Problem 10 - Calculus Early Transcendentals: Pearson New International Edition

Question

In Problems 1-16, find all first partial derivatives of each function. \(f(s, t)=\ln \left(s^{2}-t^{2}\right)\)

Step-by-Step Solution

Answer

\( \frac{\partial f}{\partial s} = \frac{2s}{s^2 - t^2} \) and \( \frac{\partial f}{\partial t} = \frac{-2t}{s^2 - t^2} \)

Step 1: Understand the Function

The given function is a two-variable function expressed as \[ f(s, t) = \ln(s^{2} - t^{2}) \]We need to find the first partial derivatives with respect to both variables \(s\) and \(t\).

Step 2: Differentiate with Respect to s

To find the partial derivative of \( f(s, t) \) with respect to \( s \), treat \( t \) as a constant.Apply the chain rule:First, differentiate the natural logarithm function: \[ \frac{d}{ds} \ln(u) = \frac{1}{u} \cdot \frac{du}{ds} \]Where \( u = s^2 - t^2 \) and \( \frac{du}{ds} = 2s \).So the partial derivative is:\[ \frac{\partial f}{\partial s} = \frac{1}{s^2 - t^2} \cdot 2s = \frac{2s}{s^2 - t^2} \]

Step 3: Differentiate with Respect to t

Now, find the partial derivative of \( f(s, t) \) with respect to \( t \), treating \( s \) as a constant.Again, use the chain rule:\[ \frac{d}{dt} \ln(u) = \frac{1}{u} \cdot \frac{du}{dt} \]Where \( u = s^2 - t^2 \) and \( \frac{du}{dt} = -2t \).So the partial derivative is:\[ \frac{\partial f}{\partial t} = \frac{1}{s^2 - t^2} \cdot (-2t) = \frac{-2t}{s^2 - t^2} \]

Key Concepts

Chain Rule

The chain rule is a fundamental concept in calculus that helps us differentiate composite functions. When we have a function that is composed of one function inside another, the chain rule allows us to take the derivative by first differentiating the outer function and then multiplying it by the derivative of the inner function. This helps us understand how changes in the variables affect the overall function.

In the context of partial derivatives, the chain rule is applied similarly. For two-variable functions, we need to account for how changing one variable influences the function while treating the other variable as constant. In the given problem, we have a function involving a natural logarithm, where the argument is another function of two variables,

• We started with a function: \[ f(s, t) = \ln(s^2 - t^2) \]
• To find the partial derivative with respect to either variable, say \(s\), the chain rule states:\[ \frac{\partial}{\partial s} \ln(u) = \frac{1}{u} \cdot \frac{du}{ds}\] Here, \(u = s^2 - t^2\).

Using this method, we found how each variable independently contributes to changes in the function by keeping the other constant. The chain rule is essential in tackling such problems as it simplifies the process of finding derivatives for more complex functions.

Natural Logarithm

The natural logarithm, denoted as \(\ln(x)\), is a logarithm to the base \(e\), where \(e\) is an irrational and transcendental number approximately equal to 2.71828. It is a critical function in mathematics due to its unique properties and utility in calculus. The derivative of the natural logarithm function is straightforward and forms the basis for differentiating more complex logarithmic expressions.

Some properties of the natural logarithm include:

• The derivative: \( \frac{d}{dx}\ln(x) = \frac{1}{x} \)
• It transforms multiplication inside the log into addition: \( \ln(xy) = \ln(x) + \ln(y) \)
• Defined only for positive arguments, since \( \ln(x) \) is undefined for \(x \leq 0\).

In the problem, we dealt with \(\ln(s^2 - t^2)\). Here, understanding the derivative of \(\ln\) was key to applying the chain rule correctly:

• We differentiated \(\ln(u)\) with respect to \(s\) and \(t\), where \(u = s^2 - t^2\), obtaining:
  • \( \frac{1}{u} \cdot \frac{du}{ds} \) for \(s\)
  • \( \frac{1}{u} \cdot \frac{du}{dt} \) for \(t\)

This illustrates the natural logarithm's role in simplifying the differentiation process of functions that have complex arguments.

Two-Variable Functions

Two-variable functions, such as \(f(s, t) = \ln(s^2 - t^2)\), are functions that depend on two independent variables. In calculus, exploring how each variable affects the function is essential, and partial derivatives are the tools for this exploration.

Partial derivatives allow us to see the rate of change of the function with respect to one variable while keeping the other constant. This can be particularly useful in fields like physics and economics, where relationships between multiple varying quantities need to be understood.

For a general function \( f(x, y) \):

• Partial derivative with respect to \(x\): \( \frac{\partial f}{\partial x} \)
• Partial derivative with respect to \(y\): \( \frac{\partial f}{\partial y} \)

In the exercise, we performed partial differentiation by treating:

• \(s\) as the variable while keeping \(t\) constant, and vice versa.
• The results showed how \(s\) and \(t\) independently affected \(f(s, t)\).

This approach is pivotal for understanding functions of multiple variables and allows one to break down complex behavior into simpler, understandable changes.