Problem 38 - Calculus Early Transcendentals: Pearson New International Edition

Question

Find the indicated derivative. \(\frac{d y}{d x}\) if \(e^{x+y}=4+x+y\)

Step-by-Step Solution

Answer

The derivative \(\frac{dy}{dx} = -1\).

Step 1: Differentiate Both Sides Implicitly

We have the equation \(e^{x+y} = 4 + x + y\). To find \(\frac{dy}{dx}\), we need to differentiate both sides of the equation with respect to \(x\). Using implicit differentiation, the derivative of \(e^{x+y}\) with respect to \(x\) is \(e^{x+y}(1 + \frac{dy}{dx})\). For the right side, the derivative of \(4\) is \(0\), \(x\) is \(1\), and \(y\) is \(\frac{dy}{dx}\).

Step 2: Set the Derivatives Equal

Equating the derivatives from both sides, we get:\[ e^{x+y}(1 + \frac{dy}{dx}) = 1 + \frac{dy}{dx} \]

Step 3: Rearrange to Solve for \(\frac{dy}{dx}\)

Subtract \(\frac{dy}{dx}\) from both sides to isolate like terms:\[ e^{x+y} + e^{x+y}\frac{dy}{dx} - \frac{dy}{dx} = 1 \] Or,\[ (e^{x+y} - 1) \frac{dy}{dx} = 1 - e^{x+y} \]

Step 4: Simplify the Equation

We can isolate \(\frac{dy}{dx}\) by dividing both sides by \(e^{x+y} - 1\):\[ \frac{dy}{dx} = \frac{1 - e^{x+y}}{e^{x+y} - 1} \]

Step 5: Simplify Further If Possible

Notice that \(1 - e^{x+y} = -(e^{x+y} - 1)\), thus simplifying the expression further:\[ \frac{dy}{dx} = \frac{-1(e^{x+y} - 1)}{e^{x+y} - 1} = -1 \]

Key Concepts

Derivative Calculation

Taking the derivative of a function is like finding the slope of a line that best fits the curve at any given point. It gives us insight into how one variable changes in relation to another. When calculating derivatives, especially when dealing with functions where variables are intermingled like in \(e^{x+y} = 4 + x + y\), implicit differentiation is a powerful tool. Unlike explicit differentiation, where one variable is solely dependent on the other, implicit differentiation deals with equations where both variables are interdependent.

• First, identify terms you need to differentiate with respect to your variable of interest.
• In our example, both sides of the equation \(e^{x+y} = 4 + x + y\) need to be differentiated with respect to \(x\).

For part of the left side, \(e^{x+y}\), this involves using the chain rule, resulting in \(e^{x+y}(1 + \frac{dy}{dx})\). The idea here is to remember not just to differentiate each part, but also account for changes in \(y\) as \(x\) changes. On the right side, we simply get constants and derivatives of simpler terms, leading to \(1 + \frac{dy}{dx}\) since the constant \(4\) differentiates to zero. By setting these derivatives equal to each other, we form a basis for solving for \(\frac{dy}{dx}\).

Exponential Functions

Exponential functions, such as \(e^{x+y}\), appear frequently in calculus. These functions involve the constant \(e\), roughly equal to 2.71828, and are considered transcendental numbers due to their non-repeating nature. The function \(e^{x}\) grows rapidly, doubling the output with every increase by one in the input.

• For instance, \(e^1\) is approximately \(2.718\) and \(e^2\) is about \(7.389\).
• Exponential functions have unique derivatives and integrals, which makes them very significant in calculus.

For differentiation, the derivative of an exponential function, such as \(e^{x}\), remains \(e^{x}\). However, with \(e^{x+y}\), due to the sum in the exponent, you must implement the chain rule—it requires multiplication by the derivative of the internal term, i.e., \(1+\frac{dy}{dx}\) for the implicit differentiation case. It's this property that underscores why, in our derivative calculation, the left-hand side expression \(e^{x+y}(1 + \frac{dy}{dx})\) arises.
Understanding how exponential functions behave and how they get differentiated is critical in solving complex calculus problems.

Equation Solving

Equation solving, especially involving derivatives, is a common task in calculus where we aim to find an unknown value. Here, the unknown is the derivative \(\frac{dy}{dx}\). Following implicit differentiation, we step into the realms of algebra to isolate our target term.

• Initially, we balance the differentiated equation derived earlier: \(e^{x+y}(1 + \frac{dy}{dx}) = 1 + \frac{dy}{dx}\).
• Through algebraic manipulation, terms involving \(\frac{dy}{dx}\) are grouped together.

The goal is to simplify this into a single statement representing \(\frac{dy}{dx}\). Subtracting \(\frac{dy}{dx}\) from both sides gives:\[(e^{x+y} - 1) \frac{dy}{dx} = 1 - e^{x+y}\]At this stage, divide each side by \(e^{x+y} - 1\) to isolate the desired term. Just remember that terms can often simplify surprisingly far—as we find here where the expression resolves to \(-1\). Equation solving requires careful handling of algebraic principles to arrive at the simplest form.