यदि `y= (sinx )^(tan x)+ (cos x)^(secx),` तब `(dy)/(dx) ` का मान ज्ञात कीजिए|

Explanation

यहाँ ` y= (sin x )^(tan x ) +(cos x)^(sec x)`
माना ` y_1 =(sin x )^(tan x )`
तथा `y_2 =(cos x)^(sec x)`
` therefore " "y=y_1 +y_2 `
chunki `y_1 =(sin x )^(tan x)`
दोनों पक्षों का log लेने पर
` log y_1 =(tan x )(log sin x )`
अब `'x'` के सापेक्ष अवकलन करने पर
`(1)/(y_1) *(dy_1)/(dx) =(d)/(dx) [(tan x)*(log sin x )] `
` = (tan x ) (d)/(dx) (log sin x)+ (logsin x ) *sec ^(2)x` `=(tan x )*(1)/(sin x ) *(cos x+ (log sin x ) *sec ^(2)x `
` rArr (dy_1)/(dx) =y_1 p[tan x*(1)/(tan x)+(log sin x) *sec^(2) x]`
` " "=y_1 [1+ (log sin x) *sec^(2) x]`
` =(sin x) ^(tan x) [1+ (log sin x )* sec^(2) x]`
`=(sin x) ^(tan x) [1+ (log sin x ) *sec^(2)x ]`
अब,` y_2 =(cos x) ^(secx) `
दोनों पक्षों का log लेने पर
` log y_2 =(secx )log (cos x ) `
x के सापेक्ष अवकलन करने पर,
` (1)/(y_2) *(dy_2)/(dx) (sec x) log ( cos x ) `
` =(secx ) *(1)/(cos x ) (-sinx )+ (log cosx) *sec x tan x `
` " "= (sec x tan x)[log (cos x )-1]`
` (dy_2)/(dx) =y_2 (secxtan x ) [log (cos x ) -1]`

` " ""=(cos x) ^(secx ) (sec x tan x )[ (log cos x ) -1]`
अब,` " "y=y_1 +y_2 `
` rArr" " (dy)/(dx) =(dy_1)/(dx) +(dy_2)/(dx) `
` (dy)/(dx) =(sin x )^(tan x)[ 1+ (log sin x) (sec^(2) x)]`
` " "+ (cos x)^(secx) (sec xtan x ) [(log cos x )-1)]`