Q.The maximum value of $\left(\dfrac{1}{x}\right)^x$ is:
(A) $e$
(B) $e^e$
(C) $e^{1/e}$
(D) $\left(\dfrac{1}{e}\right)^{1/e}$

Concept: Derivative Sign Analysis for a function of the form $f(x) = x^{-x}$.

Let $y = \left(\frac{1}{x}\right)^x = x^{-x}$. Take natural logs: $\log y = -x \log x$. Differentiate implicitly:

Set $\frac{dy}{dx}=0$: $\log x + 1 = 0 \Rightarrow \log x = -1 \Rightarrow x = e^{-1} = \frac{1}{e}$.

For $0 < x < \frac{1}{e}$, $\log x + 1 < 0$, so $\frac{dy}{dx} > 0$ (increasing). …

We want the maximum value of $f(x) = \left(\frac{1}{x}\right)^x$, where $x > 0$ (since $x$ appears in the exponent and base, negative $x$ would lead to complex values for non-integer $x$). The function is defined for positive reals.

Why derivative sign analysis?

To find a maximum, we locate where the function stops increasing and starts decreasing — that is, where its derivative changes from positive to negative. The derivative tells us the slope; a zero slope with a sign change from + to − indicates a local maximum. For a continuous function on an open interval, the global maximum (if it exists) will occur at such a critical point or at a boundary — but here the domain is $(0, \infty)$, so we check the critical point and the limits at the ends.

Step-by-step solution

1. Rewrite the function for easier differentiation.

The expression $\left(\frac{1}{x}\right)^x$ is an exponential form with a variable base and exponent. Take the natural logarithm to bring the exponent down:

Thus

This is valid for $x > 0$.

2. Differentiate $f(x)$.

Using the chain rule:

Now differentiate $-x \log x$ using the product rule:

So

Since $e^{-x \log x} > 0$ for all $x > 0$, the sign of $f'(x)$ is entirely determined by the factor $-(\log x + 1)$.

3. Find critical points.

Set $f'(x) = 0$:

So $x = 1/e$ is the only critical point in $(0, \infty)$.

4. Determine the nature of the critical point (max or min).

Check the sign of $f'(x)$ on either side of $x = 1/e$.

• For $x < 1/e$: Since $\log x$ is increasing, $\log x < -1$, so $\log x + 1 < 0$. Then $-(\log x + 1) > 0$, hence $f'(x) > 0$. The function is increasing.
• For $x > 1/e$: $\log x > -1$, so $\log x + 1 > 0$, giving $-(\log x + 1) < 0$, hence $f'(x) < 0$. The function is decreasing. …

Method: Finding the Maximum VALUE of a Variable-Exponent Function

This extends logarithmic differentiation one step further: after finding the critical $x$, the question asks for the function's value there, not the $x$-coordinate itself — so the final substitution step matters as much as the differentiation.

Steps

Step 1: Rewrite using logs (or as an exponential)

For $y = [g(x)]^{h(x)}$, take logs: $\log y = h(x)\log g(x)$, or equivalently write $y = e^{h(x)\log g(x)}$ so the chain rule applies directly.

Step 2: Differentiate and isolate $\frac{dy}{dx}$

using the product rule inside the bracket.

Step 3: Find the critical point and confirm it's a maximum …

Common Mistakes

Mistake 1: Reporting the critical $x$-value as if it were the maximum value

After correctly finding $x=\frac{1}{e}$ as the critical point, it's tempting to stop there — but the question asks for the maximum value of the function, which means substituting $x=\frac{1}{e}$ back into $y=\left(\frac{1}{x}\right)^x$ to get $e^{1/e}$, a different number from $\frac{1}{e}$ itself.

Mistake 2: Differentiating $\left(\frac{1}{x}\right)^x$ directly without logs …