Partial Derivatives

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Jun 12th 2008, 09:17 AM
vodka

Partial Derivatives

Can someone show me the steps to solving these...

f(x,y) = x ln y + y ln x

df/dx =

df/dy =

df/dx^2 =

df/dy^2 =

df/dydx =

df/dxdy =

I believe df/dx = ln y + y/x

and df/dy = x/y + lnx

how would I find the critical points from that?
Jun 12th 2008, 09:40 AM
colby2152

Quote:

Originally Posted by vodka

Can someone show me the steps to solving these...

f(x,y) = x ln y + y ln x

df/dx =

df/dx is the partial derivative of f(x, y) with respect to x; every other variable is considered a constant, so let's derive:

$\frac{df}{dx} = \ln{y}*x'+ y*(\ln{x})'$

Derivative w.r.t x of x is just 1. The derivative of ln(x) is 1/x

$\frac{df}{dx} = \ln{y} + \frac{y}{x}$

That simple? Yep. I believe you can solve the rest of the problem.
Jun 12th 2008, 10:10 AM
TheEmptySet

1 Attachment(s)

Quote:

Originally Posted by vodka

Can someone show me the steps to solving these...

I believe df/dx = ln y + y/x

and df/dy = x/y + lnx

how would I find the critical points from that?

Finding the critical points is not easy.

I don't think it can be solved analytically(I could be wrong(Wink))

I drew a graph to see what it looked like and here it is.

Attachment 6763

So we want to solve the system of equations

$\ln(y)+\frac{y}{x}=0 \iff \frac{x}{y}=-\frac{1}{\ln(y)}$

$\frac{x}{y}+\ln(x)=0$ subbing the first into the 2nd we get

$-\frac{1}{\ln(y)}+\ln(x)=0 \iff -1 +\ln(x)\ln(y)=0 \iff ln(x)\ln(y)=1$

finally we get $\ln(y)=\frac{1}{\ln(x)} \iff y=e^{\frac{1}{\ln(x)}}$

If we sub this back into the first equation we get

$\frac{1}{\ln(x)}+\frac{e^{\frac{1}{\ln(x)}}}{x}=0$

Finally we can use newtons method on the last equation

$x_{n+1}=x_n-\frac{f(x_n)}{f'(x_n)}$

Based on the graph from above I used $x_0=0.5$

After applying newtons method three times I got the values(on a calculator)

$0.5,0.3628837,0.36784455,0.36787943$

Oddly enough when I repeated the process but eliminted the x's and used newtons method again I got the same value for y.

so the critical numbers are $(\approx 0.36787943,\approx 0.36787943$ )

I hope this helps.

P.S if anyone knows another way I would love to see it

Thanks, The Empty Set.
Jun 14th 2008, 05:21 PM
TheEmptySet

I have been thinking about this and just noticed that

$\ln(y)+\frac{y}{x}=0 \iff x=-\frac{y}{\ln(y)}$ and

$\frac{x}{y}+\ln(x)=0 \iff y=-\frac{x}{\ln(x)}$

Are inverses of each other. So they are symmetric over the line y=x

This means that any of their intersections would have to be on the line y=x.

So if I sub this into the first equation I get

$x=-\frac{x}{\ln(x)} \iff \ln(x)=-1 \iff x=e^{-1}=\frac{1}{e} \approx 0.36787$

So the exact value is $x=y=\frac{1}{e}$