`sqrt(sin 3x)` को प्रथम सिद्धांत से अवलंकित कीजिएः

Explanation

माना `y=sqrt(sin 3x)`
`Rightarrow y+deltay=sqrt(sin 3(x+deltax))`
`Rightarrow y+deltay-y=sqrt(sin 3(x+8x))-sqrt(sin 3x)`
`Rightarrow (deltay)/(deltax)=(sqrt(sin 3(x+deltax))-sqrt(sin 3x))/(deltax)`
अब `(dy)/(dx)=underset(deltax to 0)lim (deltay)/(deltax)`
`=underset(deltax to 0)lim [({sqrt(sin (3x+3deltax))-sqrt(sin 3x)})/(deltax)xx({sqrt(sin 3x+3deltax)+sqrt(sin3x)})/({sqrt(sin(3x+3deltax))+sqrt(sin 3x)})]`
`=underset(deltax to 0)lim ([sin (3x+3deltax)-sin3x])/(deltax.{sqrt(sin(3x+3deltax))+sqrt(sin 3x)})`
`=underset(deltax to 0)lim (2 cos (3x+(3)/(2)deltax)sin((3)/(2)deltax))/(deltax.{sqrt(sin (3x+3deltax))+sqrt(sin 3x)})`
`=underset(deltax to 0)lim [2cos(3x+(3)/(2)deltax).(sin((3)/(2)deltax))/(((3)/(2)deltax))xx(3)/(2) (1)/(sqrt(sin (3x+3deltax))+sqrt(sin 3x))]`
`=underset(deltax to 0)lim cos (3x+(3)/(2)deltax).underset((3)/(2)deltax to 0)lim (sin((3)/(2)deltax))/(((3)/(2)deltax)).3 underset(deltax to 0)lim (1)/(sqrt(sin(3x+3deltax))+sqrt(sin 3x))`
`=3 cos 3x xx 1 xx (1)/(2sqrt(sin3x))=(3 cos 3x)/(2 sqrtsin 3x)`