If `y = (1 + (2)/(x)) (1 + (2)/(x))(1 + (3)/(x))...(1 + (n)/(x))`
` x ne 0 , "then " (dy)/(dx) ` when x = - 1 is

If y=(1+1/x)(1+2/x)(1+3/x)…(1+n/x) and x ne 0," then "(dy)/(dx) when x = -1 is :a)n! b) (n-1)! c) (-1)^(n)(n-1)! d) (-1)^(n)n!

If y = 1 +(1)/(x) +(1)/(x^(2)) + (1)/(x^(3)) + ……..+oo with |x| gt 1 then (dy)/(dx) is .

If y=1+(1)/(x)+(1)/(x^(2))+(1)/(x^(3))+… to oo with |x|gt 1 then (dy)/(dx)=

( 1 ) /(x) + (1 ) /( y) = 7 , ( 2)/(x) + ( 3) /( y) = 17 (x ne 0, y ne 0 ) .

If y = (1 + x) (1 + x^(2)) .. (1 + x^(2n)) , then ((dy)/(dx))_(x = 0) is equal to

If y=1+(1)/(x)+(1)/(x^(2))+(1)/(x^(3))+... to oo with |x|>1 then (dy)/(dx)=

2 x y+y^2-2 x^2 (dy)/(dx)=0 , y=2 when x=1

If (x)/(sqrt( 1+x)) +(y)/( sqrt(1+y))=0and x ne y ,then (dy)/(dx)=