The Slope of x to the x Is Both Wrong Answers Added

Differentiate $x^x$. The answer is

and the interesting part is that the two rules a student reaches for are both wrong, and their sum is exactly right. That is not a coincidence to be memorised. It is a theorem, and knowing which one makes a class of similar problems routine.

Two rules, neither of which applies

The power rule handles $x^a$ with a constant exponent. Applying it here, treating the exponent as if it were frozen, gives

which is a valid piece of algebra applied to an invalid premise. The exponential rule handles $a^x$ with a constant base. Applying that instead, freezing the base, gives $x^x \ln x$.

At $x = 2$ these read 4 and $4\ln 2 = 2.7726$, while the true slope is $4(1 + \ln 2) = 6.7726$. Both candidates are too shallow, and neither is a rewriting of the other, so this is a real disagreement rather than a notational one.

The route that works

Take logarithms first. The function is positive for $x > 0$, so writing $y = x^x$ and $\ln y = x\ln x$ is legitimate, and the right side is now a product of two things whose derivatives are known:

Multiplying back by $y = x^x$ gives equation (1). The move that made it easy was turning an exponent into a factor, which is the one thing logarithms are for.

A second route, if the implicit step feels uncomfortable, is to rewrite the function outright as $x^x = e^{x\ln x}$ and apply the chain rule. The derivative of the exponent is $1 + \ln x$ again, and the outer derivative reproduces $e^{x\ln x} = x^x$. Same answer, and it makes plain that there was only ever one function here, not a power and an exponential in disguise.

Why the two wrong answers add up

Consider $u^v$ as a function of two independent variables. Its partial derivatives are the two rules, each holding the other slot fixed:

Now walk along the diagonal $u = v = x$. The chain rule for a function of two variables says the total rate of change is the sum of the two partial effects, each multiplied by the rate of its own variable, which is 1 in both cases:

So the two rules were not wrong in the sense of being nonsense. Each of them correctly measured one of the two ways the function changes, and each of them silently discarded the other. This is the reason the two errors are complementary rather than independent, and it is why striking both out as worthless would misrepresent the situation.

The same argument works unchanged for $f(x)^{g(x)}$, and gives the general rule

with equation (1) recovered by $f = g = x$. Equation (6) is worth carrying because it never needs to be rediscovered under pressure, and because the two terms in it are visibly the two partial effects from equation (4).

Three points where the answer can be tested

At $x = 1/e$ the bracket in equation (1) is $1 + \ln(1/e) = 0$, so the slope is exactly zero. That is the bottom of the dip, with value $e^{-1/e} = 0.69220$. The power rule claims the slope there is $(1/e)^{1/e} = 0.69220$, which is plainly not the slope at a minimum. Checking the second derivative confirms the shape:

At $x = 1$ something awkward happens. The bracket is 1, so the true slope is 1, and the power rule gives $1 \cdot 1^0 = 1$ as well. The rule that does not apply produces the right number at that one point. Anyone who spot-checks a rule at $x = 1$ will conclude it works, which is a general hazard of testing at points where several quantities collapse to 1.

And as $x \to 0^{+}$, $x\ln x \to 0$, so $x^x \to 1$ while the slope $x^x(1 + \ln x) \to -\infty$. The curve arrives at height 1 with a vertical tangent, which is the steep left edge in figure 1.

The same method on the next floor up

The real test of a method is whether it survives being asked again. Take $y = x^{x^x}$, where the exponent is itself the function we just differentiated. The logarithmic route needs no new ideas: $\ln y = x^x \ln x$, and differentiating that product uses equation (1) for one of its factors:

No new rule was needed, and the same would be true of a tower four levels high. That is the payoff of equation (3) over memorising equation (1): the method composes, and the answer does not have to be recognised.

Where the function stops existing

Everything above assumed $x > 0$, and that is not caution for its own sake. For negative $x$, $x^x$ is real only on a scattered set: it is $1/4$ at $x = -2$ and undefined over the reals at $x = -1/2$, since that would ask for a square root of a negative number. A set with no interval in it has no derivative anywhere, so equation (1) has nothing to say there. The complex extension exists and is multivalued, and it is a different subject.

One last thing, mostly for pleasure. The function is awkward to differentiate and its integral over the unit interval has a startling closed form:

which converges absurdly fast, six terms giving five correct digits. The companion identity $\int_0^1 x^{-x}dx = \sum_{n\ge1} n^{-n} = 1.2912860\ldots$ comes from the same expansion with the signs left alone.

Sources and further reading

• The method in equation (3) — Logarithmic differentiation
• The two-variable step behind equation (5) — Total derivative
• The rule used inside equation (3) — Product rule
• Equation (9) and its companion — Sophomore's dream

Equation (1) was measured as well as derived: the slope was estimated by Richardson-extrapolated central differences at sixty-three points across $(0, 4]$, matching the formula to a worst error of $3 \times 10^{-12}$. A ternary search located the minimum at 0.367879451 against $1/e = 0.367879441$, where the measured slope was $-7 \times 10^{-13}$ while the power rule claimed 0.692201.