प्रथम सिद्धांत से f का अवकलज ज्ञात कीजिए जहा fइस प्रकार प्रदत है :
(i) ` f(x) = ( 2x+3)/(x-2)` (ii) ` f(x) = x+1/x `

Explanation

(i) ` :' f'(x) = lim _( h to 0 ) ( f(x+h)-f(x))/h`
` = lim _( h to 0 ) ((2(x+h)+3)/(x+h-2)-(2x+3)/(x-2))/h`
` = lim _( h to 0 ) (( x-2)(2x+2h+3)-(2x+3)(x+h-2))/(h( x + h - 2)(x - 2 ))`
` = lim _( h to 0 )((x-2)(2x+3)+2h(x-2)-(2x+3)(x-2)-h(2x+3))/(h( x + h - 2)(x - 2 ))`
` = lim _( h to 0) (h(2x-4-2x-3))/(h(x+h-2)(x-2))`
` = lim _( h to 0 ) (-7)/(( x+h-2)(x-2))=(-7)/(x-2)^(2)`
(ii) ` :' f'(x) = lim _( h to 0) ( f (x+h)-f(x))/h`
` = lim _( h to 0) (( x +h+1/(x+h))-(x+1/x))/h`
` = lim _( h to 0 ) 1/h[1/(x+h)+x+h-x-1/x]`
` = lim _( h to 0 ) 1/h[ 1/(x+h) +h-1/x]`
` = lim _( h to 0 ) 1/h[(x+hx(x+h)-x-h)/(x(x+h))]`
` = lim _( h to 0 ) 1/h[(h(x^(2)+hx-1))/(x(x+h))]= lim_( h to 0 )[(x(x+h))/(x(x+h))-1/(x^(2)+xh)]`
` = 1- 1/(x^(2))`