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Please check my partial diferentiation

aruwin

Member
Jul 4, 2012
121
I have calculated 3 times and I still don't get the answer. The answer should be 0.
Here's the question and my work. Which part am I wrong?


f(x,y) = 1/√(1-2xy+y^2)

Prove that ∂/∂x{(1-x^2)*∂f/∂x} + ∂/∂y{(y^2)*∂f/∂y} = 0

 

Sudharaka

Well-known member
MHB Math Helper
Feb 5, 2012
1,621
I have calculated 3 times and I still don't get the answer. The answer should be 0.
Here's the question and my work. Which part am I wrong?


f(x,y) = 1/√(1-2xy+y^2)

Prove that ∂/∂x{(1-x^2)*∂f/∂x} + ∂/∂y{(y^2)*∂f/∂y} = 0

Hi aruwin, :)

You have used the quotient rule incorrectly when calculating, \(\displaystyle\frac{\partial}{\partial x}\left[\frac{y(1-x^2)}{(1-2xy+y^2)^{\frac{3}{2}}}\right]\mbox{ and }\frac{\partial}{\partial y}\left[\frac{(x-y)y^2}{(1-2xy+y^2)^{\frac{3}{2}}}\right]\).

Kind Regards,
Sudharaka
 

aruwin

Member
Jul 4, 2012
121
Hi aruwin, :)

You have used the quotient rule incorrectly when calculating, \(\displaystyle\frac{\partial}{\partial x}\left[\frac{y(1-x^2)}{(1-2xy+y^2)^{\frac{3}{2}}}\right]\mbox{ and }\frac{\partial}{\partial y}\left[\frac{(x-y)y^2}{(1-2xy+y^2)^{\frac{3}{2}}}\right]\).

Kind Regards,

Sudharaka
Oh my gosh, yes. I have found what is wrong :)
 
Last edited by a moderator:

Fantini

"Read Euler, read Euler." - Laplace
MHB Math Helper
Feb 29, 2012
342
Let's try using the product rule: $$ \frac{\partial}{\partial x} \left[ (y-x^2y)(1-2xy+y^2)^{-\frac{3}{2}} \right] $$ $$= (-2xy)(1-2xy+y^2)^{-\frac{3}{2}} + (y-x^2y)\left( - \frac{3}{2} \right) (-2y) (1-2xy+y^2)^{-\frac{5}{2}}$$ $$= \frac{(-2xy)(1-2xy+y^2) + (y-x^2y)(3y)}{(1-2xy+y^2)^{\frac{5}{2}}} $$ $$= \frac{-2xy+4x^2y^2-2xy^3+3y^2-3x^2y^2}{(1-2xy+y^2)^{\frac{5}{2}}}$$ $$= \frac{-2xy+3y^2-2xy^3+x^2y^2}{(1-2xy+y^2)^{\frac{5}{2}}}.$$ Hope it helps.
 

aruwin

Member
Jul 4, 2012
121
Thanks,guys!I have solved this question :)