# Evaluating a First Partial at a Point

#### 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?

#### p0oint

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 $$\displaystyle z_x$$

#### p0oint

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 $$\displaystyle z_x$$

• Shananay

#### Prove It

MHF Helper
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.

• Shananay

#### 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.

#### Prove It

MHF Helper
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 $$\displaystyle (x, y, z) = (1, 2, 3)$$.

• Shananay

#### p0oint

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.

• Shananay

#### Shananay

Thank you both very much. I understand now.