1.)f(x,y) = arctan(y/x) fx(2,3)

how would I do this?

2.) f(x,y) = ln(x+ sqrt(x^2+y^2) ) fx(3,4)

1/(+ sqrt(x^2+y^2) * 1+ 1/2(x^2 + y^2) * 2x

is that correct for #2?

Thanks

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- Nov 12th 2008, 07:15 PMkhuezyneed help on partial derivatives
1.)f(x,y) = arctan(y/x) fx(2,3)

how would I do this?

2.) f(x,y) = ln(x+ sqrt(x^2+y^2) ) fx(3,4)

1/(+ sqrt(x^2+y^2) * 1+ 1/2(x^2 + y^2) * 2x

is that correct for #2?

Thanks - Nov 12th 2008, 07:21 PMChris L T521
First, partially differentiate f(x,y) with respect to x, and then evaluate it at (2,3):

$\displaystyle f(x,y)=\tan^{-1}\left(\frac{y}{x}\right)\implies\frac{\partial f}{\partial x}=\frac{1}{1+\left(\frac{y}{x}\right)^2}\cdot \left(-\frac{y}{x^2}\right)=-\frac{y}{x^2+y^2}$

Now what is $\displaystyle f_x\left(2,3\right)$??

--Chris - Nov 12th 2008, 07:23 PMkhuezythanks
i just plug in x and y into that equation right?

- Nov 12th 2008, 07:24 PMkhuezythanks
how would I do this?

2.) f(x,y) = ln(x+ sqrt(x^2+y^2) ) fx(3,4)

1/(+ sqrt(x^2+y^2) * 1+ 1/2(x^2 + y^2) * 2x

is that correct for #2? - Nov 12th 2008, 07:30 PMChris L T521
- Nov 12th 2008, 07:31 PMChris L T521
- Nov 12th 2008, 08:08 PMkhuezyhi again
Find all the second partial derivatives of

v = xy/(x-y)

could you quickly show me the steps to see if I did it correctly?

Thanks again - Nov 12th 2008, 08:28 PMChris L T521
First you need to find $\displaystyle \frac{\partial f}{\partial x}$ and $\displaystyle \frac{\partial f}{\partial y}$

$\displaystyle f(x,y)=\frac{xy}{x-y}$

$\displaystyle \frac{\partial f}{\partial x}=\frac{(x-y)y-xy(1)}{(x-y)^2}=-\frac{y^2}{(x-y)^2}$

$\displaystyle \frac{\partial f}{\partial y}=\frac{(x-y)x-xy(-1)}{(x-y)^2}=\frac{x^2}{(x-y)^2}$

Now, there are four different second partial derivatives:

$\displaystyle \frac{\partial^2 f}{\partial x^2}$, $\displaystyle \frac{\partial^2 f}{\partial y^2}$, $\displaystyle \frac{\partial^2 f}{\partial y\partial x}$ and $\displaystyle \frac{\partial^2 f}{\partial x\partial y}$

$\displaystyle \frac{\partial^2 f}{\partial x^2}=-\frac{(x-y)^2(0)-y^2(2(x-y))}{(x-y)^4}=\frac{2y^2}{(x-y)^3}$

$\displaystyle \frac{\partial^2 f}{\partial y^2}=\frac{(x-y)^2(0)-x^2(-2(x-y))}{(x-y)^4}=\frac{2x^2}{(x-y)^3}$

$\displaystyle \frac{\partial^2 f}{\partial y\partial x}=-\frac{(x-y)^2(2y)-y^2(-2(x-y))}{(x-y)^4}=-\frac{2y(x-y)+2y^2}{(x-y)^3}=-\frac{2xy}{(x-y)^3}$

$\displaystyle \frac{\partial^2 f}{\partial x\partial y}=\frac{(x-y)^2(2x)-x^2(2(x-y))}{(x-y)^4}=\frac{2x(x-y)-2x^2}{(x-y)^3}=-\frac{2xy}{(x-y)^3}$

Note, you should always see that $\displaystyle \frac{\partial^2f}{\partial x\partial y}\equiv \frac{\partial^2f}{\partial y\partial x}$

Does this make sense?

--Chris - Nov 13th 2008, 07:53 AMkhuezyi got the fxx and fyy
but I don't get how you did the fyy/fxx

aren't you suppose to the the derivative of fx

it looks like you took it from the f(x,y) instead of fx(x,y) - Nov 13th 2008, 01:32 PMMathstud28