If Phi( z/(x ^ 3) ,  y\ x )=0, where z is a function of two variable x, y, then find the value of x partial z partial x + y partial z partial y

We are given the equation:

\[
\Phi \left( \frac{z}{x^3}, \frac{y}{x} \right) = 0
\]

where \( z \) is a function of two variables \( x \) and \( y \). We need to find the value of:

\[
x \frac{\partial z}{\partial x} + y \frac{\partial z}{\partial y}
\]

### Step 1: Implicit Differentiation
To solve this, we will implicitly differentiate the equation with respect to \( x \) and \( y \). Let:

\[
u = \frac{z}{x^3}, \quad v = \frac{y}{x}
\]

So the equation becomes:

\[
\Phi(u, v) = 0
\]

Now, apply the chain rule to differentiate with respect to \( x \) and \( y \).

### Step 2: Differentiate with Respect to \( x \)

Using the chain rule:

\[
\frac{d}{dx} \Phi(u, v) = \frac{\partial \Phi}{\partial u} \cdot \frac{du}{dx} + \frac{\partial \Phi}{\partial v} \cdot \frac{dv}{dx} = 0
\]

First, find the partial derivatives of \( u \) and \( v \) with respect to \( x \):

\[
\frac{du}{dx} = \frac{1}{x^3} \frac{\partial z}{\partial x} - \frac{3z}{x^4}
\]

\[
\frac{dv}{dx} = -\frac{y}{x^2}
\]

Thus, the equation becomes:

\[
\frac{\partial \Phi}{\partial u} \left( \frac{1}{x^3} \frac{\partial z}{\partial x} - \frac{3z}{x^4} \right) + \frac{\partial \Phi}{\partial v} \left( - \frac{y}{x^2} \right) = 0
\]

### Step 3: Differentiate with Respect to \( y \)

Similarly, differentiate \( \Phi(u, v) = 0 \) with respect to \( y \):

\[
\frac{d}{dy} \Phi(u, v) = \frac{\partial \Phi}{\partial u} \cdot \frac{du}{dy} + \frac{\partial \Phi}{\partial v} \cdot \frac{dv}{dy} = 0
\]

Now, find the partial derivatives of \( u \) and \( v \) with respect to \( y \):

\[
\frac{du}{dy} = \frac{1}{x^3} \frac{\partial z}{\partial y}
\]

\[
\frac{dv}{dy} = \frac{1}{x}
\]

Substitute these into the equation:

\[
\frac{\partial \Phi}{\partial u} \cdot \frac{1}{x^3} \frac{\partial z}{\partial y} + \frac{\partial \Phi}{\partial v} \cdot \frac{1}{x} = 0
\]

### Step 4: Combine the Results
We now have two equations:

1. \[
\frac{\partial \Phi}{\partial u} \left( \frac{1}{x^3} \frac{\partial z}{\partial x} - \frac{3z}{x^4} \right) + \frac{\partial \Phi}{\partial v} \left( - \frac{y}{x^2} \right) = 0
\]

2. \[
\frac{\partial \Phi}{\partial u} \cdot \frac{1}{x^3} \frac{\partial z}{\partial y} + \frac{\partial \Phi}{\partial v} \cdot \frac{1}{x} = 0
\]

We are interested in finding \( x \frac{\partial z}{\partial x} + y \frac{\partial z}{\partial y} \).

By analyzing these equations and recognizing that they imply that the expression \( x \frac{\partial z}{\partial x} + y \frac{\partial z}{\partial y} \) is independent of \( \Phi \), the result is:

\[
x \frac{\partial z}{\partial x} + y \frac{\partial z}{\partial y} = 3z
\]

### Final Answer:
\[
x \frac{\partial z}{\partial x} + y \frac{\partial z}{\partial y} = 3z
\]

Would you like to explore this solution further or get more details on the process? Here are some follow-up questions you might find useful:

1. How does implicit differentiation apply to more complex equations?
2. What is the chain rule, and how is it used in partial differentiation?
3. Can you apply this method to different types of equations involving implicit functions?
4. How does one interpret \( \Phi \) when solving for partial derivatives implicitly?
5. What are the geometric interpretations of partial derivatives in multivariable calculus?

**Tip**: Always keep track of variables and their dependencies when applying implicit differentiation!

If Phi( z/(x ^ 3) ,  y\ x )=0, where z is a function of two variables x, y, then find the value of x partial z partial x + y partial z partial y

Explore how to apply implicit differentiation to solve for partial derivatives in the multivariable equation Phi(z/(x^3), y/x) = 0. Learn how to find the value of x ∂z/∂x + y ∂z/∂y = 3z using the chain rule and implicit function theorem. This advanced problem is ideal for college-level students.

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