We need to differentiate an implicit function. Implicit differentiation is very similar to regular differentiation. We just have to bear in mind that `y` is a function if `x.` Therefore, if we need to differentiate a function of `y` we will proceed as follows:

`d/dx f(y)=(df)/dy cdot dy/dx`

You can see more examples in the link below.

`y=log(xy)`

We differentiate the whole equation (both left and right side).

`dy/dx=1/(xy log10)cdot(y+x cdot dy/dx)`

In the line above we have applied the chain rule because `log(xy)` is a composite function. On top of that we have applied product rule for `xy` to get expression in the brackets.

`dy/dx=1/(x log 10)+1/(y log 10)dy/dx`

`dy/dx-1/(y log 10)dy/dx=1/(x log 10)`

Factor the left side.

`dy/dx(1-1/(y log 10))=1/(x log10)`

`dy/dx cdot (y log 10-1)/(y log 10)=1/(x log 10)`

`dy/dx=y/(x(ylog(10)-1))`** <-- Solution**

We could also put `log(xy)` instead of `y` but that wouldn't make the solution any simpler.

**Further Reading**