\(\frac{x^3 + 6x - 7}{x-1}\): When numerator is differentiated → 3x\(^2\) + 6 When denominator is differentiated → 1 : \(\frac{3x^2 + 6}{1}\) substitute x for 1 \(\frac{3 * 1^2 + 6}{1}\) = \(\frac{3 + 6}{1}\) = \(\frac{9}{1}\) = 9