Problem 124 - Calculus Volume 3

Question

Let \(z=\ln \left(\frac{x}{y}\right)\). Find \(\frac{\partial z}{\partial x}\) and \(\frac{\partial z}{\partial y}\).

Step-by-Step Solution

Answer

\( \frac{\partial z}{\partial x} = \frac{1}{x} \); \( \frac{\partial z}{\partial y} = -\frac{1}{y} \).

Step 1: Expression for z

Start by identifying the expression for the given function. We have \( z = \ln\left(\frac{x}{y}\right) \). This is the natural logarithm of the ratio \( \frac{x}{y} \).

Step 2: Derivative of ln function

Recall that the derivative of \( \ln(u) \) with respect to \( u \) is \( \frac{1}{u} \). Therefore, the derivative of \( \ln\left(\frac{x}{y}\right) \) with respect to \( \frac{x}{y} \) is \( \frac{1}{\frac{x}{y}} = \frac{y}{x} \).

Step 3: Partial Derivative with respect to x

Apply the chain rule to find \( \frac{\partial z}{\partial x} \). We have \( z = \ln\left(\frac{x}{y}\right) = \ln(x) - \ln(y) \). Thus, the partial derivative with respect to \( x \) is \( \frac{\partial z}{\partial x} = \frac{1}{x} \).

Step 4: Partial Derivative with respect to y

Similarly, consider the partial derivative with respect to \( y \). Again using the chain rule, the partial derivative is \( \frac{\partial z}{\partial y} = -\frac{1}{y} \). This result comes from differentiating \(-\ln(y)\) with respect to \( y \).

Key Concepts

Multivariable Calculus

Multivariable calculus involves dealing with functions of more than one variable. Typically, you work with functions like \( f(x, y) \) or \( z = \ln\left(\frac{x}{y}\right) \), where you have to consider how changes in multiple inputs affect the output.

This type of calculus extends single-variable calculus by allowing you to

• Compute partial derivatives, which are like regular derivatives but for one variable at a time.
• Understand surfaces and gradients, giving insight into the curvature of surfaces and their steepest ascent directions.

Understanding partial derivatives is crucial, as they provide information on how each variable separately affects the function. For example, in our original exercise, finding \( \frac{\partial z}{\partial x} \) and \( \frac{\partial z}{\partial y} \) helps us understand how the ratios of \( x \) and \( y \) contribute to the value of \( z \). By treating each variable independently, you can analyze the multidimensional shape's different aspects.

Chain Rule

The chain rule is a fundamental tool in calculus used for differentiating compositions of functions. It helps when you have a function nested inside another, like \( f(g(x)) \). In multivariable calculus, you frequently encounter compositions involving multiple variables.

When calculating partial derivatives for such functions, the chain rule becomes essential. In the context of our exercise, we consider the natural logarithm of a ratio, \( \ln\left(\frac{x}{y}\right) \). By applying the chain rule, we break down this complex expression into simpler parts. First, recognize that \( z = \ln\left(\frac{x}{y}\right) = \ln(x) - \ln(y) \).

• When differentiating \( \ln(x) \), we focus on how changes in \( x \) affect \( z \), resulting in \( \frac{1}{x} \) for \( \frac{\partial z}{\partial x} \).
• The chain rule provides a systematic way to find these derivatives, ensuring accuracy and clarity.

For each part of the composition, you differentiate individually and multiply by the derivative of the inner function, if necessary. This is particularly useful for capturing the nuanced dependencies in multivariable functions.

Natural Logarithm

The natural logarithm, denoted by \( \ln \), is a special logarithm with the base \( e \), where \( e \) is an irrational constant approximately equal to 2.71828. It appears frequently in calculus, mathematical modeling, and real-world scenarios, particularly involving exponential growth or decay.

In calculus, the natural logarithm is appreciated for its nice derivative properties. Specifically, the derivative of \( \ln(u) \) with respect to \( u \) is \( \frac{1}{u} \). This comes extremely handy when differentiating more complex expressions involving natural logarithms.

• For example, in \( z = \ln\left(\frac{x}{y}\right) \), to find \( \frac{\partial z}{\partial y} \), recognize that you can rewrite it as \( z = \ln(x) - \ln(y) \).
• Therefore, the derivative with respect to \( y \) is the derivative of \(-\ln(y)\), resulting in \(-\frac{1}{y} \).

This influential function simplifies exponential relations and helps interpret multiplicative processes additively. It is one of the staples in both theoretical and applied mathematics.