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

[phosgene]

Homework Statement

Given the function

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

Find the partial derivative f_xx(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?

 

[vela]

Your work looks fine to me. What was the supposed answer?

 

[HallsofIvy]

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.

 

[Char. Limit]

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?

 

[phosgene]

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...

 

[Mark44]

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.

 

[phosgene]

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 :)