Problem 14 - Calculus with Concepts in Calculus

Question

Determine \(d f\). $$ f(x, y)=y \ln \frac{1+x}{1-x} $$

Step-by-Step Solution

Answer

The differential is \( df = \left(y \frac{2}{1-x^2}\right) dx + \left(\ln \frac{1+x}{1-x}\right) dy \).

Step 1: Understanding the Function

The function given is \( f(x, y) = y \ln \frac{1+x}{1-x} \). It is a multivariable function with variables \( x \) and \( y \). To find the differential \( df \), we need to compute the partial derivatives of \( f \) with respect to each variable, then apply the formula for the total differential.

Step 2: Partial Derivative with Respect to \( x \)

Let's find the partial derivative of \( f \) with respect to \( x \), denoted as \( f_x \).\[ f_x = \frac{\partial}{\partial x}\left(y \ln \frac{1+x}{1-x}\right)\]Using the chain rule, we have:\[f_x = y \cdot \frac{d}{d x}\left(\ln \frac{1+x}{1-x}\right)\]The derivative of \( \ln \frac{1+x}{1-x} \) with respect to \( x \) is \( \frac{2}{1-x^2} \).Thus, \( f_x = y \cdot \frac{2}{1-x^2} \).

Step 3: Partial Derivative with Respect to \( y \)

Now, find the partial derivative of \( f \) with respect to \( y \), denoted as \( f_y \).\[ f_y = \frac{\partial }{\partial y}\left(y \ln \frac{1+x}{1-x}\right)\]This simplifies directly to:\[ f_y = \ln \frac{1+x}{1-x}\]

Step 4: Calculate the Total Differential

The formula for the total differential \( df \) of a function \( f(x, y) \) is:\[ df = f_x \, dx + f_y \, dy\]Substitute the partial derivatives:\[ df = \left(y \cdot \frac{2}{1-x^2}\right) dx + \left(\ln \frac{1+x}{1-x}\right) dy\]

Step 5: Present the Solution

The differential of the function \( f(x, y) = y \ln \frac{1+x}{1-x} \) is:\[df = \left(y \cdot \frac{2}{1-x^2}\right) dx + \left(\ln \frac{1+x}{1-x}\right) dy\]

Key Concepts

Partial Derivatives

Partial derivatives in multivariable calculus are like the regular derivatives you learned in single-variable calculus, but now you focus on one variable at a time. Imagine freezing all other variables except one—this allows you to see how the function changes just due to a small change in that specific variable.
In the given exercise, we have a function of two variables, \( f(x, y) = y \ln \frac{1+x}{1-x} \). To find the partial derivative with respect to \( x \), we treat \( y \) as a constant. Similarly, for \( f_y \), we keep \( x \) unchanged and differentiate concerning \( y \).
When you take these partial derivatives, you're essentially slicing the multivariable function to understand its shape in each direction. This gives us:

• \( f_x = y \cdot \frac{2}{1-x^2} \)
• \( f_y = \ln \frac{1+x}{1-x} \)

Chain Rule

The chain rule is a powerful tool in calculus used to find the derivative of composite functions. In multivariable calculus, the chain rule helps us find the partial derivative when one function is composed with another function. It's like peeling an onion, layer by layer.
In our exercise, to find the partial derivative \( f_x \) for the term \( \ln \frac{1+x}{1-x} \), we use the chain rule to differentiate it with respect to \( x \). We compute this derivative while treating \( y \) as constant.

• The derivative of \( \ln \frac{1+x}{1-x} \) with respect to \( x \) gives us \( \frac{2}{1-x^2} \). This is because we first differentiate the outer function \( \ln \) and then multiply by the derivative of the inner function \( \frac{1+x}{1-x} \).

This layering approach captures how each variable identifies change over different paths, revealing how intricate functions relate to their components.

Differential

The concept of a differential provides a way to approximate how much a multivariable function's output changes along different paths near a point. It combines partial derivatives and changes in both variables. This makes differentials flexible and powerful in calculus.
For the function \( f(x, y) = y \ln \frac{1+x}{1-x} \), the total differential \( df \) combines changes in \( x \) and \( y \) to approximate changes in \( f \). The formula is:

• \( df = f_x \, dx + f_y \, dy \)

This formula means that:\(

\)
• The change in the function \( df \), is affected by both the change in \( x \), denoted \( dx \), and the change in \( y \), denoted \( dy \).

The partial derivatives, \( f_x \) and \( f_y \), weigh these incremental changes appropriately.
This calculated differential gives us a linear approximation of the function's change, providing a good estimate of behavior for small distances from any point \((x, y)\).