Answer
$f''(x) = \frac{118}{(2 + 3 x)^3}$
Work Step by Step
a) Using WolframAlpha and inputting "limit ((2(x+h)^2-(x+h)+5)/(3(x+h)+2) - (2(x)^2-(x)+5)/(3(x)+2))/h as h approaches 0", we get $f'(x) =\frac{-17 + 8 x + 6 x^2}{(2 + 3 x)^2}$
b)$f'(x) = \frac{(4x-1)(3x+2)-(2x^2-x+5)(3)}{(3x+2)^2} = \frac{-17 + 8 x + 6 x^2}{(2 + 3 x)^2}$
c) Using WolframAlpha and inputting "derivative (-17 + 8 x + 6 x^2)/(2 + 3 x)^2", we get $f''(x) = \frac{118}{(2 + 3 x)^3}$
