Problem 90 - Advanced Calculus

Question

Let \(\mathrm{u}=\mathrm{u}(\mathrm{x}, \mathrm{y})\) be implicitly defined as a function of \(\mathrm{x}\) and \(\mathrm{y}\) by the equation \(u+\ln u \approx x y\). Find \((\partial \mathrm{u} / \partial \mathrm{x}),(\partial \mathrm{u} / \partial \mathrm{y}),\left[\left(\partial^{2} \mathrm{u}\right) /(\partial \mathrm{x} \partial \mathrm{y})\right]\) and \(\left[\left(\partial^{2} \mathrm{u}\right) /(\partial \mathrm{y} \partial \mathrm{x})\right]\)

Step-by-Step Solution

Answer

The partial derivatives are: \[\frac{\partial u}{\partial x} = \frac{yu}{u+1}\] and \[\frac{\partial u}{\partial y} = \frac{xu}{u+1}\], and the mixed partial derivatives are: \[\frac{\partial^2 u}{\partial x \partial y} = \frac{u^2}{(u+1)^2}\] and \[\frac{\partial^2 u}{\partial y \partial x} = \frac{u^2}{(u+1)^2}\].

Step 1: Differentiate with respect to x

First, we will differentiate the entire equation with respect to x. As we do so, we will treat u as a function of x and y, without finding this function explicitly. Differentiate both sides of the equation: \[\frac{d(u + \ln u)}{dx} = \frac{d(xy)}{dx}\] Using the chain rule on the left-hand side, we obtain: \[\frac{\partial u}{\partial x} + \frac{\partial (\ln u)}{\partial x} = \frac{d(xy)}{dx}\] Now we can explicitly find the expression for \(\frac{\partial u}{\partial x}\).

Step 2: Simplify and find first partial derivative

Using the chain rule and the implicit derivative of the natural logarithm function, we can simplify the equation as follows: \[\frac{\partial u}{\partial x} + \frac{1}{u} \frac{\partial u}{\partial x} = y\] Now solve for \(\frac{\partial u}{\partial x}\): \[\frac{\partial u}{\partial x}(1+\frac{1}{u}) = y \Rightarrow \frac{\partial u}{\partial x} = \frac{y}{1+\frac{1}{u}}\] Now we have found the first partial derivative: \[\frac{\partial u}{\partial x} = \frac{yu}{u+1}\]

Step 3: Differentiate with respect to y

By following a similar procedure, we will differentiate the entire equation with respect to y. Differentiate both sides of the equation: \[\frac{d(u + \ln u)}{dy} = \frac{d(xy)}{dy}\] Using the chain rule on the left-hand side, we obtain: \[\frac{\partial u}{\partial y} + \frac{\partial (\ln u)}{\partial y} = \frac{d(xy)}{dy}\] Now we can explicitly find the expression for \(\frac{\partial u}{\partial y}\).

Step 4: Simplify and find second partial derivative

Using the chain rule and the implicit derivative of the natural logarithm function, we can again simplify the equation as follows: \[\frac{\partial u}{\partial y} + \frac{1}{u} \frac{\partial u}{\partial y} = x\] Now solve for \(\frac{\partial u}{\partial y}\): \[\frac{\partial u}{\partial y}(1+\frac{1}{u}) = x \Rightarrow \frac{\partial u}{\partial y} = \frac{x}{1+\frac{1}{u}}\] Now we have found the second partial derivative: \[\frac{\partial u}{\partial y} = \frac{xu}{u+1}\]

Step 5: Compute mixed partial derivatives

Next, we will find the mixed partial derivatives of u. We can do this by differentiating \(\frac{\partial u}{\partial x}\) with respect to y and \(\frac{\partial u}{\partial y}\) with respect to x. For \(\frac{\partial^2 u}{\partial x \partial y}\), differentiate \(\frac{\partial u}{\partial x}\) with respect to y: \[\frac{\partial^2 u}{\partial x \partial y} = \frac{d(\frac{yu}{u+1})}{dy}\] For \(\frac{\partial^2 u}{\partial y \partial x}\), differentiate \(\frac{\partial u}{\partial y}\) with respect to x: \[\frac{\partial^2 u}{\partial y \partial x} = \frac{d(\frac{xu}{u+1})}{dx}\]

Step 6: Simplify and find mixed partial derivatives

Use the product rule to simplify the mixed partial derivatives: \[\frac{\partial^2 u}{\partial x \partial y} = \frac{u(1+\frac{1}{u}) - y \frac{\partial u}{\partial y}}{(u+1)^2}\] \[\frac{\partial^2 u}{\partial y \partial x} = \frac{u(1+\frac{1}{u}) - x \frac{\partial u}{\partial x}}{(u+1)^2}\] Now substituting expressions for \(\frac{\partial u}{\partial x}\) and \(\frac{\partial u}{\partial y}\) we found earlier: \[\frac{\partial^2 u}{\partial x \partial y} = \frac{u(1+\frac{1}{u}) - y \frac{xu}{u+1}}{(u+1)^2} = \frac{(u+1)u - y(xu)}{(u+1)^2}\] \[\frac{\partial^2 u}{\partial y \partial x} = \frac{u(1+\frac{1}{u}) - x \frac{yu}{u+1}}{(u+1)^2} = \frac{(u+1)u - x(yu)}{(u+1)^2}\] After simplifying them further, you can observe that they are equal: \[\frac{\partial^2 u}{\partial x \partial y} = \frac{u^2}{(u+1)^2}\] \[\frac{\partial^2 u}{\partial y \partial x} = \frac{u^2}{(u+1)^2}\] So the final answers are: \[\frac{\partial u}{\partial x} = \frac{yu}{u+1}\] \[\frac{\partial u}{\partial y} = \frac{xu}{u+1}\] \[\frac{\partial^2 u}{\partial x \partial y} = \frac{u^2}{(u+1)^2}\] \[\frac{\partial^2 u}{\partial y \partial x} = \frac{u^2}{(u+1)^2}\]

Key Concepts

Partial Derivative

When we talk about a partial derivative, we refer to the rate at which a function changes with respect to one variable, while keeping all other variables constant. In the context of implicit differentiation, partial derivatives are crucial because they help us understand how a multi-variable implicit function changes along a specific axis. For example, in the given exercise, the partial derivatives \(\frac{\partial u}{\partial x}\) and \(\frac{\partial u}{\partial y}\) express how the function \(u\) changes as \(x\) or \(y\) change respectively, while the other variable remains fixed.

To calculate \(\frac{\partial u}{\partial x}\), we differentiate the equation with respect to \(x\) treating \(u\) as a function of \(x\) and \(y\). The resulting expression allows us to see the unique effect that changes in \(x\) have on \(u\). Similarly, \(\frac{\partial u}{\partial y}\) is found by considering changes in \(y\) alone. These concepts are foundational in fields such as physics, engineering, and economics, where systems are often influenced by multiple variables that can change independently of one another.

Chain Rule

The chain rule is a fundamental tool in calculus used to find the derivative of composite functions. When differentiating an implicit function where one variable is a function of others, the chain rule allows us to disentangle these relationships and calculate derivatives with respect to individual variables.

For instance, when differentiating \(\ln u\) in our example, where \(u\) is a function of \(x\) and \(y\), we write: \(\frac{\partial (\ln u)}{\partial x} = \frac{1}{u} \frac{\partial u}{\partial x}\). Here, \(\frac{1}{u}\) comes from the derivative of the \(\ln u\) function, and \(\frac{\partial u}{\partial x}\) is the partial derivative of \(u\) with respect to \(x\), which we need to find. The chain rule enables us to break down complex derivative problems into simpler parts that are easier to manage and solve. It's like unraveling a twisted rope by following each strand one by one to understand the overall structure.

Natural Logarithm Function

The natural logarithm function, denoted as \(\ln(x)\), is the inverse of the exponential function \(e^x\). It plays a vital role in calculus, especially when dealing with growth processes or compound interest scenarios. In our exercise, the natural logarithm appears as part of the implicitly defined function involving \(u\).

The derivative of the natural logarithm function is quite elegant and simple: \(\frac{d(\ln x)}{dx} = \frac{1}{x}\). This becomes exceptionally useful when using implicit differentiation, as we leverage this property to find the derivative of \(\ln u\) with respect to other variables. In the exercise, the derivative \(\ln u\) with respect to \(x\) and \(y\) becomes pivotal in finding the required partial derivatives of \(u\). Understanding the natural logarithm not only aids in solving calculus problems but also enriches one's comprehension of continuous growth and the relationship between exponential and logarithmic functions.

Mixed Partial Derivatives

In multivariable calculus, mixed partial derivatives are second-order derivatives that involve differentiation with respect to two different variables. Consider \(\frac{\partial^2 u}{\partial x \partial y}\) from our exercise; this denotes a mixed partial derivative of \(u\) first with respect to \(x\) and then with respect to \(y\).

A key property of mixed partial derivatives, when the function is well-behaved (smooth), is that the order of differentiation does not matter—this is known as Clairaut's theorem. This means that \(\frac{\partial^2 u}{\partial x \partial y}\) is equal to \(\frac{\partial^2 u}{\partial y \partial x}\) under suitable conditions, which we see is the case in the given solution. Understanding mixed partial derivatives is essential for analyzing how functions change directionally in space and is heavily utilized in optimization problems, economics for analyzing marginal effects, and in physics for understanding concepts such as wave propagation and heat transfer.