Discussion
If \(y = tan^{-1}(x^{2} - x)\), then \(\frac{dy}{dx}\)=
Answer: B
\(\frac{dy}{dx} = \) \(\frac{2x - 1}{1 + (x^{2} - x)^{2}}\).
With \(u = x^{2} - x\), \(\frac{d}{dx}\tan^{-1}u = \frac{1}{1 + u^{2}}\cdot\frac{du}{dx}\) and \(\frac{du}{dx} = 2x - 1\).
Option A differentiates \(x^{2} - x\) as \(2x\), dropping the \(-1\); option C uses \(1 - u^{2}\), which is not the denominator in the derivative of \(\tan^{-1}u\); option D has the sign reversed, as for \(\cot^{-1}\); option E multiplies by \(1 + u^{2}\) instead of dividing.
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