# Partial integration - correct solutions?

• Sep 17th 2012, 10:37 AM
FireTheft
Partial integration - correct solutions?
Let f(x; y; z) = (x^3+3xy^2-3xyz-z^4)/(x^2 + z^2)

Calculate

a. @f/@x

b. @f/@y

c. @f/@z

I suppose that the solution for the partial derivative of the function to x is too simple as I get: 3(x^2+y^2-yz)/2x

Am I right when I have to use the quotient rule? as it is f´g - g´f / g^2......But what is then g only x^2 or x^2+z^2 and is f (x^3+3xy^2-3xyz-z^4) or just (x^3+3xy^2-3xyz) (as the unused variables are held constant?

so is this approach correct:

(3x^2+3y^2-3yz)*(x^2) - 2x(x^3+3xy^2-3xyz) / x^4

• Sep 17th 2012, 03:17 PM
topsquark
Re: Partial integration - correct solutions?
Quote:

Originally Posted by FireTheft
Let f(x; y; z) =
Calculate

a. @f/@x

so is this approach correct:

(3x^2+3y^2-3yz)*(x^2) - 2x(x^3+3xy^2-3xyz) / x^4

Your last expression is close to correct. I suspect that notation is getting in the way as well. We already have a function f(x, y, x) so using it in the quotient rule is a little confusing. So I am going to say
$f(x, y, z) = \frac{ x^3+3xy^2-3xyz-z^4}{x^2 + z^2}$

And I am going to define the quotient rule as
given a function
$u(x) = \frac{m(x)}{n(x)}$

then
$u'(x) = \frac{m'(x)\cdot n(x) - m(x) \cdot n'(x)}{(n(x))^2}$

So to get the derivative then choose m(x, y, z) = x^3+3xy^2-3xyz-z^4 and n(x) = x^2 + z^2

See what you can do with that.

-Dan
• Sep 18th 2012, 01:05 AM
FireTheft
Re: Partial integration - correct solutions??
So am I right that u(x) equals:

(3x^2+3y^2-3yz)*(x^2+z^2) - 2x(x^3+3xy^2-3xyz-z^4) / (x^2 + z^2)^2

so is the correct answer for u´(x):

(x^4+3xy^2+3x^2yz+3x^2z^2+3y^2z^2-3yz^3-6x^3y^2+2xz^4)/(x^2+z^2)^2

for u´(y) I got:

(6x^3y-3x^3z+6xyz^2-3xz^3)/(x^2+z^2)^2

for u(z) I got:

(-3x^3y-4x^2z^3-3xyz^2-2z^5-2x^3z-6xy^2z+6xyz)/(x^2+z^2)^2

I hope someone can confirm these solutions ;-)

Thx a lot