\(\frac{x^3 + 6x - 7}{x-1}\):

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