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 July 2nd, 2017, 03:29 PM #1 Senior Member     Joined: Nov 2015 From: United States of America Posts: 167 Thanks: 21 Math Focus: Calculus and Physics implicit partial differentiation Hello! This is an example problem in my textbook. I am stuck on understanding the second line. Everything makes absolutely perfect sense until they went ahead and tacked on that very last term the $+\,6xy\dfrac{\partial z}{\partial x}$. Perhaps the product rule was used? I am not sure. I would really appreciate some clarification on that line if it's possible! Thank you! Jacob Last edited by skipjack; July 2nd, 2017 at 05:23 PM.
 July 2nd, 2017, 05:01 PM #2 Member   Joined: Dec 2016 From: - Posts: 62 Thanks: 10 Hello Jacob, the result is as follows. The key is to understand that $y$ is a constant here. From the first line, call $\alpha=\partial z/\partial x$, then we have \begin{eqnarray} 3x^{2} + 3z^{2}\alpha + 6yz +6yx\alpha=0 \to \alpha(6xy + 3z^{2})=-3x^{2}-6yz \end{eqnarray} then dividing we end up with \begin{eqnarray} \alpha=\frac{\partial z}{\partial x}=-\frac{3x^{2}+6yz}{6xy + 3z^{2}}=-\frac{x^{2}+2yz}{z^{2}+2xy} \end{eqnarray} Hope it is clear now, same procedure is done for $\partial z/\partial y$ but keeping $x$ as a constant. Last edited by skipjack; July 2nd, 2017 at 05:27 PM.
July 2nd, 2017, 05:04 PM   #3
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 Originally Posted by SenatorArmstrong Hello! This is an example problem in my textbook. http://i.imgur.com/zqCtZcZ.jpg I am stuck on understanding the second line. Everything makes absolutely perfect sense until they went ahead and tacked on that very last term the $+\,6xy\dfrac{\partial z}{\partial x}$. Perhaps the product rule was used? I am not sure. I would really appreciate some clarification on that line if it's possible! Thank you! Jacob
Ok I saw now what you mean, precisely is the product rule, so that $\partial_{x}(6xyz)= 6yz + 6xy(\partial _{x}z)$

Last edited by skipjack; July 2nd, 2017 at 05:24 PM.

July 3rd, 2017, 03:01 PM   #4
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 Originally Posted by nietzsche Ok I saw now what you mean, precisely is the product rule, so that $\partial_{x}(6xyz)= 6yz + 6xy(\partial _{x}z)$
Thank you for walking me through this. Makes crystal clear sense now!

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