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?

  • $\begingroup$ 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

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.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.