If the maximum value of the function $f(x) = \frac{\log_ex}{x}$, $x > 0$ occurs at $x = a$, then $a^2f''(a)$ is equal to

We have $f(x) = \frac{\ln x}{x}$ where $x > 0$

To find where the maximum occurs, we need to find the critical points by taking the derivative and setting it equal to zero.

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Finding $f'(x)$ using the Quotient Rule:

Here: $u = \ln x$, so $u' = \frac{1}{x}$ and $v = x$, so $v' = 1$

$f'(x) = \frac{\frac{1}{x} \cdot x - \ln x \cdot 1}{x^2} = \frac{1 - \ln x}{x^2}$

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Set $f'(x) = 0$:

$\frac{1 - \ln x}{x^2} = 0$

Since $x^2 \neq 0$, we need:

$1 - \ln x = 0$

$\ln x = 1$

$x = e$

So $a = e$

When $x < e$: $\ln x < 1$, so $f'(x) > 0$ (function increasing)

When $x > e$: $\ln x > 1$, so $f'(x) < 0$ (function decreasing)

Since it goes from increasing to decreasing, $x = e$ gives a maximum.

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Finding the second derivative $f''(x)$ using the Quotient Rule:

We have $f'(x) = \frac{1 - \ln x}{x^2}$

$u = 1 - \ln x$, so $u' = -\frac{1}{x}$ and $v = x^2$, so $v' = 2x$

$f''(x) = \frac{-\frac{1}{x} \cdot x^2 - (1 - \ln x) \cdot 2x}{x^4}$

$f''(x) = \frac{-x - 2x + 2x\ln x}{x^4}$

$f''(x) = \frac{x(-3 + 2\ln x)}{x^4}$

$f''(x) = \frac{-3 + 2\ln x}{x^3}$

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Finding $f''(a)$ where $a = e$:

$f''(e) = \frac{-3 + 2\ln e}{e^3}$

Since $\ln e = 1$:

$f''(e) = \frac{-3 + 2(1)}{e^3} = \frac{-1}{e^3}$

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Calculating $a^2f''(a)$:

$a^2f''(a) = e^2 \cdot \frac{-1}{e^3} = \frac{e^2}{e^3} \cdot (-1) = -\frac{1}{e}$

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Answer: $-\dfrac{1}{e}$