Logarithmic differentiation: variable powers, products, and worked problems

Why logarithms help

When both the base and exponent depend on $x$, neither the constant-power rule nor the constant-base exponential rule applies on its own. On an interval where $f(x)>0$, let

$$h(x)=f(x)^{g(x)}.$$

Taking natural logarithms gives $\ln h=g\ln f$. Differentiate using the chain and product rules:

$$\frac{h'}h=g'\ln f+g\frac{f'}f.$$

Therefore

$$\boxed{h'=f^g\left(g'\ln f+g\frac{f'}f\right).}$$

Natural logarithms keep the derivative formula simple. Other logarithm bases are possible but introduce conversion factors. The notation $\ln$ throughout this article makes the natural base explicit.

Three useful situations

A constant base: recover the exponential derivative

For $y=a^x$ with constant $a>0$, $\ln y=x\ln a$, so

$$\frac{y'}y=\ln a,\qquad\boxed{y'=a^x\ln a.}$$

The case $a=1$ gives zero as it should. This derivation presumes that the logarithm, its derivative, and the relevant inverse-function differentiation rules have already been established. It should not be used circularly to establish those same starting facts.

Both base and exponent vary

For $f(x)=x^x$ on $x>0$,

$$\ln f=x\ln x,$$

hence

$$\boxed{f'(x)=x^x(\ln x+1).}$$

Treating the exponent as a constant and writing $x\cdot x^{x-1}$ would miss the change in the exponent.

Many products and powers

Example. Differentiate the real fifth-root function

$$y=\sqrt[5]{\frac{(x-1)(x+3)^3}{(x+2)^4}}.$$

The original real function is defined for $x\ne-2$. To take a logarithm, also work away from the zeros $x=-3,1$ and use an absolute value:

$$\ln|y|=\frac15\{\ln|x-1|+3\ln|x+3|-4\ln|x+2|\}.$$

Differentiate:

$$\frac{y'}y=\frac15\left(\frac1{x-1}+\frac3{x+3}-\frac4{x+2}\right)=\frac{12}{5(x-1)(x+3)(x+2)}.$$

Thus, for $x\notin\{-3,-2,1\}$,

$$\boxed{y'=\frac{12}{5(x-1)(x+3)(x+2)}\sqrt[5]{\frac{(x-1)(x+3)^3}{(x+2)^4}}.}$$

The points $x=-3,1$ require separate analysis rather than substitution into this formula. Locally the function behaves like a nonzero constant times a signed power of order $3/5$ or $1/5$, respectively, so it has no finite derivative there. The point $x=-2$ is outside the original domain entirely.

Taking $\ln|y|$ accommodates negative values of $y$; it does not make a logarithm of zero permissible.

An equivalent shortcut: write the power as an exponential

For a positive base,

$$f(x)^{g(x)}=e^{g(x)\ln f(x)}.$$

The chain rule immediately gives the same derivative. For instance,

$$(x^x)'=(e^{x\ln x})'=e^{x\ln x}(\ln x+1).$$

This is a shorter presentation of the same reasoning, not a different rule that avoids the domain condition.

Problem 1: minimize x to the power x

Problem. Find the minimum of $x^x$ for $x>0$.

Since $x^x>0$, the sign of its derivative is the sign of $1+\ln x$. This is negative for $0<x<e^{-1}$, zero at $x=e^{-1}$, and positive for $x>e^{-1}$.

The function therefore decreases and then increases, giving the global minimum

$$\boxed{e^{-1/e}=\left(\frac1e\right)^{1/e},\qquad x=\frac1e.}$$

Problem 2: two bounds for e

The following problem is attributed in the original article to the University of Tokyo's 2016 science-track paper.

Problem. Define $e$ by

$$e=\lim_{t\to\infty}\left(1+\frac1t\right)^t.$$

Prove that for every real $x>0$,

$$\boxed{\left(1+\frac1x\right)^x<e<\left(1+\frac1x\right)^{x+1/2}.}$$

The exponent on the right is $x+1/2$, which is important for the sharper upper bound.

Take logarithms and define

$$F(x)=x\ln\left(1+\frac1x\right),\qquad G(x)=\left(x+\frac12\right)\ln\left(1+\frac1x\right).$$

Both tend to $1$ as $x\to\infty$. Direct differentiation gives

$$F'(x)=\ln\left(1+\frac1x\right)-\frac1{x+1},\qquad F''(x)=-\frac1{x(x+1)^2}<0,$$

and

$$G'(x)=\ln\left(1+\frac1x\right)-\frac1{2x}-\frac1{2(x+1)},\qquad G''(x)=\frac1{2x^2(x+1)^2}>0.$$

Both first derivatives tend to zero at infinity. Since $F'$ is strictly decreasing toward zero, it is positive. Since $G'$ is strictly increasing toward zero, it is negative. Thus $F$ increases strictly toward $1$ and $G$ decreases strictly toward $1$:

$$F(x)<1<G(x).$$

Exponentiation, which preserves inequalities, proves both bounds.

Problem 3: a variable exponent, a graph, and two roots

This problem is attributed in the source to Ochanomizu University's 2018 paper. Let

$$f(x)=x^{x^2},\qquad x>0.$$

1. Find $f'$ and $f''$.
2. Sketch the curve, including its behavior as $x\to0^+$ and that of its derivative. You may use $x\ln x\to0$ as $x\to0^+$; the coordinates of inflection points are not required.
3. Find the real values of $k$ for which $f(x)=k$ has two distinct positive roots.

Part 1: derivatives

Since $\ln f=x^2\ln x$,

$$\boxed{f'(x)=x(2\ln x+1)x^{x^2}.}$$

Another differentiation gives

$$\boxed{f''(x)=\{3+2\ln x+x^2(2\ln x+1)^2\}x^{x^2}.}$$

Part 2: limits and monotonicity

As $x\to0^+$, $x^2\ln x=x(x\ln x)\to0$, so

$$f(x)\to1,\qquad f'(x)=(2x\ln x+x)f(x)\to0.$$

The point $(0,1)$ is approached but is not in the domain. The derivative is negative for $0<x<e^{-1/2}$ and positive for $x>e^{-1/2}$. The minimum is

$$m=f(e^{-1/2})=e^{-1/(2e)}=\left(\frac1{\sqrt e}\right)^{1/e}.$$

As $x\to\infty$, $f(x)\to\infty$. These facts give the graph's endpoint behavior, unique minimum, and increasing branch. The second derivative is negative sufficiently close to zero and positive at the minimum, consistent with a change of concavity before that minimum.

Part 3: count the intersections with y = k

The decreasing branch takes precisely the values $m<y<1$, while the increasing branch takes every value above $m$. At $y=m$ the two branches meet in one point. Therefore two distinct positive roots occur exactly when

$$\boxed{e^{-1/(2e)}<k<1.}$$

At $k=1$, only the increasing branch contributes, since the endpoint $x=0$ is excluded. At $k=m$ there is one root, and below $m$ none.

This final step is a parameter-separation and graphical root-counting problem. Logarithmic differentiation supplies the graph; domain and endpoint checks determine the answer.