1. ## Partial Derivatives

Having a little trouble with this one:

Find the first Partial Derivatives xy/x^2+y^2

fx(x,y)= My interpretation was that in the numerator, x would be 1, leaving me with y

and in the denominator x^2 would be 2x and y^2 would be zero, so I am left with

y/2x, which is wrong. Not sure where I am mistaken so any help would be greatly appreciated. Thanks

2. ## Re: Partial Derivatives

You are looking at $\displaystyle \frac{xy}{x^2+y^2}$ ?

If so you have to use the quotient rule which is if $\displaystyle y = \frac{u}{v} \implies y' = \frac{vu'-uv'}{v^2}$

Have you seen this rule before?

3. ## Re: Partial Derivatives

Yes, i am familiar with the quotient rule, but in this chapter they are only using "partial differentiation". The only example I have to go off of is f(x,y)= 3x - x²y² + 2x²y f(x) (x,y)= 3-2xy² + 4xy and f(y) (x,y) = -2x²y + 2x² So I am applying that same logic. Problem is I have no examples showing how to do a partial derivative of a quotient.

4. ## Re: Partial Derivatives

Ok, for $\displaystyle f(x,y) = \frac{xy}{x^2+y^2}$ there are two first partial derivatives $\displaystyle \frac{\partial f}{\partial x}$ and $\displaystyle \frac{\partial f}{\partial y}$

Ill give the first one a go, you do the second.

By the quotient rule $\displaystyle \frac{\partial f}{\partial x} = \frac{(x^2+y^2)\times y- xy\times 2x}{(x^2+y^2)^2}$

How was that?

5. ## Re: Partial Derivatives

I was actually close, but not all the way there. Your example I had the same denominator, but in the numerator i had (x² + y²) * y - xy * (2x + 2y)

f (y) (x,y)= x^3 -xy²
----------
(x² + y²)²

Thank you for you help it is greatly appreciated, and I do have a much better grasp of these problems now.

6. ## Re: Partial Derivatives

Looks like I need to learn to type a little better on here, sorry.

7. ## Re: Partial Derivatives

You can write equations using "tex" tags, hover your mouse over my equation to see the code.