यदि ` y=(log x)^(cos )+(x^(2)+1)/(x^(2)-1)`तब ` (dy)/(dx)` का मान ज्ञात कीजिए|

Explanation

यहाँ ` y= (log x )^(cos x )+ (x^(2) +1)/(x^(2)-1) `
माना ` y_1 =(log x )^(cos x ) ` तथा ` y_2 =(x^(2)+1)/(x^(2)-1) `
` therefore " "y= y_1 +y_2" "...(1) `
` " "y_1 =(log x ) ^(cos x ) ` लेने पर,
` log y_1 =cos x log (logx ) `
x के सापेक्ष अवकलन करने पर ,
` (1)/(y_1) *(dy_1)/(dx) =cos x (d)/(dx) [log (log x )] +[log (log x ) ](d)/(dx) cos x`
` =cos x *(1)/(log x ) *(1)/(x )-(sin x) ][log (log x )] `
` rArr " "(dy_1)/(dx) =y_1 [cos x *(1)/(xlog x ) (sin x )[log(log x ) ]]`
` " "=(log x )^(cosx) [(cos x)/(x log x ) -(sin x )* [log (log x )]]`
अब,` y_2 =(x^(2)+1)/(x^(2)-1) `
`'x'` के सापेक्ष अवकलन करने पर,
` (dy_2)/(dx) =((x^(2)-1)(d)/(dx) (x^(2) +1) -(x^(2) +1) (d)/(dx) (x^(2)-1) )/((x^(2) -1) ^(2) )`
` " "= ((x^(2) -1)*(2x -(x^(2) +1) *(2x)))/((x^(2) -1) ^(2) )= (-4x )/( (x^(2) -1)^(2))`
` therefore " "= y_=y_1 +y_2 `
` rArr(dy)/(dx) =(dy_1)/(dx)+ (dy_2)/(dx) `
` " "= (log x )^(cos x ) { (cos x )/(x log x ) -(sin ) log *(log x ) }`
` " " -(4x)/((x^(2) -1)^(2))`