Problem 24 - Calculus Early Transcendentals: Pearson New International Edition

Question

If \(f(x, y)=e^{y} \cosh x\), find \(f_{x}(-1,1)\) and \(f_{y}(-1,1)\).

Step-by-Step Solution

Answer

\(f_{x}(-1, 1) = -e \sinh 1\); \(f_{y}(-1, 1) = e \cosh 1\).

Step 1: Find the Partial Derivative with Respect to x

To find the partial derivative of the function \(f(x, y) = e^{y} \cosh x\) with respect to \(x\), we treat \(y\) as a constant. The derivative of \(\cosh x\) with respect to \(x\) is \(\sinh x\). Therefore, \(f_{x}(x, y) = e^{y} \sinh x\).

Step 2: Evaluate at the Given Point for \(f_{x}\)

Now substitute \(x = -1\) and \(y = 1\) into the partial derivative \(f_{x}(x, y) = e^{y} \sinh x\). Calculate: \[f_{x}(-1, 1) = e^{1} \sinh(-1) = e \times (-\sinh 1)\]. The hyperbolic sine of \(-1\) is \(- \sinh 1\), so \(f_{x}(-1, 1) = -e \sinh 1\).

Step 3: Find the Partial Derivative with Respect to y

To find the partial derivative of \(f(x, y) = e^{y} \cosh x\) with respect to \(y\), we treat \(x\) as a constant. The derivative of \(e^{y}\) with respect to \(y\) is \(e^{y}\). Therefore, \(f_{y}(x, y) = e^{y} \cosh x\).

Step 4: Evaluate at the Given Point for \(f_{y}\)

Substitute \(x = -1\) and \(y = 1\) into the partial derivative \(f_{y}(x, y) = e^{y} \cosh x\). Calculate: \[f_{y}(-1, 1) = e^{1} \cosh(-1) = e \cosh 1\]. The hyperbolic cosine function is even, so \(\cosh(-1) = \cosh 1\), hence \(f_{y}(-1, 1) = e \cosh 1\).

Key Concepts

Hyperbolic Functions

Hyperbolic functions are analogues of the trigonometric functions but for the hyperbola or exponential functions. The most common hyperbolic functions are the hyperbolic sine (\( \sinh \)) and hyperbolic cosine (\( \cosh \)). These functions are defined using exponential functions:

• The hyperbolic sine is given by \( \sinh x = \frac{e^x - e^{-x}}{2} \).
• The hyperbolic cosine is given by \( \cosh x = \frac{e^x + e^{-x}}{2} \).

Both functions have unique properties:

• \( \cosh x \) is an even function, which means that \( \cosh(-x) = \cosh x \).
• \( \sinh x \) is an odd function, implying \( \sinh(-x) = -\sinh x \).

These special symmetries are useful in calculus, particularly when taking derivatives and evaluating them at different points. For instance, in our exercise, knowing the even property of \( \cosh x \) simplifies the problem by understanding that the function's value remains the same when switching signs of the variable.

Multivariable Calculus

Multivariable calculus is the extension of calculus to functions of more than one variable. It involves concepts like partial derivatives and multiple integrations. When you have a function such as \( f(x, y) = e^{y} \cosh x \), both \( x \) and \( y \) act as variables. This complexity allows you to explore how changes in one variable influence the function, holding the other variable constant.Some crucial concepts in multivariable calculus include:

• **Partial Derivatives:** These measure how a function changes with respect to one variable while keeping others constant.
• **Gradient Vector:** This encompasses all partial derivatives, indicating the direction of steepest ascent of a function.

In the given exercise, we focus on finding partial derivatives \( f_x \) and \( f_y \), approaching them step by step to handle multiple variables efficiently. It's very similar to single-variable derivatives but requires treating the other variables as constants during the differentiation process.

Evaluating Derivatives

Evaluating derivatives involves calculating the actual value of the derivative at specific points. In the process, you already determined the formula for the derivative, and then you're plugging in specific values to get numerical results. Here's a quick overview:

• **Step-by-Step Evaluation:** After finding the derivative expression, replace the variables with the provided numbers to evaluate it.
• **Application:** Use properties of the functions involved, like those of hyperbolic functions, to simplify calculations.

In this exercise, we compute \( f_x(-1,1) \) and \( f_y(-1,1) \) by substituting \( x = -1 \) and \( y = 1 \) into the expressions for the partial derivatives previously found. Knowing that \( \sinh(-1) = -\sinh(1) \) and \( \cosh(-1) = \cosh(1) \), we simplify and obtain results. Evaluating derivatives at points is a practical way to understand a function's behavior at specific coordinates.