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