यदि ` y= sin [sqrt sin sqrt x 1,` तब ` (dy)/(dx) ` का मान ज्ञात कीजिए

Explanation

यहाँ ` " " y= sin [ sqrt (sin sqrtx) ]`
माना ` sqrt(x) =t, sin sqrt(x)= sin t=u ` व `sqrt (sin sqrt x) = sqrt(u) =v `
` therefore " " y= sin v `
` (dy)/(dx) =cos v, " "(du)/(dt)= cos t, " " (dv)/(du) =(1)/(2) u^(-1//2) `
व ` (dt)/(dx) =(1)/(2sqrt(x))`
हम जानते है की ` (dy)/(dx) =((dy)/(dv)xx(dv)/(du)xx(du)/(dt)xx(dt)/(dx)) " "...(1)`
सभी मान समीकरण (1 ) में रखने पर
` " "(dy)/(dx) =[ cos v*""(1)/(2sqrtu)*cos t*""(1)/(2sqrt ( x))]`
` " "= [cos sqrtu *(1)/(2sqrt(u))cost*""( 1)/(2sqrtx) ] " " जहाँ v= sqrt u `
`" "(1)/(4) cos ""( sqrt (sin t)) *(1)/(sqrt (sin sqrt x ) ) *cos sqrtx *(1)/(sqrt(x) ) `
` " "` जहाँ ` u= sin t`
` =(1)/(4) cos (sqrt sin sqrt x ) * ( 1)/(sqrt sin sqrt x )*cos sqrtx *(1)/(sqrt(x) ) `
` " " ` (जहाँ `t= sqrt x ` )
` = (cos sqrt sin sqrtx ) /( 4 sqrtx sqrt sin sqrt x ) * cos sqrt x `