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Use the chain rule to compute $\dfrac{\partial z}{\partial x}$ and $\dfrac{\partial z}{\partial y}$ for $2x^2+y^2+z^2=9$.

I got the following, however I don't think I'm using the chain rule. Where exactly do I use it and how?

$\dfrac{d}{dy}$ $(2x^2+y^2+z^2)$ = $\dfrac{d}{dy}$ $(9)$.

$\dfrac{d}{dy}$$(2x^2) + $$\dfrac{d}{dy}$(y^2) + $\dfrac{\partial z}{\partial y}$$(z^2)$ = $\dfrac{d}{dy}$(9)

0 + 2y + $\dfrac{\partial z}{\partial y}$$(z^2)$ = 0

2y + $\dfrac{\partial z}{\partial y}$(2z) = 0

$\dfrac{\partial z}{\partial y}$(2z) = -2y

$\dfrac{\partial z}{\partial y}$= $\dfrac{-2y}{2z}$ = $\dfrac{-y}{z}$

$\dfrac{d}{dx}$ $(2x^2+y^2+z^2)$ = $\dfrac{d}{dx}$ $(9)$

2$\dfrac{d}{dx}$$(x^2)$+$\dfrac{d}{dx}$$(y^2)$+$\dfrac{\partial z}{\partial x}$$(z^2)$=$\dfrac{d}{dx}$(9).

2(2x)+$\dfrac{d}{dx}$$(y^2)$+$\dfrac{\partial z}{\partial x}$$(z^2)$=$\dfrac{d}{dx}$(9)

4x+ 0 + $\dfrac{\partial z}{\partial x}$$(z^2)$ = 0

4x + $\dfrac{\partial z}{\partial x}$(2z) = 0

$\dfrac{\partial z}{\partial x}$(2z) = -4x

$\dfrac{\partial z}{\partial x}$= $\dfrac{-4x}{2z}$ = $\dfrac{-2x}{z}$

My answers of -y/z and -2x/z are correct, but how do I apply the chain rule to the problem. I also want to apologize up front for the formatting, I just am still trying to figure it out.

Tara H
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  • $\dfrac{\partial z}{\partial x}$ = $\dfrac{\partial z}{\partial x}$. $\dfrac{df}{dx}$ = $\dfrac{df}{dx}$. Use double $'s to get math centered on a new line $$\dfrac{a}{b} = c^2$$ = $$\dfrac{a}{b} = c^2$$ –  Apr 03 '17 at 02:17
  • My homework stated: Use the chain rule to compute $\dfrac{\partial z}{\partial x}$ and $\dfrac{\partial z}{\partial y}$ for $2x^2+y^2+z^2=9$. @Fawad. – Tara H Apr 03 '17 at 02:27
  • +1 For the effort on the edit, don't see why you where downvoted. As formatting goes, for example: 0 + 2y + $\dfrac{\partial z}{\partial y}$$(z^2)$ = 0 only requires $ at he beginning and at the end of a line, see: $0 + 2y + \dfrac{\partial z}{\partial y}(z^2) = 0$ looks better than 0 + 2y + $\dfrac{\partial z}{\partial y}$$(z^2)$ = 0. To put in display mode, simply use double $ at the beginning and the end of the line: $$0 + 2y + \dfrac{\partial z}{\partial y}(z^2) = 0$$ renders as $$0 + 2y + \dfrac{\partial z}{\partial y}(z^2) = 0$$ –  Apr 03 '17 at 03:12
  • For information about writing math on MSE see here, here, here and here. Also, here is a very cool little tool called Detexify where you can draw the symbol you are looking for and the system will tell you what the TeX command is. –  Apr 03 '17 at 03:16

1 Answers1

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You need to use implicit differentiation. The idea is that, suppose you can write $z = g(x,y)$. Then you can rewrite the equation about as $F(x,y, g(x,y)=z) = 0$ where $F = 2x^2+y^2+z^2-9$. To compute $z_x$ and $z_y$, we use the chain rule.

In computing these partials, we will assume $F_z \not = 0$. Thus, since $z = g(x,y)$, for a small neighborhood of the points $(x,y,g(x,y))$ we can choose a curve $c(t) \in \mathbb{R}^2$ such that:

\begin{align*} (x,y,g(x,y)) \in F( c(t)) \Rightarrow 0 &= \frac{d}{dt}(F \circ c)(t)\\&= \nabla F \cdot \frac{\partial}{\partial x} (x,y,g(x,y) \\&= (F_x, F_y, F_z) \cdot (1,0, g_x) \\& \Rightarrow F_x + F_z g_x =0 \\& \Rightarrow g_x=\frac{\partial z}{\partial x} = - \frac{F_x}{F_z} = - \frac{2x}{z} \end{align*}

I've given $F$ above, and so you have half the answer. Do a similar computation to derive $z_y$ i.e just replace $\partial/ \partial x$ with $\partial/ \partial y$.

Faraad Armwood
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  • How does implicit differentiation transfer over to the chain rule? – Tara H Apr 04 '17 at 00:55
  • The is just the name given to the type of differentiation we must use to produce $z_x,z_y$. The term implicit here simply means we know $z$ is a function of $x,y$ and so instead of algebraically trying to write $z = g(x,y)$, we just say such a function $g$ exists. The rest should be clear. – Faraad Armwood Apr 04 '17 at 01:14
  • If this answers your question, you can accept to close the question, otherwise, do you have any more questions? – Faraad Armwood Apr 05 '17 at 12:30
  • My homework stated to use the chain rule to solve for this. My question was specifically referring to the chain rule. My question to you was regarding the chain rule specifically and how it transfers to the chain rule. I have the answers to the question, however my homework specifically stated to use the chain rule. – Tara H Apr 06 '17 at 04:06
  • I did use the chain rule, isn't $\frac{d}{dt} (f \circ c) = \nabla f \cdot c'$ the chain rule? I used that above. – Faraad Armwood Apr 06 '17 at 11:35
  • is implicit differentiation the same as the chain rule? – Tara H Apr 07 '17 at 15:28
  • You use the chain rule to do implicit differentiation as seen above. – Faraad Armwood Apr 07 '17 at 20:25