प्रथम सिद्धांत से `xe^(x)` को अवलंकित कीजिएः

Explanation

माना `y=xe^(x)`
`Rightarrow y+deltay=(x+deltax)e^((x+deltax))`
`Rightarrow y+deltay-y=(x+deltax)e^((x+deltax))-xe^(x)`
`Rightarrow (deltay)/(deltax)=((x+deltax)e^(x+deltax)-xe^(x))/(deltax)`
अब `(dy)/(dx)=underset(deltax to 0)lim (deltay)/(deltax)`
`=underset(deltax to 0)lim ((x+deltax)e^(x+deltax)-xe^(x))/(deltax)`
`=underset(deltax to 0)lim (xe^(x+deltax)-xe^(x)+deltax.e^(x+deltax))/(deltax)`
`=underset(deltax to 0)lim [xe^(x)(e^(deltax-1)/(deltax))+e^(x+deltax)]`
`=xe^(x).underset(deltax to 0)lim ((e^(deltax)-1)/(deltax))+underset(deltax to 0)lim e^(x+deltax)`
`=xe^(x)xx1+e^(x)" "[therefore underset(deltax to 0)lim ((e^(deltax)-1)/(deltax))=1]`
`=xe^x+e^x`
`=e^x(x+1)`