Discussion

Let \(f(x) = \frac{1}{x^{2}}\) and let \(u = f(x)f''(x)\). then \(\frac{du}{dx} =\)

Answer: A

\(\frac{du}{dx} =\) \(-36x^{-7}\).

With \(f(x) = x^{-2}\):

\[ \begin{aligned} f''(x) &= 6x^{-4} \\ u &= x^{-2}\cdot 6x^{-4} = 6x^{-6} \\ \frac{du}{dx} &= -36x^{-7} \end{aligned} \]

\(36x^{-7}\) loses the minus sign from differentiating a negative power, and \(\pm 42x^{-7}\) and \(-30x^{-7}\) come from wrong exponents.

A general check: if \(f = x^{-2}\) then \(f f''\) is \(6x^{-6}\), and any power \(x^{k}\) differentiates to \(kx^{k-1}\), here \(-6 \times 6 = -36\).

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