यदि ` (tan ^(-1) x )^(y) +y ^(cot x ) =1 ` तब,`(dy)/(dx)` का मान ज्ञात कीजिए|

Explanation

माना ` " "y_1 =(tan ^(-1) x) ^(y) ` व ` y_2=y^(cot x )`
तब, ` " "y_1 +y_2 =1`
x के सापेक्ष अवकलन करने पर
` rArr" "(dy_1)/(dx) +(dy_2)/(dx) =0" "....(1)`
` y_1 =(tan ^(-1) x )^(y) ` लेने पर
` " " log y_1 =y log (tan ^(-1) x )`
x के सापेक्ष अवकलन करने पर,
` (1)/(y_1) *(dy_1)/(dx) =y* (1)/(tan ^(-1) x ) *(1) /((1+x^(2)))+ log (tan ^(-1) x )(dy)/(dx) `
` (dy_1)/(dx) =y_1[y* ( 1)/(tan ^(-1) x)* ( 1)/((1+x^(2) ))+ log (tan ^(-1)x )(dy)/(dx) ]`
` =(tan ^(-1)x) ^(y) { (y)/((1+x^(2))(tan ^(-1) x ))+ log (tan ^(-1) x)(dy)/(dx) }`
` " "=(y (tan ^(-1) x )^(y-1))/((1+x^(2)))+ (tan ^(-1) x )^(y) log (tan ^(-1) x ) (dy)/(dx) " "...(2)`
अब ` " "y_2 =y^(cotx )`
` " "log y_2 =cot x log y`
x के सापेक्ष अवकलन करने पर,
`(1)/(y_2) *(dy_2)/(dx)= [(cot x ) *(1)/(y) (dy)/(dx) +(log y) (-cosec ^(2)x )]`
` (dy_2)/(dx) =y_2 [(cot x )/(y ) (dy)/(dx) - cosec ^(2) x (log y) ]`
` =(cot x ) y^(cot x-1) *(dy)/(dx) -y ^(cot x )(cosec ^(2)x ) (log y ) ....(3)`
समीकरण (1 ) ,(2 ) व (3 ) से
` (y(tan ^(-1)x ) ^(y-1))/( (1+x^(2)))+ (tan ^(-1) x ) ^(y) log(tan ^(-1)x ) (dy)/(dx) `
` " "+ (cot x )y ^((cotx-1))*(dy)/(dx) `
` " "-y ^(cot x ) (cosec ^(2) x ) (logy )=0`
`[(tan ^(-1) x)^(y) log (tan ^(-1) x ) + (cot x ) y^((cotx -1))](dy)/(dx) `
`" "=( y)^(cot x) (cosec ^(2) x ) (log y ) -(y(tan ^(-1) x )^(y-1))/((1+x^(2)))`
` (dy)/(dx) =(y^(xot x)(cosec ^(2)x) (log y )- (ytan ^(-1)x) ^(y-1))/((tan ^(-1) x)^(y) log (tan ^(-1) x )+ (cot x ) y^(cotx-1))`