anonymous
  • anonymous
e^(xy)dx=ye(xy), but why doesn't it also equal xe^(xy)?
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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katieb
  • katieb
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anonymous
  • anonymous
Whoops, I left off a ^ sign in the first part of the question; it should have read e^(xy)dx=ye^(xy).
anonymous
  • anonymous
It seems to me that the order the (xy) is in is arbitrary, i.e. e^(xy)=e^(yx), which should mean that e^(xy)dx can be either ye^(xy) or xe^(xy).
anonymous
  • anonymous
\(e^{xy}\,dx\) makes very little sense written like that. Do you mean \(\frac{d}{dx}e^{xy}=ye^{xy}\)? That is only true where \(y\) is not a function of \(x\).

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anonymous
  • anonymous
Yes, that's what I mean.
anonymous
  • anonymous
So if it was (d/dy)e^(xy), the answer would then be xe^(xy)?
anonymous
  • anonymous
$$\frac{d}{dx}e^{\underbrace{xy}_u}=\frac{d}{du}e^u\cdot\frac{du}{dx}=ye^u=ye^{xy}$$
anonymous
  • anonymous
... if and only if \(\frac{du}{dx}=\frac{d}{dx}xy=y\).
anonymous
  • anonymous
I don't understand.
anonymous
  • anonymous
Give me a minute, maybe I'll get it if I stare at the screen hard enough.
anonymous
  • anonymous
You're substituting u for xy and applying the chain rule, but when you use the product rule to find the derivative of xy, you derive with respect to x, which means the x is absorbed by the derivation process, leaving only the y to drop down in front of the base.
anonymous
  • anonymous
"absorbed by the derivation process" is a technical term, btw
anonymous
  • anonymous
I don't really get why the x is 'absorbed' from a mechanical sense.
anonymous
  • anonymous
Wait, it's because the derivative of x with respect to x is one, so it disappears. y(x)'=1. Something like that.
anonymous
  • anonymous
$$\frac{d}{dx}xy=\frac{dx}{dx}y+x\frac{dy}{dx}\\\frac{dx}{dx}=1,\frac{dy}{dx}=0\implies \frac{d}{dx}xy=1y+0x=y$$
anonymous
  • anonymous
That doesn't help, I have too many questions. Thanks anyways.
anonymous
  • anonymous
Feel free to ask. The derivative of \(x\) w.r.t. \(x\) is \(1\). If \(y\) is not a function of \(x\), i.e. it's constant with respect to \(x\), then its derivative is \(0\) -- hence the \(x\dfrac{dy}{dx}\) term vanishes.
anonymous
  • anonymous
So how do we know from x(dy/dx) that y isn't a function of x?
anonymous
  • anonymous
We don't. But if it were, what you wrote wouldn't be true:$$\frac{d}{dx}e^{xy}=e^{xy}\left(y+x\frac{dy}{dx}\right)$$If \(\dfrac{dy}{dx}\neq0\) then the above will not simplify to \(ye^{xy}\).
anonymous
  • anonymous
Damn Leibniz and his notation.
zepdrix
  • zepdrix
lol
zepdrix
  • zepdrix
Still confused on this one cap'n?
anonymous
  • anonymous
Yeah...
zepdrix
  • zepdrix
So where are we at? Trying to understand this derivative?\[\large \left(e^{xy}\right)'\]
anonymous
  • anonymous
I'm trying to understand why it's ye^(xy) rather than xe^(xy). I understand that it's the latter if the derivative is taken with respect to x and vice versa, but I don't understand why.
zepdrix
  • zepdrix
Mmk let's start from the top a sec, even though the other guys have prolly covered this. :) We now that e^x when differentiated gives us back e^x right? The same rule will apply to this exponential, we'll just have an extra step after that. \[\large \left(e^{xy}\right)' \qquad = \qquad e^{xy}\color{royalblue}{\left(xy\right)'}\] So we got the same thing back, but our extra step is to apply the chain rule. ~Multiply by the derivative of the exponent. So we have to apply the product rule.\[\large e^{xy}\left(xy'+x'y\right)\]And I guess this is the part where you're confused right? Does the product rule make sense so far?
anonymous
  • anonymous
$$\frac{d}{du}e^{xy}=e^{xy}\frac{d}{du}xy=e^{xy}\left(\frac{dx}{du}y+x\frac{dy}{du}\right)$$In the case where we want the derivative w.r.t. \(x\), we let \(u=x\):$$\frac{d}{dx}e^{xy}=e^{xy}\left(\frac{dx}{dx}y+x\frac{dy}{dx}\right)=e^{xy}\left(y+x\frac{dy}{dx}\right)$$If \(y\) is not a function of \(x\), we know \(\dfrac{dy}{dx}=0\) and thus our derivative simplifies further:$$e^{xy}(y+0x)=ye^{xy}$$
anonymous
  • anonymous
If we want the derivative w.r.t \(y\), we let \(u=y\):$$\frac{d}{dy}e^{xy}=e^{xy}\left(\frac{dx}{dy}y+x\frac{dy}{dy}\right)=e^{xy}\left(\frac{dx}{dy}y+x\right)$$If \(x\) is not a function of \(y\), i.e. \(\dfrac{dx}{dy}=0\), we find it simplifies further:$$e^{xy}(0y+x)=xe^{xy}$$
anonymous
  • anonymous
zepdrix: yeah, that makes sense so far.
zepdrix
  • zepdrix
So I guess the part you're having trouble with.. is understanding the difference between \(\large x'\) and \(\large y'\). Since we're differentiating with respect to x, \(\large x'\) is an insignificant term. Here's one way to interpret it maybe: The derivative is all about change. When we see this term, \(\large \dfrac{d}{dx}y\), It's asking us, "how much does y change, as x changes an immeasurably small amount?" Well we have rules for figuring that out. But when we see this term, \(\large \dfrac{d}{dx}x\) which is the same as \(\large x'\), It's asking us, how much does x change, as x changes?" It simply changes the same amount... So this tells us, \(\large x'=1\) Again though, remember, it's because we differentiated `with respect to x`. We wanted to know how everything else changed when x changed slightly. \[\large e^{xy}\left(xy'+x'y\right) \qquad = \qquad e^{xy}\left(xy'+y\right)\] I think it's suppose to be understood that \(\large y=y(x)\). So \(\large y'\) has some significance, and we have to leave the term like that.
zepdrix
  • zepdrix
I'm sorry about the extremely long-winded response lol. Oldrin pasted some good info also, take a look at that. Hopefully you get something out of this.
anonymous
  • anonymous
Whoa, I've been busy at work, I'll have to check this out when I get home. Thanks a lot for taking the time to post this!

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