Proof: If Z = (x/y)ln(y), Then ∂z/∂x = ∂²z/∂y∂x

Hey guys! Today, we're diving deep into a super interesting mathematical problem. We're going to show that if we have a function ${ z }$ defined as ${ z = \frac{x}{y} \ln y }$, then the partial derivative of ${ z }$ with respect to ${ x }$ is equal to the second-order mixed partial derivative of ${ z }$ with respect to ${ y }$ and ${ x }$. Sounds like a mouthful, right? Don't worry, we'll break it down step by step so it's crystal clear. This is a classic example that beautifully illustrates the power of multivariable calculus, and it's something you might encounter in various fields, from physics to engineering. So, grab your calculators (or just your thinking caps!), and let's get started!

Understanding the Problem

Before we jump into the calculations, let's make sure we understand what the problem is asking. In essence, we're dealing with a function of two variables, ${ x }$ and ${ y }$, and we want to verify a specific relationship between its partial derivatives.

• Partial Derivatives: Remember, a partial derivative tells us how a function changes as we vary one variable while keeping the others constant. So, ${ \frac{\partial z}{\partial x} }$ tells us how ${ z }$ changes as ${ x }$ changes, assuming ${ y }$ stays the same.
• Mixed Partial Derivatives: These are where things get a bit more interesting. The mixed partial derivative ${ \frac{\partial^2 z}{\partial y \partial x} }$ means we first take the partial derivative of ${ z }$ with respect to ${ x }$, and then we take the partial derivative of the result with respect to ${ y }$. The order matters, so keep that in mind!

Our goal is to show that these two derivatives, ${ \frac{\partial z}{\partial x} }$ and ${ \frac{\partial^2 z}{\partial y \partial x} }$, are equal for the given function ${ z = \frac{x}{y} \ln y }$. This is a fascinating result because it touches on the concept of Clairaut's Theorem (or Schwarz's Theorem), which, under certain conditions, states that the order of differentiation doesn't matter for mixed partial derivatives.

Breaking Down the Function

To kick things off, let’s dissect the given function: ${ z = \frac{x}{y} \ln y }$. We have ${ z }$ expressed in terms of both ${ x }$ and ${ y }$. The term ${ \frac{x}{y} }$ is a simple rational function, while ${ \ln y }$ is the natural logarithm of ${ y }$. The interplay between these two terms is what makes this problem interesting. When we take partial derivatives, we'll need to apply the rules of differentiation carefully, treating each variable separately.

Why This Matters

You might be wondering, "Okay, this is a cool math problem, but why should I care?" Well, understanding partial derivatives and mixed partial derivatives is crucial in many areas of science and engineering. For instance, in physics, you might use partial derivatives to describe how temperature changes in space and time. In economics, you could use them to analyze how production changes with respect to different inputs. The concept of mixed partial derivatives is particularly important when dealing with complex systems where multiple factors interact.

Now that we've got a solid grasp of the problem, let's roll up our sleeves and dive into the actual calculations. We're going to take this step by step, so you can follow along easily. Let’s get this bread!

Calculating the Partial Derivatives

Alright, let's get our hands dirty with some actual calculus! Our mission here is to compute the partial derivatives we need to verify the given relationship. We'll start by finding ${ \frac{\partial z}{\partial x} }$, then we'll move on to calculating the mixed partial derivative ${ \frac{\partial^2 z}{\partial y \partial x} }$.

Step 1: Finding ${ \frac{\partial z}{\partial x} }$

To find the partial derivative of ${ z }$ with respect to ${ x }$, we treat ${ y }$ as a constant. Remember, our function is ${ z = \frac{x}{y} \ln y }$. We can rewrite this as:

${ z = \left(\frac{\ln y}{y}\right) x }$

Now, it’s clear that the term in the parentheses, ${ \frac{\ln y}{y} }$, is just a constant with respect to ${ x }$. So, when we differentiate with respect to ${ x }$, we simply apply the power rule:

${ \frac{\partial z}{\partial x} = \frac{\partial}{\partial x} \left[ \left(\frac{\ln y}{y}\right) x \right] }$

${ \frac{\partial z}{\partial x} = \frac{\ln y}{y} \frac{\partial}{\partial x} [x] }$

${ \frac{\partial z}{\partial x} = \frac{\ln y}{y} \cdot 1 }$

${ \frac{\partial z}{\partial x} = \frac{\ln y}{y} }$

Boom! We've found our first partial derivative. It’s actually quite neat and tidy. Notice how the result, ${ \frac{\ln y}{y} }$, is a function of ${ y }$ only, which makes sense since we treated ${ y }$ as a constant when differentiating with respect to ${ x }$.

Step 2: Finding ${ \frac{\partial^2 z}{\partial y \partial x} }$

Now, for the fun part – the mixed partial derivative. This means we'll differentiate our previous result, ${ \frac{\partial z}{\partial x} }$, with respect to ${ y }$. So, we have:

${ \frac{\partial^2 z}{\partial y \partial x} = \frac{\partial}{\partial y} \left[ \frac{\partial z}{\partial x} \right] }$

${ \frac{\partial^2 z}{\partial y \partial x} = \frac{\partial}{\partial y} \left[ \frac{\ln y}{y} \right] }$

To differentiate ${ \frac{\ln y}{y} }$ with respect to ${ y }$, we'll need to use the quotient rule. Remember the quotient rule? If we have a function ${ \frac{u(y)}{v(y)} }$, its derivative is:

${ \frac{d}{dy} \left[ \frac{u(y)}{v(y)} \right] = \frac{u'(y)v(y) - u(y)v'(y)}{[v(y)]^2} }$

In our case, ${ u(y) = \ln y }$ and ${ v(y) = y }$. So, we need to find ${ u'(y) }$ and ${ v'(y) }$:

${ u'(y) = \frac{d}{dy} [\ln y] = \frac{1}{y} }$

${ v'(y) = \frac{d}{dy} [y] = 1 }$

Now, we can plug these into the quotient rule:

${ \frac{\partial^2 z}{\partial y \partial x} = \frac{\frac{1}{y} \cdot y - \ln y \cdot 1}{y^2} }$

${ \frac{\partial^2 z}{\partial y \partial x} = \frac{1 - \ln y}{y^2} }$

Awesome! We've computed the mixed partial derivative. It's a bit more complex than our first derivative, but we got there by carefully applying the quotient rule.

Wrapping Up the Calculations

We've successfully calculated both ${ \frac{\partial z}{\partial x} }$ and ${ \frac{\partial^2 z}{\partial y \partial x} }$. Now, the moment of truth: do these results help us prove the original statement? Let's move on to the next section to compare our findings and see if we've cracked the code!

Verifying the Result

Okay, guys, we've done the heavy lifting – calculating the partial derivatives. Now comes the moment of truth: let's see if our results actually prove the relationship ${ \frac{\partial z}{\partial x} = \frac{\partial^2 z}{\partial y \partial x} }$. This is where we put everything together and see if the math gods are smiling upon us!

Recapping Our Findings

Just to keep everything fresh in our minds, let's quickly recap what we've found:

• We started with the function ${ z = \frac{x}{y} \ln y }$.
• We calculated the partial derivative of ${ z }$ with respect to ${ x }$: ${ \frac{\partial z}{\partial x} = \frac{\ln y}{y} }$
• We then calculated the mixed partial derivative of ${ z }$ with respect to ${ y }$ and ${ x }$: ${ \frac{\partial^2 z}{\partial y \partial x} = \frac{1 - \ln y}{y^2} }$

Comparing the Derivatives

Now, let's compare these two results. We want to see if ${ \frac{\partial z}{\partial x} }$ is equal to ${ \frac{\partial^2 z}{\partial y \partial x} }$. Looking at our expressions, we have:

${ \frac{\partial z}{\partial x} = \frac{\ln y}{y} }$

${ \frac{\partial^2 z}{\partial y \partial x} = \frac{1 - \ln y}{y^2} }$

Hold on a second! It seems like there's a slight discrepancy here. The expressions don't look identical at first glance. ${ \frac{\partial z}{\partial x} }$ is ${ \frac{\ln y}{y} }$, while ${ \frac{\partial^2 z}{\partial y \partial x} }$ is ${ \frac{1 - \ln y}{y^2} }$. They're not the same! So, does this mean we made a mistake somewhere? Not necessarily. It just means we need to be extra careful and double-check our work.

Spotting the Mistake

After reviewing the calculations, a subtle mistake was identified in the computation of the mixed partial derivative. The correct application of the quotient rule to ${ \frac{\partial}{\partial y} \left[ \frac{\ln y}{y} \right] }$ should yield:

${ \frac{\partial^2 z}{\partial y \partial x} = \frac{\frac{1}{y} imes y - \ln y imes 1}{y^2} = \frac{1 - \ln y}{y^2} }$

However, there was no mistake! We need to re-evaluate how we compare these derivatives.

Re-evaluating the Comparison

Let's take a step back and think about what we're trying to prove. We want to show that ${ \frac{\partial z}{\partial x} = \frac{\partial^2 z}{\partial y \partial x} }$. We found:

${ \frac{\partial z}{\partial x} = \frac{\ln y}{y} }$

And we correctly found:

${ \frac{\partial^2 z}{\partial y \partial x} = \frac{1 - \ln y}{y^2} }$

It seems there was an initial oversight in comparing these expressions directly. They are indeed different! The original statement we were trying to prove is incorrect. This is a crucial moment in problem-solving – recognizing that our initial expectation was wrong.

What Does This Mean?

So, we've shown that for the given function ${ z = \frac{x}{y} \ln y }$, the partial derivative ${ \frac{\partial z}{\partial x} }$ is not equal to the mixed partial derivative ${ \frac{\partial^2 z}{\partial y \partial x} }$. This might seem like a failure, but it's actually a valuable learning experience. It highlights the importance of carefully verifying results and being open to the possibility that a statement might be false.

This also brings up an interesting point about Clairaut's Theorem (or Schwarz's Theorem), which we mentioned earlier. This theorem states that if the second partial derivatives are continuous, then the order of differentiation doesn't matter. In other words, ${ \frac{\partial^2 z}{\partial y \partial x} }$ would be equal to ${ \frac{\partial^2 z}{\partial x \partial y} }$. However, in our case, we've shown that ${ \frac{\partial z}{\partial x} }$ is not equal to ${ \frac{\partial^2 z}{\partial y \partial x} }$, which implies that either the conditions for Clairaut's Theorem are not met, or there's something else going on.

Let's investigate further by calculating ${ \frac{\partial^2 z}{\partial x \partial y} }$ to see if it sheds any light on the situation. This will give us a more complete picture of the behavior of the partial derivatives of our function.

Further Investigation: Calculating ${ \frac{\partial^2 z}{\partial x \partial y} }$

Since we've discovered that ${ \frac{\partial z}{\partial x} }$ is not equal to ${ \frac{\partial^2 z}{\partial y \partial x} }$, it’s time to dig a little deeper. To fully understand what’s happening, let’s calculate the other mixed partial derivative, ${ \frac{\partial^2 z}{\partial x \partial y} }$. This will give us a more complete picture of the function's behavior and help us understand why Clairaut's Theorem might not apply in this case.

Step 1: Finding ${ \frac{\partial z}{\partial y} }$

First, we need to find the partial derivative of ${ z }$ with respect to ${ y }$. Recall that our function is ${ z = \frac{x}{y} \ln y }$. This time, we’re treating ${ x }$ as a constant and differentiating with respect to ${ y }$. This is where things get a bit more interesting, as we'll need to use the product rule. We can rewrite ${ z }$ as:

${ z = x \cdot \left(\frac{1}{y} \ln y\right) }$

Now, let's apply the product rule. If we have a function ${ z = u(y)v(y) }$, its derivative is:

${ \frac{dz}{dy} = u'(y)v(y) + u(y)v'(y) }$

In our case, ${ u(y) = \frac{1}{y} }$ and ${ v(y) = \ln y }$. So, we need to find ${ u'(y) }$ and ${ v'(y) }$:

${ u'(y) = \frac{d}{dy} \left[ \frac{1}{y} \right] = -\frac{1}{y^2} }$

${ v'(y) = \frac{d}{dy} [\ln y] = \frac{1}{y} }$

Now, we can plug these into the product rule:

${ \frac{\partial z}{\partial y} = x \left[ \left(-\frac{1}{y^2}\right) \ln y + \frac{1}{y} \cdot \frac{1}{y} \right] }$

${ \frac{\partial z}{\partial y} = x \left[ -\frac{\ln y}{y^2} + \frac{1}{y^2} \right] }$

${ \frac{\partial z}{\partial y} = x \left( \frac{1 - \ln y}{y^2} \right) }$

Nice! We've found the partial derivative of ${ z }$ with respect to ${ y }$. It looks a bit more complex than ${ \frac{\partial z}{\partial x} }$, but we got there by carefully applying the product rule.

Step 2: Finding ${ \frac{\partial^2 z}{\partial x \partial y} }$

Now, we'll differentiate our result, ${ \frac{\partial z}{\partial y} }$, with respect to ${ x }$. So, we have:

${ \frac{\partial^2 z}{\partial x \partial y} = \frac{\partial}{\partial x} \left[ \frac{\partial z}{\partial y} \right] }$

${ \frac{\partial^2 z}{\partial x \partial y} = \frac{\partial}{\partial x} \left[ x \left( \frac{1 - \ln y}{y^2} \right) \right] }$

Notice that the term in the parentheses, ${ \frac{1 - \ln y}{y^2} }$, is just a constant with respect to ${ x }$. So, when we differentiate with respect to ${ x }$, we simply apply the power rule:

${ \frac{\partial^2 z}{\partial x \partial y} = \left( \frac{1 - \ln y}{y^2} \right) \frac{\partial}{\partial x} [x] }$

${ \frac{\partial^2 z}{\partial x \partial y} = \left( \frac{1 - \ln y}{y^2} \right) \cdot 1 }$

${ \frac{\partial^2 z}{\partial x \partial y} = \frac{1 - \ln y}{y^2} }$

Eureka! We've computed the mixed partial derivative ${ \frac{\partial^2 z}{\partial x \partial y} }$. Interestingly, this result is the same as what we found for ${ \frac{\partial^2 z}{\partial y \partial x} }$ earlier. This is a significant observation!

Comparing the Mixed Partial Derivatives

Let's take a moment to compare our two mixed partial derivatives:

${ \frac{\partial^2 z}{\partial y \partial x} = \frac{1 - \ln y}{y^2} }$

${ \frac{\partial^2 z}{\partial x \partial y} = \frac{1 - \ln y}{y^2} }$

Look at that! The mixed partial derivatives are equal:

${ \frac{\partial^2 z}{\partial y \partial x} = \frac{\partial^2 z}{\partial x \partial y} }$

This is in line with Clairaut's Theorem, which states that if the second partial derivatives are continuous, then the order of differentiation shouldn't matter. So, while our initial assumption about ${ \frac{\partial z}{\partial x} }$ being equal to ${ \frac{\partial^2 z}{\partial y \partial x} }$ was incorrect, the mixed partial derivatives themselves do obey Clairaut's Theorem.

Putting It All Together

We've had quite the journey, guys! We started with a statement to prove, hit a snag, and then dug deeper to understand what was really going on. We've learned a lot about partial derivatives, mixed partial derivatives, and the importance of verifying results. Let's wrap up our findings in a final conclusion.

Conclusion

Alright, let's bring it all home! We started with the task of showing that if ${ z = \frac{x}{y} \ln y }$, then ${ \frac{\partial z}{\partial x} = \frac{\partial^2 z}{\partial y \partial x} }$. Through careful calculation and analysis, we've reached some interesting conclusions.

Key Findings

1. The Initial Statement is False: We found that ${ \frac{\partial z}{\partial x} = \frac{\ln y}{y} }$ and ${ \frac{\partial^2 z}{\partial y \partial x} = \frac{1 - \ln y}{y^2} }$. These are not equal, so the initial statement we were trying to prove is incorrect. This is a crucial lesson in mathematics – not every statement is true, and it's important to verify results rigorously.
2. Mixed Partial Derivatives Obey Clairaut's Theorem: We calculated the other mixed partial derivative, ${ \frac{\partial^2 z}{\partial x \partial y} }$, and found that it is equal to ${ \frac{\partial^2 z}{\partial y \partial x} }$. This confirms Clairaut's Theorem, which states that if the second partial derivatives are continuous, then the order of differentiation doesn't matter. In our case, the mixed partial derivatives are: ${ \frac{\partial^2 z}{\partial y \partial x} = \frac{\partial^2 z}{\partial x \partial y} = \frac{1 - \ln y}{y^2} }$
3. Importance of Careful Calculation: This problem highlights the importance of meticulous calculations and attention to detail. Even a small mistake in applying differentiation rules can lead to incorrect conclusions. We had to double-check our work to ensure we were on the right track.

Implications and Broader Context

This exercise is a fantastic illustration of the nuances of multivariable calculus. It shows us that while certain theorems (like Clairaut's Theorem) provide powerful tools, they come with conditions that must be met. It also underscores the importance of understanding the definitions and properties of partial derivatives.

In a broader context, understanding partial derivatives is essential in many fields. They are used extensively in physics to describe how quantities change in space and time, in engineering to optimize designs, and in economics to model complex systems. The concept of mixed partial derivatives is particularly relevant when dealing with systems where multiple variables interact.

Final Thoughts

So, there you have it, guys! We've tackled a challenging problem, learned a valuable lesson about verifying mathematical statements, and reinforced our understanding of partial derivatives. Remember, in mathematics (and in life), it's okay to be wrong sometimes. The key is to learn from our mistakes and keep exploring. Keep your mind sharp and your curiosity piqued, and you'll be amazed at what you can discover!