For the function $ln(4x^2+4y^2)$ when taking the derivative with respect to $x$, do you essentially leave the $y$ terms alone? I received the answer of $\frac{2x+y^2}{x^2+y^2}$, however the books solution deems that to be wrong. Looking for some insight on what I'm doing incorrectly.
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Yes, you "ignore" the $y$ terms just as you would constant terms. Note that this also means if there are any terms with only $y$ in them (no $x$'s), then after you take the derivative, those terms disappear just as constant terms do.
When you differentiate $4x^{2} + 5$, for example, you get $8x$ because the constant $5$ disappears.
Similarly, when you differentiate $\ln{(4x^{2} + 4y^{2})}$, you should get:
$\dfrac{1}{4x^{2} + 4y^{2}}(8x)$. This is because when you apply the chain rule and differentiate $4x^{2} + 4y^{2}$, remember that you are treating any $y$'s as constants. So the derivative of $4x^{2} + 4y^{2}$ should be identical to the derivative of $4x^{2} + 5$. In the latter, the $5$ goes away since it is a constant. Therefore, in the former the $4y^{2}$ should also go away because it is a constant.
layman
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Thank you for the detailed explanation. Much appreciated! – user153509 Sep 20 '14 at 18:46
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@user153509 You're welcome! – layman Sep 20 '14 at 18:47