How does logarithmic differentiation give the derivative of f(x) = x^x?

### The obstacle Two standard rules handle powers: - **Power rule**: `d/dx x^n = n*x^(n-1)` — valid when the **exponent** `n` is a constant. - **Exponential rule**: `d/dx a^x = a^x*ln a` — valid when the **base** `a` is a constant. The function `x^x` satisfies neither hypothesis: the base and the exponent are both the variable. Applying either rule blindly gives a wrong answer — `x*x^(x-1)` from the first, `x^x*ln x` from the second — and quoting one of those is the classic failure on this question. It is also worth noting that `x^x` is only defined in the ordinary sense for `x > 0`, so that domain restriction belongs in the answer. ### The log-derivative identity For a positive differentiable function `f`, ``` d/dx ln f(x) = f'(x) / f(x) ``` Differentiating the logarithm produces the reciprocal of its argument, multiplied by the derivative of that argument. The quantity `f'/f` is called the **logarithmic derivative** and has a direct meaning: it is the *relative* rate of change of `f`, the proportional change per unit of `x`, as opposed to the absolute change measured by `f'`. ### Applying it to x^x Write `y = x^x` with `x > 0` and take natural logs of both sides. The log of a power brings the exponent down as a factor: ``` ln y = x * ln x ``` Now differentiate both sides with respect to `x`. **Left side.** By the log-derivative identity applied to `y` as a function of `x`, it becomes `y'/y`. **Right side.** This is a product, so the product rule `(uv)' = u'v + uv'` applies with `u = x` and `v = ln x`: ``` 1*ln x + x*(1/x) = ln x + 1 ``` Setting the two sides equal: ``` y'/y = ln x + 1 ``` Multiplying through by `y` and substituting the original expression back: ``` y' = y*(ln x + 1) = x^x * (ln x + 1) ``` The last step — multiplying back by `y` — is the one candidates most often forget, leaving the answer stuck as a relative rate. **A sanity check.** At `x = 1` the formula gives `1^1*(0 + 1) = 1`, so the curve passes through the point `(1, 1)` with slope 1. That is consistent with the function being close to `x` near that point. ### The equivalent exponential route Any positive base can be rewritten through the exponential: `x^x = e^(x*ln x)`. Differentiating that form gives `e^(x*ln x)` multiplied by the derivative of the exponent, which is again `ln x + 1`, and `e^(x*ln x)` is `x^x`. The two routes are the same computation dressed differently; the log route avoids having to rewrite the function at all. ### Why the technique is worth knowing **It recovers the other rules.** For `y = x^n` with constant `n`: `ln y = n*ln x`, so `y'/y = n/x` and `y' = n*x^n/x = n*x^(n-1)` — the power rule. For `y = a^x` with constant `a`: `ln y = x*ln a`, so `y'/y = ln a` and `y' = a^x*ln a` — the exponential rule. Both fall out of a single method. **It turns products into sums.** If `f` is a product of many factors, `ln f` is a sum of logs, and the logarithmic derivative of the whole is simply the sum of the individual logarithmic derivatives: ``` f = g*h implies f'/f = g'/g + h'/h f = g/h implies f'/f = g'/g - h'/h ``` Differentiating a product of six factors directly is an error-prone cascade of product rules; differentiating its log is six easy terms added together, after which one multiplication by `f` recovers the derivative. **It measures growth in relative terms.** Because `f'/f` is the proportional rate of change, it is the natural quantity when comparing things of different scales — a slope of 5 units per unit means something quite different for a quantity near 10 than for one near 10000, while the logarithmic derivative is scale-free. ### Pitfalls - Forgetting `x > 0`, without which the logarithm and the expression itself are not defined in the ordinary reals. - Differentiating `ln y` as `1/y` and dropping the `y'` factor, which loses the entire left-hand side of the equation. - Stopping at `y'/y = ln x + 1` and calling it the derivative. - Applying the power rule or the exponential rule directly and producing `x*x^(x-1)` or `x^x*ln x`. Notice that the correct answer is the sum of a term resembling each of those wrong answers, which is a memorable way to keep it straight.