`cos^(2) x^(2)` का `x` के सपेक्ष प्रथम सिद्धांत से का अवकलन गुणांक ज्ञात कीजिएः

Explanation

माना `y=cos^(2) x^2` ......(i)
`therefore y+deltay=(x+deltax)^2` ......(ii)
`therefore y+deltay-y=cos^2(x+deltax)^(2) cos^(2)x^(2)` ........(iii)
`Rightarrow deltay=[cos (x+deltax)^2+cos x^(2)][cos(x+deltax)^2-cos x^(2)]`
`deltay={2 cos [((x+deltax)^(2)+x^(2))/(2)]cos [((x+deltay)^(2)-x^(2))/(2)]} xx{2sin [((x+deltax)^(2)+x^(2))/(2)].sin[(x^(2)-(x+deltax)^(2))/2]}`
`Rightarrow deltay={2 cos (x^(2)+x deltax+(deltax^(2))/(2))cos (deltax(2x+deltax)/(2))}.2 sin [x^(2)+xdeltax +((deltax)^(2))/(2)] {-sin [deltax(2x+deltax)/(2)]}`
`=-4cos [x^(2)+xdeltax+(deltax^(2))/(2)]sin[x^(2)+xdeltax+((deltax)^(2))/(2)]cos""(deltax(2x+deltax))/(2)sin""(delta(2x+deltax))/(2)`
दोनों पक्षों को `deltax` से भाग देकर `underset(deltax to 0)` लेने पर
`underset(deltax to 0)lim (deltay)/(deltax)=(dy)/(dx)=-4underset(deltax to 0)lim ((2x+deltax)/(2))cos (x^(2)+xdeltax+(deltax^(2))/(2))sin (x^(2)+x deltax+(deltax)^(2)/(2)) cos {([deltax(2x+deltax)])/(2)} ((sin(deltax(2x+deltax)))/(2))/(deltax((2x+deltax))/(2))`
`=-2 underset(deltax to 0)lim (2x+deltax)cos [x^2+xdeltax+(deltax^(2))/(2)]sin[x^(2)+xdeltax+(deltax^(2))/2]cos[(deltax(2x+deltax))/(2)] ((sin(deltax(2x+deltax)))/(2))/(deltax((2x+deltax))/(2))`
`=-2(2x+0)cos (x^(2)+0+0)sin(x^(2)+0+0)xx1xx1`
`-4xcos^(2)sin^(2)=-2x sin2x^(2)`