Question

Find \(\partial w / \partial t\) by using the Chain Rule. Express your final answer in terms of \(s\) and \(t\) $$ w=\ln (x+y)-\ln (x-y) ; x=t e^{s}, y=e^{s t} $$

Short answer

\(\partial w / \partial t = \frac{1}{t e^s + e^{st}}(e^s + se^{st}) - \frac{1}{t e^s - e^{st}}(e^s - se^{st})\)

Step-by-step solution

Understanding the Problem

We need to find the partial derivative \(\partial w / \partial t\) using the chain rule, given \(w = \ln(x+y) - \ln(x-y)\), where \(x = t e^s\) and \(y = e^{st}\). We will express the final answer in terms of \(s\) and \(t\).

Apply Chain Rule to w

Let's apply the chain rule to find \(\partial w / \partial t\). We'll differentiate \(w\) with respect to \(t\), treating \(s\) as a constant. The function is composed of terms involving \(x\) and \(y\), so we need to apply the chain rule to each term separately.

Differentiate Inside Logarithms

The derivative of \(\ln(u)\) is \(1/u\). Therefore, for \(w = \ln(x+y) - \ln(x-y)\), we have:\(\frac{d}{dt} \ln(x+y) = \frac{1}{x+y} \cdot \frac{d}{dt}(x+y)\)\(\frac{d}{dt} \ln(x-y) = \frac{1}{x-y} \cdot \frac{d}{dt}(x-y)\)

Find Derivatives of (x+y) and (x-y)

Compute \(\frac{d}{dt}(x+y)\) and \(\frac{d}{dt}(x-y)\):Since \(x = t e^s\) and \(y = e^{st}\):\[\frac{d}{dt}(x + y) = \frac{d}{dt}(t e^s + e^{st}) = e^s + se^{st}\]\[\frac{d}{dt}(x - y) = \frac{d}{dt}(t e^s - e^{st}) = e^s - se^{st}\]

Combine Derivatives Using Chain Rule

Substitute the derivatives obtained into the expressions for \(\partial w/\partial t\):\[\partial w / \partial t = \frac{1}{x+y}(e^s + se^{st}) - \frac{1}{x-y}(e^s - se^{st})\]Substituting back \(x = t e^s\) and \(y = e^{st}\), we maintain in terms of \(s\) and \(t\) as:\[\partial w / \partial t = \frac{1}{t e^s + e^{st}}(e^s + se^{st}) - \frac{1}{t e^s - e^{st}}(e^s - se^{st})\]

Simplification (if necessary)

At this point, the expression is in terms of \(s\) and \(t\). Simplify further if needed, ensuring the answer remains in terms of \(s\) and \(t\), but since the expression is already simplified regarding variables, this step might be optional based on the clarity needed.

Key concepts

Chain Rule

The Chain Rule is a fundamental tool in calculus, particularly useful when dealing with complex functions composed of simpler functions. In essence, it helps differentiate composite functions by linking various layers of a function to their derivatives.
To understand the chain rule, imagine we have a function defined as:
- A composite structure like \(w(x, y)\), - Where \(x\) and \(y\) are themselves functions of another variable like \(t\).
The Chain Rule allows us to express the derivative of \(w\) with respect to \(t\) in terms of the derivatives of \(x\) and \(y\) with respect to \(t\).

• If the function \(w\) depends on multiple variables, and each variable is a function of another variable, the Chain Rule helps trace all these dependencies and computes the overall rate of change.
• For our problem, use the chain rule to find \(\partial w / \partial t\) by differentiating \(\ln(x+y)\) and \(\ln(x-y)\) functions, which are expressed in terms of \(x(t)\).

This layered differentiation is what makes the Chain Rule a powerful tool in calculus.

Logarithmic Functions

Logarithmic functions, represented as \(\ln(x)\), play a crucial role in mathematical analysis because they offer a way to reverse exponential functions. They also provide a means to transform multiplicative processes into additive ones, simplifying differentiation and integration.
Here are some key points to remember:

• The derivative of \(\ln(u)\) with respect to its argument \(u\) is \(1/u\).
• In our exercise, the function \(w = \ln(x+y) - \ln(x-y)\) is a difference of two logarithms. To find \(\partial w / \partial t\), one must differentiate each logarithmic term individually using the chain rule.
• The quotient inside the logarithm translates multiplicative complexities into a sum or difference, which is easier to handle during differentiation.

Thus, understanding logarithmic functions simplifies tasks such as solving exponential growth problems and differentiating nested functions, which are quite common in calculus.

Multivariable Calculus

Multivariable Calculus extends the principles of regular calculus into more dimensions, dealing with functions involving multiple variables. It's especially pertinent in fields like physics, engineering, and computer science, where systems are seldom constrained to a single variable.
In multivariable calculus, we get familiar with:

• Partial Derivatives: Calculating the derivative of a function with respect to one variable while holding others constant, like \(\partial w / \partial t\) in our problem.
• The Chain Rule for Multiple Variables: An extended version of the chain rule, allowing differentiation of composite multivariable functions.
• Interpreting changes and rates in multi-dimensional contexts, offering insights into how change in one variable affects a system overall.

By using these concepts, you gain the ability to analyze complex systems both mathematically and intuitively, which is a significant leap in problem-solving capabilities.