Logarithmic differentiation to find dy/dx

[BlackMamba]

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: x^y = y^x

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: \frac{ylnx}{x}=lny

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 / x^2. So I then used the product rule instead. However I'm still left with ylnx 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.

 

[Jameson]

Start from here: y\ln(x)=x\ln(y)

Now use implicit differentiation to solve for \frac{dy}{dx}

 

[BlackMamba]

Ok. I had that, and as usual I went too far. I'll try that and see what I get. Thanks

 

[BlackMamba]

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


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:

\frac{d}{dx} [xlny]

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

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

x (lny (\frac{dy}{dx})) + lny

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:

\frac{d}{dx}[ylnx]

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

y(\frac{1}{x}) + lnx (\frac{dy}{dx})


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

y(\frac{1}{x}) + lnx (\frac{dy}{dx}) = x (lny (\frac{dy}{dx})) + lny

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

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

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

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

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

Am I making this more complicated then it should be?

 

[whozum]

\frac{d}{dx} [xlny]

You need to use the product rule first to separate, the two, then the chain rule on \ln y

 

[BlackMamba]

Yeah I figured that out and did that later in my post above.

 

[BlackMamba]

OK I've just read somewhere that the constant \frac{d}{dx}[lny] = 0.

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

\frac{dy}{dx} = \frac{lny - y(\frac{1}{x})}{lnx}


Is that correct?

 

[Jameson]

No.

\frac{d}{dx}[lny] = \frac{1}{y}*\frac{dy}{dx}

 

[BlackMamba]

Alright well, I'm completely lost.

My first answer is obviously wrong then. This sucks.

 

[BlackMamba]

So Jameson, couldn't I just use then the Chain rule here:

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

Then that would give me the answer:

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

But I still am unsure how to substitute for y^x = x^y Would I just simply put x^y where y is?

 

[BlackMamba]

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.

 

[BlackMamba]

How about this:

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

Is this looking any closer to a plausible answer??

 

[BlackMamba]

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.

 

[whozum]

y\ln(x)=x\ln(y)
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:

\frac{dy}{dx}\ln x + \frac{y}{x} = \ln y + \frac{x}{y}\frac{dy}{dx}

After that solving for \frac{dy}{dx} shouldn't be too tough.

 

[BlackMamba]

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.

\frac{y}{x} + \frac{dy}{dx}lnx = \frac{x}{y}\frac{dy}{dx} + lny

\frac{y}{x} - lny = \frac{x}{y}\frac{dy}{dx} - \frac{dy}{dx}lnx

\frac{y}{x} - lny = (\frac{x}{y} - lnx)\frac{dy}{dx}

\frac{\frac{y}{x} - lny}{\frac{x}{y} - lnx} = \frac{dy}{dx}

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?

 

[dextercioby]

There's another solution which doesn't require logarithmic differentiation: using the theorem on implicit functions.

Daniel.

 

[BlackMamba]

Thanks dextercioby. Unfortunately, I'm required to use logarithmic differentiation for this problem.

 

[whozum]

\frac{\frac{y}{x} - lny}{\frac{x}{y} - lnx} = \frac{dy}{dx}
Start by doing this step:
... = \frac{\frac{y-x\ln y}{x}}{\frac{x-y\ln x}{y}}, from there you can simplify the nested fraction a bit more.

 

[HallsofIvy]

\frac{\frac{y}{x} - lny}{\frac{x}{y} - lnx} = \frac{dy}{dx}
looks to me like a perfectly valid answer, but if the fractions are bothering you , multiply both numerator and denominator by xy:
\frac{dy}{dx}= \frac{y^2- xylny}{x^2- xylnx}.

 

[BlackMamba]

Thank you whozum and HallsofIvy. I tend to make things more difficult then they really need to be. Thanks again. :)