Question

How do you differentiate `g(x)=sqrtxe^x`?

Answer 1

`g'(x)=e^x(1/(2sqrt(x))+sqrt(x))`

Explanation:

`g(x)=sqrt(x)e^x=(x)^(1/2)e^x`

`u=(x)^(1/2), v=e^x`

`g'(x)=vu'+uv'`

`u'=1/2x^(-1/2)=1/(2sqrt(x)), v'=e^x`

`g'(x)=e^x/(2sqrt(x))+sqrt(x)e^x=e^x(1/(2sqrt(x))+sqrt(x))`