Question #99f2e
2 Answers
Explanation:
We will use the following:

The product rule
#d/dx f(x)g(x) = f(x)g'(x)+g(x)f'(x)# 
The sum rule
#d/dx f(x)+g(x) = f'(x)+g'(x)# 
#d/dxx^n = nx^(n1)# 
#d/dx cf(x) = cf'(x)# 
#d/dxe^x = e^x#
With those:
#=(x^3+2x)e^x+e^x((d/dxx^3)+(d/dx2x))#
#=(x^3+2x)e^x+e^x(3x^2+2)#
#=e^x(x^3+3x^2+2x+2)#
Explanation:
There's a fun little trick for differentiating functions that are multiplied by
If we have some function:
#f(x)=g(x)e^x#
Then, through the product rule:
#f'(x)=g'(x)e^x+g(x)e^x=e^x(g(x)+g'(x))#
In other words, if a function is multiplied by
Since said inner function here is
#f'(x)=e^x(x^3+2x+3x^2+2)=e^x(x^2+3x^2+2x+2)#