Find partial z partial x at (0,0,0) f * x ^ 3 + z ^ 3 + x * e ^ (yz) + z * cos x = 0

We are asked to find the partial derivative of \( z \) with respect to \( x \) at the point \( (0, 0, 0) \) for the implicit function:

\[
f(x, y, z) = x^3 + z^3 + x e^{yz} + z \cos x = 0.
\]

To solve this, we will apply **implicit differentiation** with respect to \( x \), treating \( z \) as a function of \( x \), i.e., \( z = z(x, y) \). Let's go step by step.

### Step 1: Differentiate implicitly with respect to \( x \)
We differentiate both sides of the equation \( f(x, y, z) = 0 \) with respect to \( x \):

\[
\frac{\partial}{\partial x} \left( x^3 + z^3 + x e^{yz} + z \cos x \right) = \frac{\partial}{\partial x} (0).
\]

On the left-hand side, we apply the chain rule, remembering that \( z \) is a function of \( x \), so terms involving \( z \) will require us to include \( \frac{\partial z}{\partial x} \).

### Step 2: Differentiate each term
1. \( \frac{\partial}{\partial x} \left( x^3 \right) = 3x^2 \).
2. \( \frac{\partial}{\partial x} \left( z^3 \right) = 3z^2 \cdot \frac{\partial z}{\partial x} \).
3. \( \frac{\partial}{\partial x} \left( x e^{yz} \right) \) requires the product rule:
   \[
   \frac{\partial}{\partial x} \left( x e^{yz} \right) = e^{yz} + x \cdot e^{yz} \cdot \left( y \cdot \frac{\partial z}{\partial x} \right).
   \]
4. \( \frac{\partial}{\partial x} \left( z \cos x \right) \) also requires the product rule:
   \[
   \frac{\partial}{\partial x} \left( z \cos x \right) = \frac{\partial z}{\partial x} \cdot \cos x - z \cdot \sin x.
   \]

### Step 3: Combine the derivatives
Putting everything together, we get the following equation:

\[
3x^2 + 3z^2 \cdot \frac{\partial z}{\partial x} + e^{yz} + x \cdot e^{yz} \cdot \left( y \cdot \frac{\partial z}{\partial x} \right) + \frac{\partial z}{\partial x} \cdot \cos x - z \cdot \sin x = 0.
\]

### Step 4: Evaluate at \( (0, 0, 0) \)
Now, substitute \( x = 0 \), \( y = 0 \), and \( z = 0 \) into the equation:

1. \( 3x^2 = 3(0)^2 = 0 \),
2. \( 3z^2 \cdot \frac{\partial z}{\partial x} = 0 \cdot \frac{\partial z}{\partial x} = 0 \),
3. \( e^{yz} = e^{0 \cdot 0} = 1 \),
4. \( x \cdot e^{yz} \cdot ( y \cdot \frac{\partial z}{\partial x} ) = 0 \cdot 1 \cdot ( 0 \cdot \frac{\partial z}{\partial x} ) = 0 \),
5. \( \frac{\partial z}{\partial x} \cdot \cos x = \frac{\partial z}{\partial x} \cdot \cos(0) = \frac{\partial z}{\partial x} \cdot 1 \),
6. \( - z \cdot \sin x = - 0 \cdot \sin(0) = 0 \).

Thus, the equation simplifies to:

\[
1 + \frac{\partial z}{\partial x} = 0.
\]

### Step 5: Solve for \( \frac{\partial z}{\partial x} \)
From the simplified equation, we find:

\[
\frac{\partial z}{\partial x} = -1.
\]

### Final Answer:
The value of \( \frac{\partial z}{\partial x} \) at the point \( (0, 0, 0) \) is \( \boxed{-1} \).

Let me know if you'd like further details or have any questions.

Here are some follow-up questions:
1. How does the chain rule apply in implicit differentiation?
2. What happens when \( y \) is not 0 in this problem?
3. Can we generalize this method to find \( \frac{\partial z}{\partial y} \)?
4. How would the solution change if the equation involved trigonometric functions of \( y \)?
5. Can we use this process to find second-order partial derivatives?

**Tip:** When differentiating implicitly, remember to treat the dependent variable (in this case, \( z \)) as a function of the independent variables and apply the chain rule accordingly.

Learn how to find the partial derivative of z with respect to x at the point (0, 0, 0) for the implicit function x^3 + z^3 + x * e^(yz) + z * cos(x) = 0. This solution demonstrates step-by-step implicit differentiation, chain rule, and product rule techniques. Suitable for undergraduate students studying multivariable calculus.

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