Answer:
h'(x) = (-3x^3 +2x)/e^(3x)
Step-by-step explanation:
The formula for the derivative of a quotient is ...
(u/v)' = (vu' -uv')/v²
Using u = (x^3 +x^2) and v = e^(3x), this becomes ...
h'(x) = (e^(3x)(3x^2 +2x) -(x^3 +x^2)(3)(e^(3x))/e^(6x)
h'(x) = (-3x^3 +2x)/e^(3x)
