Evaluating a First Partial at a Point

Jan 2010
62
7
Evaluate \(\displaystyle z_x\) at \(\displaystyle (1, 2, 3)\) if \(\displaystyle x^3+y^3+z^3+6xyz=72\)

Part of me says I must solve this in terms of z in order to find \(\displaystyle z_x\) but I can't see how.

The other part thinks I can just find \(\displaystyle z_x\) implicitly and plug in the point but I am getting an answer that doesn't make any sense.

What should I do?
 
Nov 2009
130
24
You need to find:
\(\displaystyle \frac{\partial z}{\partial x}\)

So differentiate implicitly with respect to x, assuming that y is constant and you will get \(\displaystyle z_x\)
 
Nov 2009
130
24
You need to find:
\(\displaystyle \frac{\partial z}{\partial x}\)

So differentiate implicitly with respect to x, assuming that y is constant and you will get \(\displaystyle z_x\)
 
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Prove It

MHF Helper
Aug 2008
12,897
5,001
Evaluate \(\displaystyle z_x\) at \(\displaystyle (1, 2, 3)\) if \(\displaystyle x^3+y^3+z^3+6xyz=72\)

Part of me says I must solve this in terms of z in order to find \(\displaystyle z_x\) but I can't see how.

The other part thinks I can just find \(\displaystyle z_x\) implicitly and plug in the point but I am getting an answer that doesn't make any sense.

What should I do?
You're not going to be able to define \(\displaystyle z\) explicitly in terms of \(\displaystyle x\) and \(\displaystyle y\).

So you will need to differentiate both sides implicitly, treating \(\displaystyle y\) as constant.
 
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Jan 2010
62
7
How do I treat z?

I have something like \(\displaystyle 3x^2+3z^2(z_x)+6y(z_x)=0\) then factor and solve for \(\displaystyle z_x\)?

I don't feel like that's right.
 

Prove It

MHF Helper
Aug 2008
12,897
5,001
Evaluate \(\displaystyle z_x\) at \(\displaystyle (1, 2, 3)\) if \(\displaystyle x^3+y^3+z^3+6xyz=72\)

Part of me says I must solve this in terms of z in order to find \(\displaystyle z_x\) but I can't see how.

The other part thinks I can just find \(\displaystyle z_x\) implicitly and plug in the point but I am getting an answer that doesn't make any sense.

What should I do?
We assume that \(\displaystyle z\) is a function of \(\displaystyle x\), so you need to use the Chain Rule...


\(\displaystyle x^3 + y^3 + z^3 + 6xyz = 72\)

\(\displaystyle \frac{\partial}{\partial x}(x^3 + y^3 + z^3 + 6xyz) = \frac{\partial}{\partial x}(72)\)

\(\displaystyle \frac{\partial}{\partial x}(x^3) + \frac{\partial}{\partial x}(y^3) + \frac{\partial}{\partial x}(z^3) + \frac{\partial}{\partial x}(6xyz) = 0\)

\(\displaystyle 3x^2 + 0 + \frac{\partial z}{\partial x}\,\frac{\partial}{\partial z}(z^3) + 6y\,\frac{\partial}{\partial x}(xz) = 0\) (Chain Rule)

\(\displaystyle 3x^2 + 3z^2\,\frac{\partial z}{\partial x} + 6y\left[x\,\frac{\partial}{\partial x}(z) + z\,\frac{\partial}{\partial x}(x)\right] = 0\) (Product Rule)

\(\displaystyle 3x^2 + 3z^2\,\frac{\partial z}{\partial x} + 6y\left[x\,\frac{\partial z}{\partial x} + z(1)\right] = 0\)

\(\displaystyle 3x^2 + 3z^2\,\frac{\partial z}{\partial x} + 6xy\,\frac{\partial z}{\partial x} + 6yz = 0\)

\(\displaystyle 3z^2\,\frac{\partial z}{\partial x} + 6xy\,\frac{\partial z}{\partial x} = -(3x^2 + 6yz)\)

\(\displaystyle (3z^2 + 6xy)\frac{\partial z}{\partial x} = -(3x^2 + 6yz)\)

\(\displaystyle \frac{\partial z}{\partial x} = -\frac{3x^2 + 6yz}{3z^2 + 6xy}\)

\(\displaystyle \frac{\partial z}{\partial x} = -\frac{x^2 + 2yz}{z^2 + 2xy}\).


Now substitute \(\displaystyle (x, y, z) = (1, 2, 3)\).
 
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Nov 2009
130
24
How do I treat z?

I have something like \(\displaystyle 3x^2+3z^2(z_x)+6y(z_x)=0\) then factor and solve for \(\displaystyle z_x\)?

I don't feel like that's right.
Treat z as a composition which contains variables x,y. So at compositions you must apply the chain rule.

Now you did well except for one mistake:

\(\displaystyle 3x^2+3z^2z'+(6xyz)'=0\)

\(\displaystyle 3x^2+3z^2z'+6y(xz)'=0\)

\(\displaystyle 3x^2+3z^2z'+6y(x'z+xz')=0\)

Now factor for z' and substitute (1,2,3) for x,y,z or first substitute and then factor z'. :D

Regards.
 
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Jan 2010
62
7
Thank you both very much. I understand now.