Question

I have to find y' if ln(xy) = x + y.

Where do I start? I'm not sure what to do with that ln(xy)

Answer 1

What you do with ln(xy) is just separate it into 2 pieces: ln(x) + ln(y)


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

Answer 2

use the chain rule on the left:
d/dx(ln(u)) = 1/u du/dx
u = xy, so
du/dx = dx/dx*y + x*dy/dx = y + xy'

now plug and chug:

1/(xy) * (y + xy') = 1 + y'
y/xy + xy'/xy = 1 + y'
1/x + y'/y = 1 + y'
y'(1/y - 1) = 1 - 1/x
or
y' * (1-y)/y = (x-1)/x
y' = y(x-1)/[x(1-y)]

Answer 3

So then that could be moved around, and it would be (1-x)/(x-xy) ?

Answer 4

Oh geez two different answers.

Answer 5

That's right. Do you see where Raf left out the 1/y?

ln(x) + ln(y) = 1 + y'
1/x + y'/y = 1 + y'
1/x - 1 = y' (1 - <b>1/y</b>)

Answer 6

tnx for the correction Steve, my bad Sarah