https://r2.erweima.ai/mathbot/image/1726382372934ikmwdsd6.jpg

The given expression is:

\[
f(zx, y^2) + g(z^2) = 5
\]

We are asked to find \( z_x \) and \( z_y \), where \( z \) is an implicit function of \( x \) and \( y \).

### Step 1: Differentiate implicitly with respect to \( x \)
We will differentiate both sides of the equation with respect to \( x \). Recall that \( z = z(x, y) \), so we will apply the chain rule.

1. First term: \( f(zx, y^2) \)
   - Differentiate with respect to \( x \):
     \[
     \frac{d}{dx} f(zx, y^2) = f_1(zx, y^2) \cdot \frac{d}{dx}(zx) + f_2(zx, y^2) \cdot \frac{d}{dx}(y^2)
     \]
     Where \( f_1 \) is the partial derivative of \( f \) with respect to its first argument, and \( f_2 \) is the partial derivative with respect to its second argument.
     \[
     = f_1(zx, y^2) \cdot \left(z + x \frac{dz}{dx}\right) + 0
     \]
     (since \( y^2 \) does not depend on \( x \)).

2. Second term: \( g(z^2) \)
   - Differentiate with respect to \( x \):
     \[
     \frac{d}{dx} g(z^2) = g'(z^2) \cdot \frac{d}{dx}(z^2) = g'(z^2) \cdot 2z \cdot \frac{dz}{dx}
     \]

3. The right-hand side:
   \[
   \frac{d}{dx} 5 = 0
   \]

Now, putting it all together:
\[
f_1(zx, y^2) \cdot (z + x z_x) + g'(z^2) \cdot 2z \cdot z_x = 0
\]

### Step 2: Solve for \( z_x \)
Now, solve the equation for \( z_x \):

\[
f_1(zx, y^2) \cdot (z + x z_x) + g'(z^2) \cdot 2z z_x = 0
\]

Group the terms with \( z_x \):
\[
f_1(zx, y^2) \cdot z + z_x \left( f_1(zx, y^2) \cdot x + g'(z^2) \cdot 2z \right) = 0
\]

Solve for \( z_x \):
\[
z_x = \frac{-f_1(zx, y^2) \cdot z}{f_1(zx, y^2) \cdot x + g'(z^2) \cdot 2z}
\]

### Step 3: Differentiate implicitly with respect to \( y \)
Now, differentiate the equation with respect to \( y \):

1. First term: \( f(zx, y^2) \)
   - Differentiate with respect to \( y \):
     \[
     \frac{d}{dy} f(zx, y^2) = f_1(zx, y^2) \cdot \frac{d}{dy}(zx) + f_2(zx, y^2) \cdot \frac{d}{dy}(y^2)
     \]
     \[
     = f_1(zx, y^2) \cdot x z_y + f_2(zx, y^2) \cdot 2y
     \]

2. Second term: \( g(z^2) \)
   - Differentiate with respect to \( y \):
     \[
     \frac{d}{dy} g(z^2) = g'(z^2) \cdot 2z \cdot \frac{dz}{dy}
     \]

3. The right-hand side:
   \[
   \frac{d}{dy} 5 = 0
   \]

Now, putting it all together:
\[
f_1(zx, y^2) \cdot x z_y + f_2(zx, y^2) \cdot 2y + g'(z^2) \cdot 2z z_y = 0
\]

### Step 4: Solve for \( z_y \)
Group the terms with \( z_y \):
\[
z_y \left( f_1(zx, y^2) \cdot x + g'(z^2) \cdot 2z \right) = -f_2(zx, y^2) \cdot 2y
\]

Solve for \( z_y \):
\[
z_y = \frac{-f_2(zx, y^2) \cdot 2y}{f_1(zx, y^2) \cdot x + g'(z^2) \cdot 2z}
\]

### Final Results
- \( z_x = \frac{-f_1(zx, y^2) \cdot z}{f_1(zx, y^2) \cdot x + g'(z^2) \cdot 2z} \)
- \( z_y = \frac{-f_2(zx, y^2) \cdot 2y}{f_1(zx, y^2) \cdot x + g'(z^2) \cdot 2z} \)

Let me know if you need further details or explanations! 

Here are 5 related questions to explore:
1. What is implicit differentiation, and when is it used?
2. How does the chain rule work in the case of implicit differentiation?
3. What are some common applications of implicit functions in physics?
4. How does partial differentiation differ from ordinary differentiation?
5. What techniques can be used to solve nonlinear implicit equations?

**Tip:** Always ensure that all variables are treated as functions of the implicit variables before differentiating implicitly.

If f(zx, y^2) + g(z^2) = 5 defines z as an implicit function of x and y, calculate z_x and z_y.

Learn how to solve an implicit differentiation problem involving f(zx, y^2) + g(z^2) = 5. This calculus problem covers key concepts such as the chain rule, partial derivatives, and implicit differentiation. Suitable for advanced high school or undergraduate students.

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