प्रथम सिद्धांत से log ax का अवकल गुणांक ज्ञात कीजिएः

Explanation

माना `y=log ax` ......(i)
`therefore y+deltay=log a(x+deltax)` ......(ii)
`therefore y+deltay-y=log a(x+deltax)-log ax`
`Rightarrow deltay=log a((x+deltax))/(ax)=log(1+(deltax)/(x))(therefore log M-log N=log""(M)/(N))`
`Rightarrow deltay=(deltax)/(x)-(1)/(2) ((deltax)/(x))^(2)+(1)/(3) ((deltax)/(x))^(3)-(1)/(4) ((deltax)/(x))^(4)_..... (therefore log (1+x)=x-(1)/(2)x^(2)+(1)/(3)x^(3)-(1)/(4)x^(4)+.....oo)`
`Rightarrow deltay=(deltax)/(x)[1-(1)/(2)((deltax)/(2))+(1)/(3)((deltax)/(x))^(2)-......oo]`
दोनों पक्षों को `deltax` से भाग देकर `underset(deltax to 0)` लेने पर
`underset(deltax to 0)lim (deltay)/(deltax)=(dy)/(dx)=underset(deltax to 0)lim (1)/(deltax).(deltax)/(x)[1-(1)/(2)((deltax)/(2))+(1)/(3)((deltax)/(x))^(2)-.....oo]`
`underset(deltax to 0)lim (1)/(x)[1-(1)/(2)((deltax)/(2))+(1)/(3)((deltax)/(x))^(2)-.....oo]`
`=(1)/(x)[1-0+0-0+....oo]=(1)/(x)`
`therefore (d)/(dx) log ax=(1)/(x)`