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