Problem 37 - Calculus Early Transcendentals: Pearson New International Edition

Question

$$ \underline{\phantom{xxx}} , \text { find the indicated derivative. } $$ $$ \frac{d y}{d x} \text { if } e^{x y}+x y=2 \text { Hint: Use implicit differentiation. } $$

Step-by-Step Solution

Answer

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

Step 1: Understand Implicit Differentiation

Implicit differentiation is used when it is difficult to solve for one variable in terms of another. In the given problem, the equation \(e^{xy} + xy = 2\) involves both \(x\) and \(y\), making it hard to express \(y\) as a function of \(x\).

Step 2: Differentiate Both Sides with Respect to x

Start by differentiating both sides of the equation \(e^{xy} + xy = 2\) with respect to \(x\). The derivative of the right-hand side, which is a constant, is 0.

Step 3: Apply Chain Rule and Product Rule

To differentiate \(e^{xy}\), use chain rule: \(e^{xy}(x \frac{dy}{dx} + y)\), because you must consider both \(x\) and \(y\) as functions of \(x\). For \(xy\), use the product rule: \(y + x \frac{dy}{dx}\).

Step 4: Set Up the Necessary Equation

After applying the differentiation rules, the expression becomes: \[ e^{xy}(x \frac{dy}{dx} + y) + y + x \frac{dy}{dx} = 0 \]

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

Rearrange the terms to isolate \(\frac{dy}{dx}\): \[ (e^{xy} x + x)y' = -e^{xy} y - y \] Factor out \(\frac{dy}{dx}\): \[ y'(e^{xy} x + x) = -e^{xy} y - y \] Solve for \(y'\) or \(\frac{dy}{dx}\): \[ y' = \frac{-e^{xy} y - y}{e^{xy} x + x} \]

Step 6: Simplify the Expression

Simplify the expression for \(\frac{dy}{dx}\) if possible. In this case, the expression: \[ \frac{dy}{dx} = \frac{-y(e^{xy} + 1)}{x(e^{xy} + 1)} \] may not simplify further depending on the values of \(x\) and \(y\) but represents the derivative in terms of \(x\) and \(y\).

Key Concepts

Chain Rule

In calculus, the chain rule is an essential tool for finding the derivative of composite functions. Essentially, it helps break down complex functions into simpler parts, making differentiation manageable.
When implicit differentiation is involved, as in the given problem, the chain rule becomes crucial because it deals with expressions where one variable is a function of another, but is not isolated.
In the equation given, \( e^{xy} + xy = 2 \), the expression \( e^{xy} \) is more complex because it involves two variables, \(x\) and \(y\).
To differentiate \( e^{xy} \), you consider \(xy\) as the inner function, and \(e^{xy}\) as the outer function. Applying the chain rule, you differentiate the outer function with respect to the inner function, then multiply it by the derivative of the inner function.
This results in:

• \( e^{xy} \cdot (x \frac{dy}{dx} + y) \)
  • \( e^{xy} \) remains the same from the exponential derivative.
  • \( (x \frac{dy}{dx} + y) \) is accounted for the derivative of \(xy\), using both \(x\) and \(y\) as functions of each other.

Product Rule

The product rule is another fundamental concept in calculus used whenever you need to differentiate a product of two functions. In implicit differentiation, it frequently appears because functions can have interdependent variables.
In our exercise, the term \( xy \) requires the product rule. This rule states that if you have two functions, \( u(x) \) and \( v(x) \), the derivative of their product \( u(x)v(x) \) is given by:

• \( u'(x) v(x) + u(x) v'(x) \)

Applying this to \( xy \):

• \( y + x\frac{dy}{dx} \)

• \( y \) stays as it is because the derivative of \( x \) with respect to \( x \) is 1.
• \( x\frac{dy}{dx} \) represents the derivative of \( y \) with respect to \( x \), showcasing the need for implicit differentiation.

This product rule simplifies deriving terms that cannot be easily separated into single-variable functions.

Calculus Differentiation

Calculus differentiation refers to the process of finding the rate at which a function changes at any given point.
This concept forms the backbone of many topics in calculus, including implicit differentiation, which we used in this exercise.
To understand calculus differentiation, it's important to grasp a few key points:

• Derivatives provide information on how functions behave and interact with variable changes.
• Differentiation techniques like the chain rule and product rule help unravel complex functions into manageable parts.

In the context of our exercise, differentiation allows us to find \( \frac{dy}{dx} \), illustrating how \( y \) changes with respect to \( x \) under the constraint provided by the equation \( e^{xy} + xy = 2 \).
By breaking down each component using implicit differentiation, you solve for the derivative, even when \( y \) and \( x \) are interdependent and cannot be explicitly separated. This highlights the power of differentiation in handling equations where variables are tangled in a multifaceted relationship.