We are given the equation:

$xyz + x^2y + y^4z = 0$

We need to find $\frac{\partial z}{\partial x}$ using implicit differentiation. Here's the step-by-step process.

Step 1: Differentiate both sides of the equation with respect to $x$.

Since $z = z(x, y)$, we must apply the product rule when differentiating terms involving $z$.

1. First term: $xyz$: Using the product rule: $\frac{d}{dx}(xyz) = \frac{d}{dx}(x) \cdot yz + x \cdot \frac{d}{dx}(yz)$

   • The first part is $yz$, since $\frac{d}{dx}(x) = 1$.
   • For the second part, use the product rule again: $\frac{d}{dx}(yz) = y \cdot \frac{d}{dx}(z) = y \cdot \frac{\partial z}{\partial x}$ So, the derivative of $xyz$ becomes: $yz + xy \frac{\partial z}{\partial x}$

2. Second term: $x^2y$: Again, apply the product rule: $\frac{d}{dx}(x^2y) = \frac{d}{dx}(x^2) \cdot y + x^2 \cdot \frac{d}{dx}(y)$ Since $y$ is independent of $x$, this simplifies to: $2xy$

3. Third term: $y^4z$: Here, only $z$ depends on $x$, so we use the chain rule: $\frac{d}{dx}(y^4z) = y^4 \frac{\partial z}{\partial x}$

Step 2: Differentiate the right-hand side (constant $0$).

$\frac{d}{dx}(0) = 0$

Step 3: Combine the results.

Putting everything together:

$yz + xy \frac{\partial z}{\partial x} + 2xy + y^4 \frac{\partial z}{\partial x} = 0$

Step 4: Solve for $\frac{\partial z}{\partial x}$.

Group the terms involving $\frac{\partial z}{\partial x}$:

$xy \frac{\partial z}{\partial x} + y^4 \frac{\partial z}{\partial x} = -yz - 2xy$

Factor out $\frac{\partial z}{\partial x}$:

$\frac{\partial z}{\partial x}(xy + y^4) = -yz - 2xy$

Finally, solve for $\frac{\partial z}{\partial x}$:

$\frac{\partial z}{\partial x} = \frac{-yz - 2xy}{xy + y^4}$

This is the derivative $\frac{\partial z}{\partial x}$ using implicit differentiation.

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Would you like more details on any part of the solution?

Here are 5 related questions for further exploration:

1. How would you calculate $\frac{\partial z}{\partial y}$ for the same equation?
2. Can you explain the product rule in more detail?
3. What happens if we substitute specific values for $x$, $y$, and $z$ into this result?
4. How does implicit differentiation differ from regular differentiation?
5. What if the equation involved more complex functions like trigonometric terms?

Tip: Always apply the product rule carefully when differentiating terms that involve products of functions of different variables.