Если f(x)=(2+x)/(3x-1)
то
f'(x) =( (2+х)' * (3х-1) - (2+х) * (3х-1)' ) / (3x-1)^2 =
( ((2)' + (х)') * (3х-1) - (2+х) * ((3х)' - (1)') ) / (3x-1)^2 =
( (0 + 1) * (3х-1) - (2+х) * (3 - 0) ) / (3x-1)^2 =
( 1 * (3х-1) - (2+х) * 3 ) / (3x-1)^2 =
( 3х-1 - 6 - 3х ) / (3x-1)^2 =
( -1 - 6 ) / (3x-1)^2 =
-7 / (3x-1)^2
f'(1) = -7 / (3*1-1)^2 = -7 / (3-1)^2 = -7 / 2^2 = -7/4