Differentiate `x e^x` from first principles.

Explanation

Let `y = xe^(x)`.
Let `deltay` be an increment in y, corresponding to an increment `deltax` in x.
Then, `y + deltay = (x+deltax)e^(x+deltax)`
`rArr deltay = (x+deltax)e^(x+deltax) -xe^(x)`
`rArr (deltay)/(deltax) = ((x+deltax)e^(x+deltax) - xe^(x))/(deltax)`
`rArr (dy)/(dx) = underset(deltax rarr 0)("lim")(deltay)/(deltax)`
`= underset(deltax rarr 0)("lim")((x+deltax)e^(x+deltax)-xe^(x))/(deltax)`
`=underset(deltaxrarr0)("lim")(xe^(x+deltax)-xe^(x)+deltax.e^(x+deltax))/(deltax)=underset(deltaxrarr0)("llim")[xe^(x).((e^(deltax)-1)/(deltax))+e^(x+deltax)]`
`xe^(x).underset(deltaxrarr0)("lim")((e^(deltax)-1)/(deltax))+underset(deltaxrarr0)("lim")e^(x+deltax)`
`=(xe^(x) xx 1) + e^(x)` , `[ :' underset(deltaxrarr0)("lim")((e^(deltax)-1)/(deltax)) = 1]`
`= (xe^(x)+e^(x))= (x+1)e^(2)`.
Hence `d/dx (xe^(x)) = (x+1)e^(x)`.