प्रथम सिद्धांत से `e^(x^(2))` का अवकल गुणांक ज्ञात कीजिएः

Explanation

माना `y=e^(x^(2))`
`Rightarrow y+delta=e^((x+deltax)^(2))`
`therefore y+deltay-y=e^((x+deltax)^(2))-e^(x^(2))`
`(deltay)/(deltax)=((e^(x+deltax))^(2)-e^(x^(2)))/(deltax)`
अब `(dy)/(dx)=underset(deltax to 0)lim (deltay)/(deltax)`
`=underset(deltax to 0)lim (e^((x+deltax)^(2))-e^(x^2))/(deltax)`
`=underset(deltax to 0)lim (e^(x^2)[e^(2x deltax^(2))-1])/(deltax)`
`=e^(x^(2))underset(deltax to 0)lim ((e^(2x deltax^(2)+deltax^(2))-1))/(deltax).(2x deltax+deltax^(2))/(deltax)`
`=e^(x^(2)).underset(t to 0)lim ((e^(t)-1)/t).underset(t to 0)lim (2x+deltax)"जहाँ" t=2xdeltax+deltax^(2), t to 0, "जबकि "deltax to 0)`
`=e^(x^2).1.2x`
`=2x.e^(x^(2))`