July 2nd, 2017, 03:29 PM  #1 
Senior Member Joined: Nov 2015 From: Alabama Posts: 140 Thanks: 17 Math Focus: Geometry, Trigonometry, Calculus  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: 54 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  
Member Joined: Dec 2016 From:  Posts: 54 Thanks: 10  Quote:
Last edited by skipjack; July 2nd, 2017 at 05:24 PM.  
July 3rd, 2017, 03:01 PM  #4 
Senior Member Joined: Nov 2015 From: Alabama Posts: 140 Thanks: 17 Math Focus: Geometry, Trigonometry, Calculus  

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differentiation, implicit, partial 
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