Finding the 2nd Partial Derivative of f(x,y) = 1/(2x^2 + y)

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Homework Help Overview

The discussion revolves around finding the second partial derivative of the function f(x,y) = 1/(2x^2 + y). Participants are exploring the differentiation process and addressing potential errors in their calculations.

Discussion Character

  • Mathematical reasoning, Problem interpretation, Assumption checking

Approaches and Questions Raised

  • One participant attempts to differentiate the function twice while treating y as a constant but encounters difficulties with the results. Others suggest alternative methods, such as rewriting the function to apply the product rule instead of the quotient rule. There are questions regarding the correctness of the supposed answer and the notation used in the differentiation process.

Discussion Status

The discussion includes various attempts to differentiate the function correctly, with some participants providing alternative approaches and questioning the initial setup. There is acknowledgment of potential errors in the calculations, and one participant expresses confidence in their method while seeking confirmation from others.

Contextual Notes

Participants mention the use of a program to check answers, which only indicates correctness or incorrectness without providing specific feedback. This has led to uncertainty regarding the validity of their solutions.

phosgene
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Homework Statement



Given the function

[itex]f(x,y)=\frac{1}{2x^2 + y}[/itex]

Find the partial derivative fxx(x,y)

Homework Equations




The Attempt at a Solution



Seems pretty straight forward, just treat y as a constant and differentiate twice. But I keep getting the answer wrong and I have no idea why. Here's what I did:

[itex]\frac{∂f(x,y)}{∂x}=\frac{-4x}{(2x^2+y)^2}[/itex]

Then I differentiate with respect to x again using the quotient rule

[itex]\frac{∂^{2}f(x,y)}{∂x^{2}}=\frac{-4(2x^2+y)^2 + 4x(2(4x(2x^2 + y))}{(2x^2+ y)^4}[/itex]

I've also tried to do it by re-arranging and using the product rule, but this fails also. It's driving me mad. Have I done something wrong, or could the supposed correct answer actually be wrong?
 
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Your work looks fine to me. What was the supposed answer?
 
Rather than use the quotient rule, you could also write [itex]f(x,y)= (2x^2+y)^{1/2}[/itex] so that [itex]f_x= (1/2)(2x^2+y)^{-1/2}(2x)= x(2x+ y)^{-1/2}[/itex]. Now take the derivative, with respect to x, again, using the product rule.
 
HallsofIvy said:
Rather than use the quotient rule, you could also write [itex]f(x,y)= (2x^2+y)^{1/2}[/itex] so that [itex]f_x= (1/2)(2x^2+y)^{-1/2}(2x)= x(2x+ y)^{-1/2}[/itex]. Now take the derivative, with respect to x, again, using the product rule.

Um, shouldn't that be [itex]f(x,y) = (2 x^2 + y)^{-1}[/itex], and not 1/2 as you put?
 
Thanks for the replies, as stated, I've tried rearranging into a form where I can apply the product rule, but the answer is also wrong. I've also tried expanding the brackets and then differentiation, but wrong again. I don't know the correct answer as it's entered onto a program which only tells me if my answer is correct or incorrect. I've e-mailed the course coordinator about it, as at this point I'm almost certain that my answer is correct and that there's a problem with the program...
 
Try this version:
$$ \frac{4(6x^2 - y)}{(2x^2 + y)^3}$$

All I did was find the common factor for the terms in the numerator, and then simplify. Many times when the quotient rule is involved, the textbook answers will to this kind of simplification.
 
I finally figured it out, I was accidentally placing the minus sign in front of the bracketed numerator. There were so many brackets that it slipped by me. Thanks for all the input though guys :)
 

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