Problem 17 - The Calculus with Analytic Geometry

Question

In Exercises 15 through 18 , find the total derivative \(d u / d t\) by using the chain rule; do not express \(u\) as a function of \(t\) before differentiating. $$ u=\frac{x+t}{y+t^{\prime}} x=\ln t ; y=\ln \frac{1}{t} $$

Step-by-Step Solution

Answer

\( \frac{du}{dt} = \frac{y + 2t}{t(y + t)^2} \)

Step 1 - Understand the given functions

Given: \( u = \frac{x + t}{y + t'} \) \( x = \ln t \) \( y = \ln \frac{1}{t} \) Note: \( t' \) (prime) represents \( t \).

Step 2 - Compute the partial derivatives of \(u\)

Using \( u = \frac{x + t}{y + t} \): \( \frac{\partial u}{\partial x} = \frac{1}{y + t} \) \( \frac{\partial u}{\partial y} = -\frac{x + t}{(y + t)^2} \) \( \frac{\partial u}{\partial t} = \frac{1}{y + t} - \frac{(x + t)}{(y + t)^2} \)

Step 3 - Compute the derivatives of \(x\) and \(y\) with respect to \(t\)

\( \frac{dx}{dt} = \frac{d}{dt}(\ln t) = \frac{1}{t} \) \( \frac{dy}{dt} = \frac{d}{dt}(\ln \frac{1}{t}) = -\frac{1}{t} \)

Step 4 - Apply the chain rule

The chain rule formula: \( \frac{du}{dt} = \frac{\partial u}{\partial x} \frac{dx}{dt} + \frac{\partial u}{\partial y} \frac{dy}{dt} + \frac{\partial u}{\partial t} \)

Step 5 - Substitute the partial derivatives and simplify

Substitute into the chain rule formula: \( \frac{du}{dt} = \frac{1}{y + t} \cdot \frac{1}{t} + \left( -\frac{x + t}{(y + t)^2} \right) \cdot \left( -\frac{1}{t} \right) + \left( \frac{1}{y + t} - \frac{x + t}{(y + t)^2} \right) \) Simplify each term: \( \frac{du}{dt} = \frac{1}{t(y + t)} + \frac{x + t}{t(y + t)^2} + \left( \frac{1}{y + t} - \frac{x + t}{(y + t)^2} \right) \)

Step 6 - Combine and simplify the terms

Combine all terms under a common denominator: \( \frac{du}{dt} = \frac{(y + t + x + t) + t - (x + t)}{t(y + t)^2} \) Simplify further: \( \frac{du}{dt} = \frac{y + 2t}{t(y + t)^2} \)

Key Concepts

Chain Rule

The Chain Rule is essential in calculus for finding the derivative of a composite function. This rule is particularly useful when dealing with functions of multiple variables.

Consider a function \(u = u(x, y, t)\), where each variable, such as \(x\) and \(y\), is itself a function of \(t\). Instead of rewriting \(u\) as an explicit function of \(t\) and then differentiating, we use the chain rule.

It is formulated as:

\[ \frac{du}{dt} = \frac{\text{∂}u}{\text{∂}x} \frac{dx}{dt} + \frac{\text{∂}u}{\text{∂}y} \frac{dy}{dt} + \frac{\text{∂}u}{\text{∂}t} \]

Each partial derivative term \( \frac{\text{∂}u}{\text{∂}x} \), \( \frac{\text{∂}u}{\text{∂}y} \), and \( \frac{\text{∂}u}{\text{∂}t} \) requires careful computation, followed by multiplication with the respective derivatives \( \frac{dx}{dt} \) and \( \frac{dy}{dt} \). This approach simplifies finding \( \frac{du}{dt} \) without explicitly rewriting \ u \ in terms of \( t \).

Partial Derivatives

Partial Derivatives involve differentiating a multivariable function with respect to one variable while holding the others constant. This is crucial when dealing with functions like \(u = \frac{x + t}{y + t}\).

Here:

• \( \frac{\text{∂}u}{\text{∂}x} = \frac{1}{y + t} \)
• \( \frac{\text{∂}u}{\text{∂}y} = -\frac{x + t}{(y + t)^2} \)
• \( \frac{\text{∂}u}{\text{∂}t} = \frac{1}{y + t} - \frac{(x + t)}{(y + t)^2} \)

Each of these partials explains how \(u\) changes as we tweak just one variable, keeping the other variables constant. These insights are vital for applying the chain rule effectively and calculating the total derivative \( \frac{du}{dt} \).

Differentiation

Differentiation is the backbone of calculus, assisting us in understanding how a function changes. It involves taking derivatives, which measure the rate of change of a variable.

In simple terms:

• The derivative of \(x = \text{ln}(t)\) with respect to \(t\) is \( \frac{dx}{dt} = \frac{1}{t} \)
• The derivative of \(y = \text{ln}\frac{1}{t} \) with respect to \(t\) is \( \frac{dy}{dt} = -\frac{1}{t} \)

Equipped with these derivatives, we can use the chain rule to determine the total derivative \( \frac{du}{dt} \). This method simplifies our understanding of how complex functions navigate changes in their variables. With these basics, calculating the total derivative becomes a clear and systematic process.