Question Detail

We use the product rule for differentiation: (\frac{d}{dx}(uv) = u'v + uv').Let (u = 3x^2) and (v = \cos^{-1}x).Then (u' = \frac{d}{dx}(3x^2) = 6x).And (v' = \frac{d}{dx}(\cos^{-1}x) = -\frac{1}{\sqrt{1-x^2}}).Applying the product rule:(\frac{d}{dx} (3x^2 \cos^{-1}x) = (6x)(\cos^{-1}x) + (3x^2)\left(-\frac{1}{\sqrt{1-x^2}}\right))(= 6x \cos^{-1}x - \frac{3x^2}{\sqrt{1-x^2}}).