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

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

Casey
 
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 said:
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
 
Saladsamurai said:
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? :smile:

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.
 
Oh i keep forgeting u can't distribute ln :rolleyes:

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

e^{xy}=2-xy

now take the ln of both sides and differentiate
 
  • #10
cristo said:
Is that aimed at my British-ness? :smile:

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

Casey
 
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