How do you differentiate #f(x)=xe^(x-x^2/2)# using the product rule?

1 Answer
Jul 11, 2016

Answer:

#e^(x-(x^2/2))(1+x-x^2)#

Explanation:

The product property of differentiate is stated as follows:
#f(x)=u(x)*v(x)#
#color(blue)(f'(x)=u'(x)v(x)+v'(x)u(x))#

In the given expression take
#u=x and v=e^(x-(x^2/2))#

We have to evaluate #u'(x)# and #v'(x)#
#u'(x)=1#

Knowing the derivative of exponential that says:
#(e^y)'=y'e^y#

#v'(x)=(x-(x^2/2))'e^(x-(x^2/2))#
#v'(x)=(1-x)e^(x-(x^2/2))#

#color (blue)(f'(x)=u'(x)v (x)+v'(x)u(x))#
#f'(x)=1 (e^(x-(x^2/2)))+x (1-x)(e^(x-(x^2/2)))#
Taking #e^(x-(x^2/2))# as common factor:
#f'(x)=e^(x-(x^2/2))(1+x(1-x))#
#f'(x)=e^(x-(x^2/2))(1+x-x^2)#