`tan x^(2)` का प्रथम सिद्धांत द्वारा x के सापेक्ष अवकल गुणांक ज्ञात कीजिए ।

Explanation

माना ` f(x) = tan x ^(2)`
तब `f( x+h) = tan ( x+ h)^(2)`
` :. d/(dx) tan x^(2) = lim_( h to0) ( tan ( x+h)^(2) -tan x^(2))/(h)`
` = lim _( h to 0 ) ((sin(x+h)^(2))/(cos ( x+h)^(2)) -(sinx^(2))/(cosx^(2)))/h`
`= lim _( h to 0) ( sin ( x +h)^(2) cos x^(2) - cos ( x+ h)^(2). sin x^(2))/( h cos ( x+h)^(2) cos x^(2))`
` = lim _ ( h to 0) ( sin [ ( x + h)^(2) - x^(2)])/(hcos(x+h)^(2) cos x^(2))`
`= lim_( h to 0) ( sin [ ( x+h)^(2) - x^(2)])/( h cos ( x + h)^(2) . cos x^(2) )`
` = lim _( h to 0) (sin [ h^(2) + 2xh])/(h cos ( x+ h)^(2) . cos x^(2))`
` = lim _( h to 0) ( sin [h^(2) + 2xh])/( h cos (x +h)^(2) . cos x^(2))`
` = lim _( h to 0) ( sin h( h + 2x))/( h ( h + 2x)).((h+2x))/( cos ( x+h)^(2) . cos x^(2))`
`= lim _( h to 0) ( sin h(h + 2x))/(h ( h + 2x)).lim _ ( x to 0) ( h + 2x)/( cos ( x+ h)^(2) . cos x^(2))`
` = 1. (2x)/( cos x^(2). cos x^(2))= (2x)/( cos^(2)x)`
` = 2 sec ^(2) x^(2).`