Log(x/y)=x+y find dy/dx... As this is of differentiation chapter
log(x/y) = x+y
logx - logy = x+y
1/x + 1/y y' = 1 + y'
(1/y - 1)y' = 1 - 1/x
(1-y)/y y' = (x-1)/x
y' = (x-1)y / x(1-y)
logx - logy = x+y
1/x + 1/y y' = 1 + y'
(1/y - 1)y' = 1 - 1/x
(1-y)/y y' = (x-1)/x
y' = (x-1)y / x(1-y)