# Dx ln problem

1. Dec 5, 2007

### Rhduke

1. The problem statement, all variables and given/known data

e^(xy) + xy = 2, find y'

3. The attempt at a solution

xy + lnxy = ln2
xy' + y + 1/x + y'/y = 1/2
xy' + 1/y (y') = 1/2 - y -1/x
y'(x + 1/y) = 1/2 - y - 1/x
y' = (1/2 - y - 1/x)/(x+1/y)

i simplified it further but could not get the right answer. help would be appreciated

2. Dec 5, 2007

I see you are using implicit diferentiation, good. Now the derivative of
ln(2) is NOT 1/2.

Casey

3. Dec 5, 2007

### cristo

Staff Emeritus
Yea, there's a few mistakes here. Firstly, you can't just take the logarithm of each term individually; you need to take the logarithm of each side. That's if you even want to take logs in the first place.

To be honest, I can't see how to do this; the implicit derivative of e^xy is bothering me. I'm blaming it on the fact that it's late!

4. Dec 5, 2007

Cristo he did take ln of both sides. It has just been simplified since the lne^(xy) is just xy. Have some tea!

Then we have product rule+chain rule=0

5. Dec 5, 2007

### cristo

Staff Emeritus
Is that aimed at my British-ness?

However, he didn't. If I have an expression $e^{xy}+a=b$ then taking logarithm of both sides yields $\ln(e^{xy}+a)=\ln(b)$. Now, $\ln(v+w)\neq\ln(v)+\ln(w)$ so the equation cannot be simplified as he has done in the OP.

6. Dec 5, 2007

### Rhduke

Oh i keep forgeting u cant distribute ln

ok i c what your saying so then i get:

ln(e^xy(x)(y)) = ln2
1/[e^xy(x)(y)] (e^xy) (xy' + y) + e^xy (xy' + y) = 0
(xy' + y)(1/xy + e^xy) = 0
xy' + y = 0
y' = -(y/x)

thank u all for your help!!

7. Dec 5, 2007

### cristo

Staff Emeritus
How did you get your first line? Taking the logarithm of your expression gives $\ln(e^{xy}+xy)=\ln 2$

8. Dec 5, 2007

### Rhduke

er oh.... i just realized i did that wrong, strangely it worked, ok i fixed it:

1/(e^xy + xy) (e^xy) (xy' + y) + (xy' + y) = 0
(xy' + y)[(e^xy)/(e^xy + xy) + 1] = 0
xy' + y = 0
y' = -(y/x)

my brain gets big picture but misses the small details

9. Dec 5, 2007

### rocomath

$$e^{xy}+xy=2$$

$$e^{xy}=2-xy$$

now take the ln of both sides and differentiate

10. Dec 5, 2007