If y=\ln^{2}(\frac{x^{2}-x+1}{x^{2}+x+1}), then what is \frac{dy}{dx} at x=0 equal to ?

Explanation

Let u = \frac{x^2-x+1}{x^2+x+1}. At x=0, u = \frac{1}{1} = 1. Then y = (\ln u)^2. Using the chain rule, \frac{dy}{dx} = 2(\ln u) \cdot \frac{1}{u} \cdot \frac{du}{dx}. Since u=1 at x=0, we have \ln 1 = 0. Since \frac{du}{dx} is finite at x=0, the entire derivative evaluates to 0.

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