# Thread: need help on partial derivatives

1. ## need 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

2. Originally Posted by khuezy
f(x,y) = arctan(y/x) fx(2,3)

how would I do this?
Thanks
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

3. ## thanks

i just plug in x and y into that equation right?

4. ## thanks

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?

5. Originally Posted by khuezy
how would I do this?

2.) f(x,y) = ln(x+ sqrt(x^2+y^2) ) fx(3,4)
1/(x+ sqrt(x^2+y^2)) * 1+ (1/2(x^2 + y^2) * 2x)
is that correct for #2?
Note the slight modifications I made.

It should be $\displaystyle f_x(x,y)=\frac{1}{x+\sqrt{x^2+y^2}}\cdot\left(1+\f rac{x}{\sqrt{x^2+y^2}}\right)$

Then plug in 3 for x and 4 for y.

--Chris

6. Originally Posted by Chris 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
Originally Posted by khuezy
i just plug in x and y into that equation right?
Yes! What did you get for the answer?

--Chris

7. ## hi 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

8. Originally Posted by khuezy
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
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

9. ## i 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)

10. Originally Posted by khuezy
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)
$\displaystyle \frac{\partial{f(x,y)}}{\partial{y}^2}=\frac{\part ial}{\partial{y}}\left\{\frac{\partial{f(x,y)}}{\p artial{y}}\right\}$

Is that what you are asking?

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# partial derivative of ln(x sqrt(x^2 y^2)) with respect to y

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