# What is the derivative y' of x + y + ln(xy) = 0

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Differentiate x + y + ln(xy) = 0

`dx + dy + 1/(xy) d(xy) = 0` Using the chain rule

d(xy) = xdy+ydx using the product rule

`dx+dy+1/(xy)(xdy+ydx)=0`

Distributing `1/(xy)` we get

`dx + dy + dy/y + dx/x = 0`

Moving the dx terms to the right side

`dy + 1/y dy = - (dx + 1/x dx)`

Factoring dx and dy we get

`dy(1+1/y) = -(1+1/x)dx`

`dy/dx = -(1+1/x)/(1+1/y) = -((x+1)/x)/((y+1)/y)=-(y(x+1))/(x(y+1)) `

We have to find the derivative of x + y + ln(xy) = 0. This can be done using implicit differentiation.

x + y + ln(xy) = 0

=> [x + y + ln(xy)]' = 0

=> x' + y' + [ln x]' + [ln y]' = 0

=> 1 + y' + 1/x + (1/y)y' = 0

=> y'(1 + 1/y) = -1 - 1/x

=> y' = (-1 - 1/x)*y/(1 + y)

**The derivative of the given expression is y' = (-1 - 1/x)*y/(1 + y)**