If `x^m.y^n=(x+y)^(m+n)`, show that `dy/dx =y/x`

Explanation

We have, `x^(m).y^(n) = (x+y)^(m+n)`
(i) Differentiating Eq. (i) w.r.t.x, we get
` d/dx (x^(m).y^(n)) = (d)/(dx)(x+y)^(m+n)`
`rArr x^(m).(d)/(dy)y^(n).(d)/(dx)x^(m)=(m+n)(x+y)^(m+n-1)(d)/(dx)(x+y)`
`rArr x^(m).ny^(n-1)'(dy)/(dx)+y^(n).mx^(m-1)=(m+n)(x+y)^(m+n-1)(1+(dy)/(dx))`
`rArr (dy)/(dx)[x^(m).ny^(n-1)-(m+n).(x+y)^(m+n-1)]=(m+n)(x+y)^(m+n-1)-y^(n)mx^(m-1)`
`rArr (dy)/(dx)[nx^(m)y^(n-1)-(m+n)(x+y)^(m+n-1)]=(m+n).(x+y)^(m+n-1)-(y^(n-1).y.mx)/(x)`
`:.(dy)/(dx)(((m+n)(x+y)^(m+n))/(x+y)-(y^(n-1).y.mx^(m))/(x))/((nx^(m)y^(n))/(y)-(m+n)(x+y)^(m+n)(1)/((x+y)))`
`=((x(m+n)(x+y)^(m+n)-(x+y).y.^(n-1)y.mx^(m))/((x+y).x))/(((x+y)nx^(m)y^(n)-y(m+n)(x+y)^(m+n))/((x+y).y))`
`=(((x(m+n).x^(m).y^(n)-m(x+y)y^(n)x^(m))/((x+y).x))/((x+y)nx^(m).y^(n)-y(m+n).x^(m).y^(n)))/((x+y).y)`, `[:'(x+y)^(m+n)=x^(m).y^(n)]`
`= (x^(m)y^(n)[mx+nx-mx-my].(x+y)y)/(x^(m)y^(n)[nx+ny-my-ny].(x+y).x)`
`= (y)/(x)`
(ii) Furthr, differentiating Eq. (ii) i.e., `(dy)/(dx) = (y)/(x)` on both the sides w.r.t. x, we get
`(d^(2)y)/(dx^(2)) = (x.(dy)/(dx)-y.1)/(x^(2))`
`= (x.(y)/(x)-y)/(x^(2)) , [:' (dy)/(dx) = (y)/(x)]`
`= 0`