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
thank you in advance....
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
thank you in advance....
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
 = \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
 = \frac{m(x)}{n(x)})
then
 = \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
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