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यदि ` y = x ^ x ` हो , तो सिद्ध कीजिये कि -
` ( d^2 y ) / ( dx ^ 2 ) - ( 1 ) /( y ) ( ( dy ) / ( dx )) ^ 2 - ( y ) /( x ) = 0 `

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` y = x ^ x `
दोनों पक्षों का लघुगणक लेने पर,
` logy = x log x `
दोनों पक्षों का x के सापेक्ष अवकलन करने पर,
` ( 1 ) /( y ) * ( dy ) / ( dx ) = x * ( 1 ) /( x ) + log x `
` rArr ( 1 ) /( y ) ( dy ) / ( dx ) = 1 + log x `
` rArr ( d y ) /( dx ) = y ( 1 + log x )" " `...(1)
दोनों पक्षों का x के सापेक्ष पुनः अवकलन करने पर,
` ( d^ 2 y ) /( dx ^ 2 ) = y xx ( 0 + ( 1 )/( x )) + ( dy ) / (dx ) ( 1 + log x ) `
` rArr ( d ^ 2 y ) /( dx ^ 2 ) = ( y ) /( x ) + ( 1 + log x ) ( dy ) /( dx ) " " `...(2)
अब, ` ( d^ 2 y ) / ( dx ^ 2 ) - ( 1 ) /( y ) (( dy ) / ( dx ) ) ^ 2 - ( y ) / ( x ) `
` = ( y ) / ( x ) + ( 1 + log x ) ( dy ) / ( dx ) - ( 1 ) /( y ) [ y ( 1 + log x ) ] ^ 2 - ( y ) / ( x ) `
[ समी. (1 ) और (2 ) से ]
` = ( y ) / (x ) + ( 1 + log x ) y ( 1 + log x ) - y ( 1 + log x ) ^ 2 - ( y ) / ( x ) `
` = ( y ) / ( x ) + y ( 1 + logx ) ^2 - y ( 1 + log x ) ^ 2 - ( y ) / ( x ) `
` = 0 ` .
अतः ` ( d ^2 y ) / ( dx ^ 2 ) - ( 1 ) /( y ) ((dy ) /( dx) ) ^ 2 - ( y ) / ( x ) = 0 `

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