How Do You Derive y' from e^(xy) + xy = 2?

[Rhduke]

Homework Statement

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

the actual answer is -(y/x)

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

 

[Saladsamurai]

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

Casey

 

[cristo]

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!

 

[Saladsamurai]

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!

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

 

[cristo]

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

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.

 

[Rhduke]

Oh i keep forgeting u can't 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!

 

[cristo]

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

 

[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

 

[rocomath]

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

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

now take the ln of both sides and differentiate

 

[Saladsamurai]

Is that aimed at my British-ness?

Yes. And apparently, I need the tea...oh wait, here is some over here

Casey