Q.$f(x) = x^x$ has a stationary point at:
(A) $x = e$
(B) $x = \dfrac{1}{e}$
(C) $x = 1$
(D) $x = \sqrt{e}$

Concept: Derivative Sign Analysis – find where $f'(x)=0$ by logarithmic differentiation.

1. Write $f(x)=x^x$. Take $\log f(x)=x\log x$. Differentiate:

   $\frac{f'(x)}{f(x)} = \log x + 1$.

2. So $f'(x)=x^x(\log x+1)$. Set $f'(x)=0$. Since $x^x>0$ for $x>0$, we require $\log x+1=0$. …

The key to solving this lies in understanding what a stationary point means: it’s where the derivative of the function is zero. For a function like $x^x$, which is neither a simple power nor an exponential in the usual sense, we can’t just apply the power rule or the exponential rule directly. Instead, we need a technique that handles a variable both in the base and the exponent — that’s where logarithmic differentiation shines.

Why logarithmic differentiation?

If you take the natural log of both sides, you turn the exponent into a product: $\log f(x) = x \log x$. Now the right side is a product of two familiar functions, and we can differentiate it using the product rule. Then we multiply through by $f(x)$ to recover $f'(x)$. This is a standard trick for functions of the form $[g(x)]^{h(x)}$.

Let’s walk through it step by step.

1. Set up the function and take logs.

   Let $y = x^x$. Then $\log y = \log(x^x) = x \log x$.

   This is valid for $x > 0$, which is the domain we care about (since $x^x$ is real for positive $x$).

2. Differentiate both sides with respect to $x$.

   On the left, by the chain rule: $\frac{d}{dx} \log y = \frac{1}{y} \cdot \frac{dy}{dx}$.

   On the right, use the product rule: $\frac{d}{dx} (x \log x) = 1 \cdot \log x + x \cdot \frac{1}{x} = \log x + 1$.

   So we have:

3. Solve for $\frac{dy}{dx}$. Multiply both sides by $y$:

Method: Logarithmic Differentiation for Variable-Base, Variable-Exponent Functions

Use this whenever a function has the form $[g(x)]^{h(x)}$ — the variable appears in BOTH the base and the exponent — so neither the power rule nor the exponential rule alone applies.

Steps

Step 1: Take the natural log of both sides

For $y = [g(x)]^{h(x)}$, write

This converts the troublesome variable exponent into an ordinary product, which the product rule can handle.

Step 2: Differentiate implicitly

Differentiate both sides with respect to $x$. The left side needs the chain rule ($\frac{d}{dx}\log y = \frac{1}{y}\frac{dy}{dx}$); the right side needs the product rule:

Step 3: Solve for $\frac{dy}{dx}$ …

Common Mistakes

Mistake 1: Applying the ordinary power rule to a variable exponent

Treating $x^x$ as if the exponent were a fixed constant $n$ and writing $\frac{d}{dx}(x^x) = x\cdot x^{x-1} = x^x$ is wrong — the power rule only works when the exponent is a constant, and here the exponent is $x$ itself, which also needs to be differentiated via the log step.

Mistake 2: Missing that $x^x$ itself is never zero …