Differentiating implicitly with the natural log?

Question

2lnx+lny=x-y.

I'm stuck because I don't know the derivative of lny?

Answer 1

the derivative of lny is (1/y)(dy/dx) (you need the dy/dx at the end because it is not x, it is a function of x)
so just try that, well, you can do:
(2/x)+(1/y)(dy/dx)=1-(dy/dx)
and just solve for dy/dx by getting it on one side.
(dy/dx)+(1/y)(dy/dx)=1-(2/x)
(dy/dx)(1+(1/y))=1-(2/x)
(dy/dx)=(1-(2/x))/(1+(1/y))

i think that is correct.

((FYI: (1/y)(dy/dx) is the same as y'/y (imagine just multiplying the dy/dx into the numerator of 1/y---its like you have (dy/dx)/y dy/dx is same thing as y'))

you are then supposed to substitute the original function in for y, i believe, to get rid of y, if that is what you want.

Answer 2

derivating both sides both sides
2/x+1/y *dy/dx =1 -dy/dx
so ((1/y)+1) dy/dx = 1- 2/x==> dy/dx= (x-2)/x * y/((y+1)

Answer 3

2/x+y'/y = 1-y'
Solve for y'
y'
= (1-2/x)/(1/y+1)
= y(x-2)/[x(1+y)]