Question

find dy/dx where ln(xy) = e

Answer 1

\[\frac{1}{xy}\times (xy'+y)=0\]solve for \(y'\)

Answer 2

don't know how you got there though...

Answer 3

i know that \[\ln(xy) = e \] is the same as \[e^e = xy\] but i dont' know what to do from there.

Answer 4

ln (xy) = lnx + lny

so lnx + lny = e

the first derivative is 1/x + 1/y dy/dx = 0

1/x = -1/y dy/dx
dydx = (1/x)/ (-1/y) = -y/x

Answer 5

you are thinking that \(y\) is a function of \(x\) even though you do not know hat that function is. so it is like taking the derivative of
\[\ln(xf(x))\] and you need the chain rule for this

Answer 6

the derivative of \[\ln(whatever)\] is
\[\frac{1}{whatever}\times \frac{d}{dx}(whatever)\] in your case it is
\[\frac{1}{xf(x)}\frac{d}{dx}[xf(x)]\] for the second part you need the product rule

Answer 7

the derivative of
\[xf(x)\] is
\[f(x)+xf'(x)\]

Answer 8

what seitys says makes a little more sense to me.

i get that lnx = 1/x but when i'm just sitting there with 1/x + 1/y = 0 i dont know how or why it gets moved to the other side.

Answer 9

now we rewrite this replacing \(f(x)\) by \(y\) and \(f'(x)\) by \(y'\) which gives us
\[\frac{1}{xy}\left(y+xy'\right)\]

Answer 10

i have to say his way is much better although you will wind up with the same thing

Answer 11

well i don't know why -1/ydy/dx gets moved over to the right side of the equal sign...

Answer 12

nothing gets moved to the other side
if you start with
\[\ln(x)+\ln(y)=e\] taking derivatives give
\[\frac{1}{x}+\frac{y'}{y}=0\] now your job is to solve for \(y'\) in terms of \(x\) and \(y\)
that part is algebra

Answer 13

it is like solving
\[\frac{1}{5}+\frac{z}{7}=0\] for \(z\)

Answer 14

\[\frac{1}{x}+\frac{y'}{y}=0\]subtract \(\frac{1}{x}\) get
\[\frac{y'}{y}=-\frac{1}{x}\]mutiply by \(y\) get
\[y'=-\frac{y}{x}\]

Answer 15

ooooh. thanks. i get it (i think lol)