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)
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)
6 answers
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)]
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)]
So then that could be moved around, and it would be (1-x)/(x-xy) ?
Oh geez two different answers.
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 - 1/y)
ln(x) + ln(y) = 1 + y'
1/x + y'/y = 1 + y'
1/x - 1 = y' (1 - 1/y)
tnx for the correction Steve, my bad Sarah