Question

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

Short answer

\(f_x(-1,1)=-e \sinh(1)\), \(f_y(-1,1)=e \cosh(1)\).

Step-by-step solution

Understand Partial Derivatives

The partial derivative with respect to a variable describes how a function changes as that specific variable changes, while keeping other variables constant. We need to compute both the partial derivatives \(f_x\) and \(f_y\) for \(f(x, y)=e^{y} \cosh x\).

Compute \(f_x(x, y)\)

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

Evaluate \(f_x(-1, 1)\)

Now that we have \(f_x(x, y) = e^y \sinh x\), substitute \(x = -1\) and \(y = 1\) into the expression: \(f_x(-1, 1) = e^1 \sinh(-1)\). Since \(\sinh(-1) = -\sinh(1)\), this becomes \(-e \sinh(1)\).

Compute \(f_y(x, y)\)

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

Evaluate \(f_y(-1, 1)\)

Substitute \(x = -1\) and \(y = 1\) into the expression: \(f_y(-1, 1) = e^1 \cosh(-1)\). Since \(\cosh(-1) = \cosh(1)\), this becomes \(e \cosh(1)\).

Key concepts

Calculus

Calculus is a branch of mathematics that focuses on the study of change and motion. It provides tools for understanding complex dynamics in various mathematical models. One of the key concepts in calculus is differentiation, which involves finding the derivative of a function. This tells us how a function changes as its input changes. Differentiation is particularly important when dealing with physical phenomena where change is continuous.

Partial derivatives are an essential part of multivariable calculus. A partial derivative of a multivariable function is its derivative with respect to one of the variables, holding the other variables constant.

• Partial derivatives are used to analyze functions of two or more variables.
• In applications, they describe how a system changes when altering one parameter at a time.

In the given exercise, you learned how to compute partial derivatives of a function of two variables, specifically turning to calculus for finding rates of change with respect to each of the variables separately.

Function Evaluation

Function evaluation involves substituting specific values into a function to determine its output. This process is crucial for verifying and understanding the results of mathematical problems. For the function \(f(x, y)=e^{y} \cosh x\), evaluating means substituting particular values of \(x\) and \(y\) into the function to find its exact value at that point.

• Determine expressions for partial derivatives \(f_x(x, y)\) and \(f_y(x, y)\).
• Substitute given values to find \(f_x(-1, 1)\) and \(f_y(-1, 1)\).

For example, by substituting \(x = -1\) and \(y = 1\) into \(f_x = e^y \sinh x\), one can calculate \(f_x(-1, 1)\) as \(-e \sinh(1)\). Similarly, for \(f_y(-1, 1)\), one replaces \(x\) and \(y\) in \(f_y = e^y \cosh x\) to get \(e \cosh(1)\). Understanding evaluation not only confirms the calculation but also clarifies the behavior of complex functions at specific points.

Hyperbolic Functions

Hyperbolic functions are analogues of the trigonometric functions but for a hyperbola rather than a circle. They are defined through exponential functions and often arise in contexts involving hyperbolas, such as in certain types of differential equations.

In the function \(f(x, y)=e^{y} \cosh x\), the hyperbolic cosine function, \(\cosh x\), is used. The basic definitions for hyperbolic functions are:

• \( \cosh x = \frac{e^x + e^{-x}}{2} \)
• \( \sinh x = \frac{e^x - e^{-x}}{2} \)

These functions have similar properties to their trigonometric counterparts but with differences in periodicity and symmetry. For example, whereas cosine is periodic and oscillates, \(\cosh x\) does not oscillate and grows exponentially as \(x\) moves away from zero.

Understanding how to differentiate hyperbolic functions is essential, as seen in the partial derivatives of \(f(x, y)\). Knowing that \(\frac{d}{dx}\cosh x = \sinh x\) and \(\frac{d}{dx}\sinh x = \cosh x\) allows for correct computation of derivatives when solving calculus problems involving hyperbolic functions.