Question

Check my work and answer.
Consider the function h(x)= a(−2x +1)5−b , where a ≠ 0 and b ≠ 0 are constants.



A. Find h′(x) and h′′(x). (6 points)
H’(x): Chain rule: a*D(-2x+1)* 5(-2x+1)^5-1 +Db-->a*-2x* 5(-2x+1)^4 + 0 -->
-10a(-2x+1)^4

H’’(x): Chain rule: -10a*D(-2x+1)* 4 (-2x+1)^4-1--> -10a*-2* 4 (-2x+1)^3--> 80a(-2x+1)^3