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I need to find the second-order partial derivative $\frac{\partial^2z}{\partial x^2}$ of the following problem:

$$z(x,y) = \frac{1}{2}\ln(x²+y²)$$

I got the first part:

First derivative is $$\frac{x}{x² + y²}$$


I then thought I would use the quotient rule to find the second derivative of x.

So this would mean: $$\frac{(x²+y²)-2x²}{(x²+y²)²}$$ right? But the correct answer is: $\dfrac{y² - x²}{(x²+y²)²}$

Anyone care to explain why and what I did wrong?

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  • $\begingroup$ You did everything right. :) The solution just computed $(x^2+y^2)-2x^2=x^2+y^2-2x^2=y^2-x^2$ in the nominator. $\endgroup$ – Piwi Oct 7 '15 at 17:03
  • $\begingroup$ Ah! Ofcourse. I have been doing too much maths today I guess. Thanks anyway! $\endgroup$ – Richard Oct 7 '15 at 17:09
  • $\begingroup$ The correct derivative should be $[(x^2 + y^2) - 2x^2]/(x^2 + y^2)^2$ (i.e., the first $(x^2 + y^2)$ term should be in the denominator of the fraction, not its own term.) $\endgroup$ – Michael Seifert Oct 7 '15 at 17:30
  • $\begingroup$ Also, the standard English phrase would be "the second derivative with respect to $x$". $\endgroup$ – Michael Seifert Oct 7 '15 at 17:31
  • $\begingroup$ @MichaelSeifert The edit misinterpreted OPs notation (which technically was wrong, but it was pretty clear from context...). I'll suggest an edit to fix that if not already done. $\endgroup$ – Piwi Oct 7 '15 at 17:39
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Community wiki answer so the question can be marked as answered:

As Piwi noted in a comment, your result is correct and you merely failed to simplify the numerator.

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