Logarithmic differentiation to find dy/dx

  • Thread starter BlackMamba
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  • #1
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Hello,

I know how I need to do the problem, I have figured out an answer as best as I could but I have a feeling the answer is not correct. In fact I'm pretty sure it's not. Anyway, I'm supposed to use the logarithmic differentiation to find dy/dx for the equation: [tex]x^y = y^x[/tex]

I know I need to take the ln of both sides first before finding dy/dx. My question is do I need to get y alone first then take the ln? As I've done the problem I just took the ln of each side isolating y that way. So after taking the log of both sides my answer at that point was: [tex]\frac{ylnx}{x}=lny[/tex]

So if I've done that right, I went on to find the derivative dy/dx for the above equation. I used the quotient rule and the chain rule. However when I used the chain rule, I somehow got 0 / [tex]x^2[/tex]. So I then used the product rule instead. However I'm still left with [tex]ylnx[/tex] in part of my answer but I didn't think that was possible since y is a function of x. I'm so confused, any help would be greatly appreciated.
 
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Answers and Replies

  • #2
793
4
Start from here: [tex]y\ln(x)=x\ln(y)[/tex]

Now use implicit differentiation to solve for [tex]\frac{dy}{dx}[/tex]
 
  • #3
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Ok. I had that, and as usual I went too far. I'll try that and see what I get. Thanks
 
  • #4
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Alrighty then. Well I'm confused as to which would be the proper rule to use. I tried both and I'm getting stuck with both. I've used the product rule first, then started over and used the chain rule. In any case I'll have lny and again I didn't think that was possible since y is a function of x.

Perhaps I'm not applying the chain rule correctly.

If I have [tex]\frac{d}{dx} [xlny][/tex] then using chain rule that would be written as: [tex]xlny \frac{d}{dx}[lny][/tex]


Looking at it more, I don't think I can use the Chain rule like that. The equation would read as the variable x (times) the ln of y so I am thinking that the product rule is needed here in which case I would work it out like such:

[tex]\frac{d}{dx} [xlny][/tex]

[tex]x \frac{d}{dx} [lny] + lny \frac{d}{dx} [x][/tex]

[tex]x (lny \frac{d}{dx}[y]) + lny[/tex]

[tex]x (lny (\frac{dy}{dx})) + lny[/tex]

Not too sure where to go from here.... (This would be the right side of the equation.)

The left side side would look something like this:

[tex]\frac{d}{dx}[ylnx][/tex]

[tex]y \frac{d}{dx}[lnx] + lnx \frac{d}{dx} [y][/tex]

[tex]y(\frac{1}{x}) + lnx (\frac{dy}{dx})[/tex]


So those two sides set equal to one another would look something like this:

[tex]y(\frac{1}{x}) + lnx (\frac{dy}{dx}) = x (lny (\frac{dy}{dx})) + lny[/tex]

[tex]y(\frac{1}{x}) + lnx (\frac{dy}{dx}) = xlny (\frac{dy}{dx}) + lny[/tex]

[tex]lnx (\frac{dy}{dx}) - xlny (\frac{dy}{dx}) = lny - y(\frac{1}{x})[/tex]

[tex]\frac{dy}{dx} (lnx - xlny) = lny - y(\frac{1}{x})[/tex]

[tex]\frac{dy}{dx} = \frac{lny - y(\frac{1}{x})}{(lnx - xlny)}[/tex]

Ok well I think I've isolated [tex]\frac{dy}{dx}[/tex] the right way, but normally we are supposed to substitute the original equation for y into the equation above. I don't have a [tex]y^x[/tex] above so do I just subsititue anyway. Like put [tex]x^y[/tex] where y is currently above? Or would it just be left as is?

Am I making this more complicated then it should be?
 
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  • #5
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[tex]\frac{d}{dx} [xlny][/tex]

You need to use the product rule first to separate, the two, then the chain rule on [itex] \ln y[/itex]
 
  • #6
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Yeah I figured that out and did that later in my post above.
 
  • #7
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OK I've just read somewhere that the constant [tex]\frac{d}{dx}[lny] = 0[/tex].

In which case that would change my answer above to something like this:

[tex]\frac{dy}{dx} = \frac{lny - y(\frac{1}{x})}{lnx}[/tex]


Is that correct?
 
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  • #8
793
4
No.

[tex]\frac{d}{dx}[lny] = \frac{1}{y}*\frac{dy}{dx}[/tex]
 
  • #9
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Alright well, I'm completely lost.

My first answer is obviously wrong then. This sucks.
 
  • #10
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So Jameson, couldn't I just use then the Chain rule here:

[tex]\frac{d}{dx}[ylnx] = \frac{d}{dx}[xlny] [/tex]

Then that would give me the answer:

[tex][\frac{ylnx}{x^2lny}]y = \frac{dy}{dx}[/tex]

But I still am unsure how to substitute for [tex]y^x = x^y[/tex] Would I just simply put [tex]x^y[/tex] where y is?
 
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  • #11
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Uhmm ok no that wouldn't be the answer. LOL Now I'm just trying anything causing I'm getting desperate. But I know better than that. Disregard what I posted above ^.

This is getting highly frustrating.
 
  • #12
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How about this:

[tex][\frac{y(\frac{1}{x}) - lny}{xlnx}]y = \frac{dy}{dx}[/tex]

Is this looking any closer to a plausible answer??
 
  • #13
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Ok, I see that the above answer isn't a possibility either. I'm through with it for tonight. It seems like I'm willing to just throw anything out there for an answer and that's not going to get me anywhere.
 
  • #14
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[tex]y\ln(x)=x\ln(y)[/tex]
Implicit differentiation, differentiate each term on both sides with respect to x given that y is a function of x:
Left hand side needs the product rule first then the chain rule. Right hand side needs the same thing.
Heres a big step if you want to cheat but I recommend doing it yourself:

[tex] \frac{dy}{dx}\ln x + \frac{y}{x} = \ln y + \frac{x}{y}\frac{dy}{dx} [/tex]

After that solving for [itex]\frac{dy}{dx}[/itex] shouldn't be too tough.
 
  • #15
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Well that's just it whozum. I have that exact line in my work here on my paper but I must just be going in the wrong direction to solve.

I'll just show you the way I am solving it and you can tell me where I'm going wrong. I'll start off from your line above, but I have it written slightly different on my paper.

[tex]\frac{y}{x} + \frac{dy}{dx}lnx = \frac{x}{y}\frac{dy}{dx} + lny[/tex]

[tex]\frac{y}{x} - lny = \frac{x}{y}\frac{dy}{dx} - \frac{dy}{dx}lnx[/tex]

[tex]\frac{y}{x} - lny = (\frac{x}{y} - lnx)\frac{dy}{dx}[/tex]

[tex]\frac{\frac{y}{x} - lny}{\frac{x}{y} - lnx} = \frac{dy}{dx}[/tex]

Now I know fractions left in an answer such as this one is not favorable, but I'm not entirely sure how to get rid of it.

Anyway, where am I going wrong my work above?
 
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  • #16
dextercioby
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There's another solution which doesn't require logarithmic differentiation: using the theorem on implicit functions.

Daniel.
 
  • #17
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Thanks dextercioby. Unfortunately, I'm required to use logarithmic differentiation for this problem.
 
  • #18
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[tex]\frac{\frac{y}{x} - lny}{\frac{x}{y} - lnx} = \frac{dy}{dx}[/tex]
Start by doing this step:
[tex] ... = \frac{\frac{y-x\ln y}{x}}{\frac{x-y\ln x}{y}} [/tex], from there you can simplify the nested fraction a bit more.
 
  • #19
HallsofIvy
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[tex]\frac{\frac{y}{x} - lny}{\frac{x}{y} - lnx} = \frac{dy}{dx}[/tex]
looks to me like a perfectly valid answer, but if the fractions are bothering you , multiply both numerator and denominator by xy:
[tex]\frac{dy}{dx}= \frac{y^2- xylny}{x^2- xylnx}[/tex].
 
  • #20
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Thank you whozum and HallsofIvy. I tend to make things more difficult then they really need to be. Thanks again. :)
 

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