limits, derivative - must be done in this "e to logarithm" form. What do we get?

By looking at the definition of f g, we see that in order for eln[f (x)] g(x) to make sense, the following conditions have to be satisfied: f (x) must make sense, g(x) must make sense, and ln[f (x)] must make sense, that is, we need f (x) > 0.

Thus we get the following:

Df g) = {xDf ) ∩ D(g); f (x) > 0}.

Similarly, differentiating f g means differentiating eln[f (x)] g(x). Using the chain rule and the product rule we get the following:

Df g) = eln( f )g ⋅ [ln( f )g]′ = f g ⋅ [(f ′/f )g + ln( f )g′].

Usually people do not remember these formulas, because they are not needed. The only thing that is needed is this basic rule: When working with general powers, always rewrite them to the "e to logarithm" form.

Example: Find the domain and derivative of xx.

Solution: We have xx = ex⋅ln(x). Since the exponential can take any number, the domain is determined by the logarithm:

D(xx) = (0,∞).

We also easily find the derivative:

[xx]′ = [exln(x)]′ = exln(x) ⋅ [xln(x)]′ = xx ⋅ [ln(x) + x(1/x)] = xx ⋅ [ln(x) + 1].

For more information about this function see appropriate problem in Derivatives - Exercises - Graphing functions.

Remark: One mistake students often make is that they try to weasel out of the (admittedly more complicated) "e to logarithm" way. Typically they would try to pretend that the base or the exponent is just some number and use appropriate formulas. Of course this is wrong, as we can see:

We pretend that xx is a power and apply the appropriate rule for derivative:

[xx]′ = xxx−1 = xx.

We pretend that xx is a an exponential with base x and apply the appropriate rule for derivative:

[xx]′ = ln(x)⋅xx.

If I got a penny every time I see one of these two incorrect calculations on a test, I would, like, have all those pennies. Take this as a warning and try to do it right.

Trigonometric functions
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