If `y=1/(1+x^(n-m)+x^(p-m))+1/(1+x^(m-n)+x^(p-m))+1/(1+x^(m-p)+x^(n-p))` the `(dy)/(dx)` is

If y=1/(1+x^(n-m)+x^(p-m))+1/(1+x^(m-n)+x^(p-n))+1/(1+x^(m-p)+x^(n-p)) , then (dy)/(dx) is equal to

If y=1/(1+x^(n-m)+x^(p-m))+1/(1+x^(m-n)+x^(p-n))+1/(1+x^(m-p)+x^(n-p)) then (dy)/(dx) at e^m^n^p is equal to: e^(m n p) (b) e^(m n//p) (c) e^(n p//m) (d) none

if x^(m)*y^(n)=1 then (dy)/(dx)

Simplify : (1)/(1+x^(m-n)+x^(m-p))+(1)/(1+x^(n-p)+x^(n-m))+(1)/(1+x^(p-m)+x^(p-n)) .

If x^m y^n = (x+y)^(m+n) then find (dy)/(dx) at x=1, y=2

If I(m,n)=int_(0)^(1)x^(m-1)(1-x)^(n-1)dx, then

Simplify the following: 1/(1+x^(m-n)+x^(m-p))+(1)/(1+x^(n-p)+x^(n-m))+(1)/(1+x^(n-m)+x^(p-n))

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