Answer 1

Let y = e^x log x
Differentiating w.r.t. x, we get

`dy/dx = d/dx ("e"^xlog x)`

= `"e"^xd/dx(logx) + (logx) d/dx ("e"^x)`

=`"e"^x(1/x) + (logx)("e"^x)`

= `"e"^x(1/x+logx)`