`e^(x) cos x ` का x के सापेक्ष अवकल - गुणांक ज्ञात कीजिए ।

Explanation

माना `f(x)=e^(x)cos x`
` f( x +h) = e^(x+h) cos ( x+h)`
` (dy)/(dx)= lim_( h to 0)(f(x+h)-f(x))/h`
` d/(dx) (e^(x) cos x) = lim _( h to 0 ) (e^( x+h) cos ( x +h) -e^(x)cos x )/h`
` = lim _( h to 0) (e^(x).e^(x)cos( x+h)-cosx)/h`
`= lim _( h to 0 ) (e^(x)[ e^(h)cos ( x +h) - cos x))/h`
` = lim _( h to 0) (e^(x) [(1+ h+(h^(2))/(2!)+. . . . . )cos ( x +h) - cos x ] )/h`
` = lim _( h to 0 ) (e^(x)[ cos ( x +h) h cos ( x+h) + (( h^(2))/(2!)+. . . . )cos ( x +h) - cos x ])/h`
` lim _( h to 0) e^(x) [(cos( x+h)-cos x)/h+(h cos ( x+h))/h+h((h/(2!)+. . . . )cos( x +h))/h]`
` lim _( h to 0 ) e^(x) [(-2 sin ((2x+h)/h)sin. h/2)/(h//2)+cos ( x +h) (h/(2!)+. . . . )cos ( x +h)]`
` = e^(x) [ -sin x . 1 + cos x]`
` d/(dx)( e^(x) cos x) = e^(x) [ -sin x + cos x].`