If \(y = \frac{1}{{{x^2}}}\), then \(\frac{{dy}}{{dx}}\) at \(x = - 1\) is:

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6 + 3 =
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If \(y = \frac{1}{{{x^2}}}\), then \(\frac{{dy}}{{dx}}\) at \(x = - 1\) is:

2

\(\frac{1}{3}\)

3

4