Problem 14 - The Calculus with Analytic Geometry

Question

In Exercises 11 through 14 , find the total derivative \(d u / d t\) by two methods: (a) Use the chain rule; (b) make the substitutions for \(x\) and \(y\) or for \(x, y\), and \(z\) before differentiating. $$ u=\frac{t+e^{x}}{y-e^{t}} ; x=3 \sin t ; y=\ln t $$

Step-by-Step Solution

Answer

Using the chain rule and substitutions both yield: \[ \frac{d u}{d t} = \frac{(\ln t - e^t)(1 + 3e^{3 \sin t} \cos t) - (t + e^{3 \sin t})(\frac{1}{t} - e^t)}{(\ln t - e^t)^2} \].

Step 1: Identify the functions

The given functions are: \[u = \frac{t + e^x}{y - e^t}, \, x = 3 \, \sin t, \, y = \ln t\]

Step 2: Use the chain rule

To find \(\frac{d u}{d t}\) using the chain rule, we use the formula: \[\frac{d u}{d t} = \frac{\partial u}{\partial t} + \frac{\partial u}{\partial x} \cdot \frac{d x}{d t} + \frac{\partial u}{\partial y} \cdot \frac{d y}{d t}\] First, compute the partial derivatives: \[ \frac{\partial u}{\partial t} = \frac{1}{y - e^t} + \frac{t + e^x}{(y - e^t)^2} e^t\] \[ \frac{\partial u}{\partial x} = \frac{e^x}{y - e^t}\] \[ \frac{\partial u}{\partial y} = -\frac{t + e^x}{(y - e^t)^2}\] Then, compute the derivatives of \(x\) and \(y\) with respect to \(t\): \[ \frac{d x}{d t} = 3 \cos t\] \[ \frac{d y}{d t} = \frac{1}{t}\] Finally, substitute into the chain rule formula: \[ \frac{d u}{d t} = \frac{1}{y - e^t} + \frac{t + e^x}{(y - e^t)^2} e^t + \frac{e^x}{y - e^t} \cdot 3 \cos t - \frac{t + e^x}{(y - e^t)^2} \cdot \frac{1}{t}\]

Step 3: Make substitutions before differentiating

Substitute the given functions \(x = 3 \sin t\) and \(y = \ln t\) into \(u\): \[ u = \frac{t + e^{3 \sin t}}{\ln t - e^t} \] Differentiate directly with respect to \(t\) using quotient rule: Let \(u = \frac{f(t)}{g(t)}\), then the derivative is given by: \[\frac{d u}{d t} = \frac{g(t) \frac{d f(t)}{d t} - f(t) \frac{d g(t)}{d t}}{(g(t))^2}\] Where \(f(t) = t + e^{3 \sin t}\) and \(g(t) = \ln t - e^t\). First, compute \(\frac{d f(t)}{d t}\) and \(\frac{d g(t)}{d t}\): \[\frac{d f(t)}{d t} = 1 + e^{3 \sin t} \cdot 3 \cos t\] \[\frac{d g(t)}{d t} = \frac{1}{t} - e^t\] Substitute these into the quotient rule formula: \[\frac{d u}{d t} = \frac{(\ln t - e^t)(1 + 3 e^{3 \sin t} \cos t) - (t + e^{3 \sin t})(\frac{1}{t} - e^t)}{(\ln t - e^t)^2}\]

Step 4: Simplify the results

Simplify the expressions obtained in both methods to confirm they are equal. After simplifying the expressions provided above: Using either method will yield the same result, confirming the accuracy of differentiation.

Key Concepts

Chain Rule

The chain rule is a fundamental concept in calculus used to differentiate compositions of functions. Essentially, it allows you to find how a function changes with respect to one variable when that function is composed with other functions of another variable.

Here's the chain rule formula we'll use for partial derivatives:
\[\frac{d u}{d t} = \frac{\partial u}{\partial t} + \frac{\partial u}{\partial x} \cdot \frac{d x}{d t} + \frac{\partial u}{\partial y} \cdot \frac{d y}{d t}\]

When applying the chain rule to the given function, we break it down into smaller, more manageable parts. First, we'll find the partial derivatives of \(u\) concerning each variable. Then, we need to determine the regular derivatives of \(x\) and \(y\) concerning \(t\). By substituting these derivatives back into our chain rule formula, we get the full derivative of \(u\) with respect to \(t\).

Key things to remember:

• Find the partial derivatives with respect to each variable.
• Compute the regular derivatives for the intermediate variables like \(x\) and \(y\).
• Plug everything back into the chain rule formula to get the final answer.

Using the formulas and derivatives computed, the final step lies in substitution and simplifying the expressions to find the derivative \(\frac{d u}{d t}\). Understanding the chain rule thoroughly can help simplify complex expressions and multi-variable functions.

Partial Derivatives

Partial derivatives are an extension of regular derivatives when functions have more than one variable. Instead of finding the rate of change of a function with just one variable, partial derivatives focus on how a function changes with respect to one specific variable while keeping all other variables constant.

In our exercise, we need to find partial derivatives of \(u\) with respect to \(t\), \(x\), and \(y\).

• \(\frac{\partial u}{\partial t}\) captures how \(u\) changes directly with \(t\).
• \(\frac{\partial u}{\partial x}\) calculates the change in \(u\) with respect to \(x\).
• \(\frac{\partial u}{\partial y}\) explains how \(u\) varies as \(y\) changes.

Some key features of finding partial derivatives include:

• Identifying the dependent and independent variables in the function.
• Differentiating the function while treating other variables as constants.
• Using these partial derivatives in the chain rule formula to eventually find the total derivative.

In our steps, we found:

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

These partials help us understand the contribution of each variable to the overall change in \(u\) when computing the total derivative.

Quotient Rule

The quotient rule is a method for differentiating functions that are expressed as the ratio of two differentiable functions. It's incredibly useful when the function you need to differentiate involves division. The quotient rule states:
\[\frac{d}{dx}\left(\frac{f(x)}{g(x)}\right) = \frac{g(x)f'(x) - f(x)g'(x)}{[g(x)]^2}\]

In our exercise, after substituting \(x\) and \(y\) with their respective functions of \(t\), we simplify \(u\) to a form that can be differentiated using the quotient rule. This is because \(u\) is given by the ratio of \(t+e^{3\sin t}\) over \(\ln t - e^t\).

Let’s break this down step-by-step:
1. Identify the numerator \(f(t)\) as \(t + e^{3\sin t}\) and the denominator \(g(t)\) as \(\ln t - e^t\).2. Compute \(f'(t)\), the derivative of the numerator, and \(g'(t)\), the derivative of the denominator.

• \(f'(t) = 1 + e^{3\sin t} \cdot 3 \cos t\)
• \(g'(t) = \frac{1}{t} - e^t\)

3. Apply the quotient rule formula to find \(\frac{d u}{d t}\).

Substituting our functions into the formula, we get:
\[\frac{g(t)f'(t) - f(t)g'(t)}{(g(t))^2} = \frac{(\ln t - e^t)(1 + 3 e^{3 \sin t} \cos t) - (t + e^{3 \sin t})(\frac{1}{t} - e^t)}{(\ln t - e^t)^2}\]

The quotient rule simplifies the differentiation process by providing a clear formula to handle divisions of functions.