Problem 9 - Calculus with Concepts in Calculus

Question

Compute \(\partial z / \partial u\) and \(\partial z / \partial v\). $$ z=\ln \left(x^{2}-y^{2}\right) ; x=u-v, y=u^{2}+v^{2} $$

Step-by-Step Solution

Answer

\(\frac{\partial z}{\partial u}\) and \(\frac{\partial z}{\partial v}\) can be found using chain and derivative rules.

Step 1: Substitute Expressions for x and y

Start by expressing \(x\) and \(y\) in terms of \(u\) and \(v\). Given \(x = u - v\) and \(y = u^2 + v^2\), substitute these expressions into the function \(z = \ln(x^2 - y^2)\), yielding:\[ z = \ln((u-v)^2 - (u^2 + v^2)^2) \]

Step 2: Simplify the Expression Inside ln

Next, simplify the expression inside the logarithm:\( (u-v)^2 = u^2 - 2uv + v^2 \) and \( (u^2 + v^2)^2 = u^4 + 2u^2v^2 + v^4 \). Thus, we have: \[ z = \ln((u^2 - 2uv + v^2) - (u^4 + 2u^2v^2 + v^4)) \]

Step 3: Use Chain Rule for Partial Derivative of z with Respect to u

Now use the chain rule to compute \(\frac{\partial z}{\partial u}\):The chain rule states \(\frac{\partial z}{\partial u} = \frac{1}{x^2 - y^2} \left(2x\frac{\partial x}{\partial u} - 2y\frac{\partial y}{\partial u}\right).\)Given \(x = u - v\), \(\frac{\partial x}{\partial u} = 1\), and \(y = u^2 + v^2\), \(\frac{\partial y}{\partial u} = 2u\).Therefore,\[ \frac{\partial z}{\partial u} = \frac{1}{x^2 - y^2} \left(2(u-v)(1) - 2(u^2+v^2)(2u) \right)\]

Step 4: Compute the Partial Derivative of z with Respect to v

Similarly, apply the chain rule to compute \(\frac{\partial z}{\partial v}\). Using a similar approach and the derivatives:\[\frac{\partial x}{\partial v} = -1, \quad \frac{\partial y}{\partial v} = 2v\]The partial derivative \(\frac{\partial z}{\partial v}\) is:\[ \frac{\partial z}{\partial v} = \frac{1}{x^2 - y^2} \left(2(u-v)(-1) - 2(u^2 + v^2)(2v) \right)\]

Step 5: Final Expressions for Partial Derivatives

The expressions for \(\frac{\partial z}{\partial u}\) and \(\frac{\partial z}{\partial v}\) are:\[ \frac{\partial z}{\partial u} = \frac{1}{x^2 - y^2} \left(2(u-v) - 4u(u^2 + v^2)\right) \]\[ \frac{\partial z}{\partial v} = \frac{1}{x^2 - y^2} \left(-2(u-v) - 4v(u^2 + v^2)\right) \]

Key Concepts

Chain Rule

The chain rule is a essential tool in calculus that helps us solve problems related to rates of change and derivatives. When dealing with complex functions that are compounded with other functions, the chain rule allows us to differentiate composite functions. In the realm of partial differentiation, this becomes particularly important when working with functions of multiple variables.

To apply the chain rule in partial differentiation, consider a function like:

• For a function like \( z = f(x,y) \), where \( x \) and \( y \) themselves are functions of \( u \) and \( v \), the chain rule offers a way to compute \( \frac{\partial z}{\partial u} \) and \( \frac{\partial z}{\partial v} \).
• The formula becomes: \[ \frac{\partial z}{\partial u} = \frac{\partial z}{\partial x} \cdot \frac{\partial x}{\partial u} + \frac{\partial z}{\partial y} \cdot \frac{\partial y}{\partial u}\]
• And, \( \frac{\partial z}{\partial v} = \frac{\partial z}{\partial x} \cdot \frac{\partial x}{\partial v} + \frac{\partial z}{\partial y} \cdot \frac{\partial y}{\partial v}\)

Applying these formulas allows one to break down complex derivatives into manageable parts by identifying the inner and outer functions. It's a powerful method to simplify the differentiation process, especially when functions involve nested layers of dependencies.

Natural Logarithm

The natural logarithm, denoted as \( \ln \), is a fundamental function in mathematics, particularly due to its unique properties and relationship with exponential functions. The natural logarithm is the inverse of the exponential function \( e^x \), and it is defined such that \( \ln(e^x) = x \).

Some key characteristics of the natural logarithm include:

• The natural logarithm of a product: \( \ln(ab) = \ln(a) + \ln(b) \).
• The natural logarithm of a quotient: \( \ln\left( \frac{a}{b} \right) = \ln(a) - \ln(b) \).
• The natural logarithm of a power: \( \ln(a^b) = b \cdot \ln(a) \).

These rules make them especially useful in solving derivations involving multiplicative processes or exponential growth.

In the exercise, the function \( z = \ln(x^2 - y^2) \) requires differentiation. When differentiating logarithmic functions, especially with additional composite arguments, it's essential to utilize these properties effectively. Understanding these rules helps in both simplifying complex logarithmic expressions and correctly applying derivative rules, which often involves using the chain rule, as seen in the exercise solutions.

Functions of Multiple Variables

Functions of multiple variables, or multivariable functions, are functions with more than one input or independent variable. They can be visualized as surfaces or landscapes, where each variable represents a different dimension.

In a typical function \( z = f(x, y) \), \( z \) is the dependent variable depending on the two independent variables \( x \) and \( y \). When these independent variables themselves depend on other variables, such as \( u \) and \( v \), the complexity of such functions increases, which is where partial differentiation and the chain rule become valuable tools.

Here are some fundamental concepts:

• Partial derivatives measure how the function changes as each individual variable changes while holding the others constant. This is crucial for understanding the relative influence of each variable on the function's output.
• Interdependencies, where multiple variables interact, can often lead to compounded effects on the function's behavior and require a breakdown approach to solve.

Understanding functions of multiple variables is key in fields ranging from physics to economics, as they model systems with many influencing factors. Mastery of these concepts is fundamental for solving real-world problems where variables are interlinked and affect outcomes in complex ways.