Problem 38 - Calculus Early Transcendentals: Pearson New International Edition

Question

$$ \underline{\phantom{xxx}} , \text { find the indicated derivative. } $$ $$ \frac{d y}{d x} \text { if } e^{x+y}=4+x+y $$

Step-by-Step Solution

Answer

\( \frac{dy}{dx} = \frac{1 - e^{x+y}}{e^{x+y} - 1} \)

Step 1: Implicit Differentiation Setup

We are given the equation \( e^{x+y} = 4 + x + y \). To find \( \frac{dy}{dx} \), we will differentiate both sides of this equation with respect to \( x \), treating \( y \) as a function of \( x \) (implicit differentiation).

Step 2: Differentiate the Left Side

For the left side, \( e^{x+y} \):Differentiate with respect to \( x \):\( \frac{d}{dx} [e^{x+y}] = e^{x+y} (1 + \frac{dy}{dx}) \). This is because we differentiate \( e^{x+y} \) normally and use the chain rule for \( y \).

Step 3: Differentiate the Right Side

For the right side, \( 4 + x + y \):Differentiate with respect to \( x \):The derivative is \( 1 + \frac{dy}{dx} \) since \( 4 \) is constant and \( y \) depends on \( x \).

Step 4: Set Equations Equal and Simplify

Equate the derivatives from both sides:\( e^{x+y} (1 + \frac{dy}{dx}) = 1 + \frac{dy}{dx} \).Re-arrange:\( e^{x+y} + e^{x+y} \frac{dy}{dx} = 1 + \frac{dy}{dx} \).

Step 5: Solve for \( \frac{dy}{dx} \)

To find \( \frac{dy}{dx} \):1. Rearrange the equation to isolate terms involving \( \frac{dy}{dx} \).2. \( e^{x+y} \frac{dy}{dx} - \frac{dy}{dx} = 1 - e^{x+y} \).3. Factor out \( \frac{dy}{dx} \):\( \frac{dy}{dx} (e^{x+y} - 1) = 1 - e^{x+y} \).4. Solve for \( \frac{dy}{dx} \):\( \frac{dy}{dx} = \frac{1 - e^{x+y}}{e^{x+y} - 1} \).

Key Concepts

Chain Rule

The Chain Rule is a crucial tool in calculus for finding derivatives of complex functions. It helps us differentiate functions that are composed of two or more simpler functions. When performing implicit differentiation, as in the given problem, the Chain Rule becomes even more important because it allows us to handle functions where one of the variables is dependent on another.

In simpler terms, if you have a function inside another function, like in the case of the expression \( e^{x+y} \), you use the Chain Rule to differentiate them step by step. Here's how the Chain Rule is applied:

• First, differentiate the outer function normally. For \( e^{x+y} \), that gives us \( e^{x+y} \) since the derivative of \( e^u \) is \( e^u \).
• Next, multiply this by the derivative of the inner function, which is \( x+y \). The derivative of \( x+y \) with respect to \( x \) is \( 1 + \frac{dy}{dx} \) because \( y \) is a function of \( x \).

This process allows us to account for the hidden dependencies between variables when differentiating, especially with exponential functions. Understanding the Chain Rule is essential for solving problems where variables are interdependent.

Derivative

A derivative represents how a function changes as its input changes. It's like assessing the rate of change, similar to how speed tells us how position changes with time. In calculus, the derivative can be found using different rules, one of which is the implicit differentiation we've applied in this exercise.

Implicit differentiation is particularly useful when dealing with equations involving two variables, like \( e^{x+y} = 4 + x + y \). Here, finding the derivative \( \frac{dy}{dx} \) requires treating \( y \) as a function of \( x \).

• Start by differentiating both sides of the equation with respect to \( x \). Treat \( y \) as a function of \( x \) using implicit differentiation, which acknowledges that any change in \( x \) affects \( y \).
• This involves using the Chain Rule on the left side to handle \( e^{x+y} \), resulting in terms that include \( \frac{dy}{dx} \).
• Then differentiate the right side where the derivative of \( 4 + x + y \) is \( 1 + \frac{dy}{dx} \).

By equating both derivatives and solving for \( \frac{dy}{dx} \), you can find how \( y \) changes with \( x \), which is the essence of derivatives.

Exponential Functions

Exponential functions have unique properties that make them vital in calculus. An exponential function, like \( e^{x} \), consists of a constant base raised to the power of a variable. When differentiating exponential functions, particularly those embedded in other expressions like \( e^{x+y} \), their properties are key.

This exercise revolves around the differentiation of such an exponential function with respect to one of its components. These functions follow distinct rules:

• The derivative of \( e^u \) is \( e^u \cdot \frac{du}{dx} \). This means that exponential functions retain their form when differentiated. The additional factor \( \frac{du}{dx} \) originates from the Chain Rule, capturing the change inside the exponent.
• In \( e^{x+y} \), since both \( x \) and \( y \) are in the exponent, differentiating requires considering how both variables might change and applying the Chain Rule. Thus, you end up multiplying \( e^{x+y} \) by the derivative of \( x+y \), which includes \( 1 + \frac{dy}{dx} \).

Understanding exponential functions and how they interact through calculus aids in mastering derivatives across various mathematical problems.