निम्न फलनों का अवकलन कीजिये|
`(i) x^(x)`
` (ii) x^(sin-1)x`
`(iii) (sin x )^(tanx )`

Explanation

(i ) माना `" "y =x^(x)`
दोनों पक्षों का लघुगणक लेने पर
` " "log y= log (x^(x))=x log x `
दोनों पक्षों का x के सापेक्ष अवकलन करने पर
` " "(d)/(dy) logy(dy)/(dx) =(d)/(dx) [x log x]`
` rArr " "(1)/(y)(dy)/(dx) =x(d)/(dx) log x +log x (d)/(dx) x `
` rArr " "(1)/(y)(dy)/(dx) =x*(1)/(x) +log x`
`rArr" "(dy)/(dx) =y (1+log x)`
अतः ` " "(dy)/(dx) =x^(3) (1+log x ) " "(because y= x^(x))`
`(ii)` माना ` " "y=x ^(sin -1)x `
दोनों पक्षों का लघुगुणक लेने पर,
` " "log y =log x ^(sin-1x)`
` " "log y =sin ^(-1) xlog x `
` therefore (d)/(dy) log y (dy)/(dx)=sin ^(-1) x(d)/(dx) log x +log x(d)/(dx) sin ^(-1)x`
` rArr " "(1)/(y)(dy)/(dx) =(sin ^(-1)x)/(x) +(log x)/(sqrt(1-x^(2)))`
` rArr " "(dy)/(dx)= y[(sin ^(-1) x)/(x)+ (log x)/(sqrt(1-x^(2))]]`
` therefore" "(dy)/(dx) =x^(sin -1_x)[(sin ^(-1)x)/(x)+(log x)/(sqrt(1-x^(2)))]`
` " " (because y= x^(sin ^(-1_x)))`
`(iii)` माना ` " " y= (sinx )^(tan x)`
दोनों पक्षों का लघुगुणक लेने पर
`" "log y=log (sinx )^(tan x ) `
` " "=tan x log (sinx ) `
दोनों पक्षों का x के सापेक्ष अवकलन करने पर
` (d)/(dy) log y (dy)/(dx) =tan x (d)/(dx) log sin x+ log sin x (d)/(dx) tan x `
` rArr " "(1)/(y) (dy)/(dx) =tan x cot x +sec ^(2) x log sin x `
` " "= 1+sec ^(2) x log sin x `
` rArr " "(dy)/(dx) =y [1+sec^(2) x log sin x ]`
` " "= (sin x ) ^(tan x ) [1+sec^(2)x log sin x ]`