Problem 13 - College Algebra and Calculus: An Applied Approach

Question

Find the first partial derivatives with respect to \(x\) and with respect to \(y\). $$ z=\ln \frac{x+y}{x-y} $$

Step-by-Step Solution

Answer

The first partial derivatives of the function \(z = \ln \left(\frac{x + y}{x - y}\right)\) are \(\frac{dz}{dx} = \frac{1}{x + y} - \frac{1}{x - y}\) and \(\frac{dz}{dy} = \frac{1}{y + x} + \frac{1}{y - x}\)

Step 1: Expression for Differentiation

The given function \(z = \ln \left(\frac{x+y}{x-y}\right)\), which is the natural logarithm of a quotient, could be rewritten as \(z= \ln(x + y) - \ln(x - y)\). This formulation will make it easier to differentiate the function.

Step 2: Differentiation with respect to \(x\)

To get the first partial derivative of \(z\) with respect to \(x\), differentiate the rewritten version of the function with respect to \(x\). Using the chain rule, the derivative \(\frac{dz}{dx} = \frac{1}{x + y} - \frac{1}{x - y}\).

Step 3: Differentiation with respect to \(y\)

Repeat a similar process for the first partial derivative of \(z\) with respect to \(y\). Use the chain rule once again to find, \(\frac{dz}{dy} = \frac{1}{y + x} + \frac{1}{y - x}\).

Key Concepts

Chain Rule

When you deal with differentiation, the chain rule is your handy toolbox when functions are nested within each other. It lets you differentiate more complex compositions by peeling back each layer of the function. For the given function of the natural logarithm, - The original function is expressed as a natural log of a quotient: \( z = \ln \frac{x+y}{x-y} \). - This can be more flexibly rewritten as \( z = \ln(x+y) - \ln(x-y) \), allowing differentiation to become simpler.By understanding that the function is composed of such parts, the chain rule helps us find the derivatives by breaking down the components- For example, to find \( \frac{dz}{dx} \) or \( \frac{dz}{dy} \), the derivative of a natural log can be utilized.- This differentiating trick is crucial to handle functions like natural logs efficiently when they are composed of sums or differences inside their arguments.

Natural Logarithm

The natural logarithm function, denoted as \( \ln \), is a special kind of logarithm to the base \( e \), where \( e \approx 2.71828 \). Understanding the natural log is vital for solving differentiation tasks involving logarithmic functions, including the one in this exercise.- The function given in the exercise was originally \( z = \ln \frac{x+y}{x-y} \). - This logarithm of a quotient can be seamlessly transformed using the property of logarithms: \( \ln(a) - \ln(b) = \ln \frac{a}{b} \).Thus, it simplifies to two separate natural log functions, \( z = \ln(x+y) - \ln(x-y) \). Differentiating becomes more straightforward. Additionally:- Remember the logarithmic differentiation rule: the derivative of \( \ln(u) \) is \( \frac{1}{u} \cdot \frac{du}{dx} \). - This is crucial when you apply the chain rule and differentiate each logarithmic part independently.

Differentiation

Differentiation involves finding the rate at which a function changes, which is fundamental in calculus. In this exercise, you need to determine the first partial derivatives with respect to both \( x \) and \( y \).- Start by expressing the given function in a more manageable form, \( z = \ln(x+y) - \ln(x-y) \).- Differentiating with respect to \( x \), and keeping \( y \) constant, you find \( \frac{dz}{dx} = \frac{1}{x+y} - \frac{1}{x-y} \). - To differentiate with respect to \( y \), the process is similar but this time keeping \( x \) constant gives \( \frac{dz}{dy} = \frac{1}{y+x} + \frac{1}{y-x} \).These derivatives, called partial derivatives, specifically show how the function \( z \) changes as one variable changes while the other remains constant. Remember that partial differentiation focuses on one variable at a time, which is instrumental in multivariable calculus.