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

Options

A. `1/2`
B. `2`
C. `(2m)/n`
D. `m/(2n)`

Answer

B

Explanation

Given `x^(m)y^(n)=(x+y)^(m+n)`
`mlogx+nlogy=(m+n)log(x+y)`
`m/x+n/y(dy)/(dx)=((m+n))/((x+y))[1+(dy)/(dx)]`
`implies(dy)/(dx)[n/y-((m+n))/((x+y))]=(m+n)/(x+y)-m/x`
`implies(dy)/(dx)=h/x`
`implies((dy)/(dx))_(x=1,y=2)=2`