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If #y=1+x+(x^2/(2!))+(x^3/(3!))+(x^4/(4!))+..... oo# so prove that ? #(dy)/(dx) = y#

If #y=1+x+(x^2/(2!))+(x^3/(3!))+(x^4/(4!))+..... oo# so prove that #(dy)/(dx) = y#

If #y=1+x+(x^2/(2!))+(x^3/(3!))+(x^4/(4!))+..... oo# so prove that #(dy)/(dx) = y#

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Jun 17, 2018

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Please see below.

Explanation:

Here,

#y=1+x+x^2/(2!)+x^3/(3!)+x^4/(4!)+x^5/(5!)+...oo#

#(dy)/(dx)=0+1+(2x)/(2!)+(3x^2)/(3!)+(4x^3)/(4!)+(5x^4)/(5!)+...oo#

#(dy)/(dx)=1+(2x)/(2xx1!)+(3x^2)/(3xx2!)+(4x^3)/(4xx3!)+ (5x^4)/(5xx4!)+...oo#

#(dy)/(dx)=1+x+x^2/(2!)+x^3/(3!)+x^4/(4!)+...oo#

#(dy)/(dx)=y#

Note :

#(i)1! =1#

#(ii)2! =2xx1!#

#(iii)3! =3xx2!#

#(iv)4! =4xx3!#
...............
................

#n! =nxx(n-1)!#

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