`e^(ax)` का प्रथम सिद्धांत से अवकल गुणांक ज्ञात कीजिए ।

Explanation

माना ` f(x) = e^(ax) , f( x+h) = e^(a(x+h)`
` (dy)/(dx) = lim_( h to 0) ( e^(a(x+h))-e^(ax))/h`
` = lim _ ( h to 0) ( e^(ax) (e^(ah)-1))/h`
` = lim _( h to 0) ( e^(ax)(1+ ah + (ah^(2))/(2!) + . . . . . . -1))/h`
` = lim_ ( h to 0) (e^(ax)(1+ ah+ (ah)^(2)/(2!)+ . . . . . -1))/(2!)`
` = lim _ ( h to 0) ( he^(ax) (a+ (ah^(2))/(2!) . . . . . . ))/h`
` = lim _( h to 0) e^(ax) ( a+ (ah^(2))/(2!)+ . . . . )`
` = e^(ax) xx a`
` = ae^(ax)`