Question

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

Answer 1

`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)`