# Thread: Evaluating a First Partial at a Point

1. ## Evaluating a First Partial at a Point

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?

2. 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 $\displaystyle z_x$

3. 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 $\displaystyle z_x$

4. Originally Posted by Shananay
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.

5. 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.

6. Originally Posted by Shananay
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 $\displaystyle (x, y, z) = (1, 2, 3)$.

7. Originally Posted by Shananay
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'.

Regards.

8. Thank you both very much. I understand now.