Discussion

If \(y = (x - 1)\log_{e}(x - 1)\), then \(\frac{d^{2}y}{dx^{2}}\) at \(x = 3\) is

Answer: D

\(\frac{d^{2}y}{dx^{2}}\) at \(x = 3\) is \(\frac{1}{2}\).

By the product rule:

\[ \begin{aligned} \frac{dy}{dx} &= \log_{e}(x - 1) + 1 \\ \frac{d^{2}y}{dx^{2}} &= \frac{1}{x - 1} = \frac{1}{2} \end{aligned} \]

at \(x = 3\). \(\frac{1}{4}\) uses \(\frac{1}{(x - 1)^{2}}\), and \(e\), \(e^{2}\) and 3 confuse the derivative with the function's values.

The derivative of \((x - 1)\log(x - 1)\) is a standard product-rule result: the '+1' comes from \((x - 1)\cdot\frac{1}{x - 1}\).

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