If ( x^ m y^ n =(x+y)^ m+n ) then ( d y d x ) is equal to

Step-by-step solution

Given that, ( x^ m y^ n =(x+y)^ m+n ) Taking log both the sides, we have ( m x+n y=(m+n) (x+y) ) Differentiating both the sides with respect to ( x ), we get [ array l m x + ( n y ) d y d x =(m+n) ( 1+ d y d x x+y ) m x - m+n x+y = ( m+n x+y ) d y d x - ( n y ) d y d x m x+m y-m x-n x x(x+y) = ( m y+n y-n x-n y (x+y) y ) d y d x 1 x = 1 y d y d x d y d x = y x array ]