Problem 9 - Calculus Early Transcendentals: Pearson New International Edition

Question

In Problems 1-16, find all first partial derivatives of each function. \(g(x, y)=e^{-x y}\)

Step-by-Step Solution

Answer

\(\frac{\partial g}{\partial x} = -y e^{-xy}, \frac{\partial g}{\partial y} = -x e^{-xy}\).

Step 1: Identify the Partial Derivatives

Given the function \(g(x, y) = e^{-xy}\), we need to find the partial derivatives \(\frac{\partial g}{\partial x}\) and \(\frac{\partial g}{\partial y}\). Partial derivatives represent the change of the function with respect to one variable, keeping the other variable constant.

Step 2: Calculate the Partial Derivative with Respect to x

To find \(\frac{\partial g}{\partial x}\), treat \(y\) as a constant and differentiate \(g(x, y)\) with respect to \(x\). Using the chain rule, \(\frac{\partial}{\partial x}[e^{-xy}] = e^{-xy} \cdot \frac{\partial}{\partial x}[-xy] = e^{-xy} \cdot (-y) = -y e^{-xy}\).

Step 3: Calculate the Partial Derivative with Respect to y

To find \(\frac{\partial g}{\partial y}\), treat \(x\) as a constant and differentiate \(g(x, y)\) with respect to \(y\). Using the chain rule, \(\frac{\partial}{\partial y}[e^{-xy}] = e^{-xy} \cdot \frac{\partial}{\partial y}[-xy] = e^{-xy} \cdot (-x) = -x e^{-xy}\).

Step 4: Conclusion

The partial derivatives of the function \(g(x, y) = e^{-xy}\) are \(\frac{\partial g}{\partial x} = -y e^{-xy}\) and \(\frac{\partial g}{\partial y} = -x e^{-xy}\).

Key Concepts

Chain Rule

The chain rule is an essential tool in calculus, particularly in the context of multivariable functions. It helps us differentiate composite functions, which are functions within other functions. In our exercise, we are working with the function \( g(x, y) = e^{-xy} \). Here, \(-xy\) is a function inside the exponential function \( e^u \), where \(u = -xy\).

• First, differentiate the outer function, \( e^u \), with respect to \(u\). The derivative is simply \( e^u \) itself.
• Second, differentiate the inside function \(-xy\) with respect to the variable you are interested in, i.e., \(x\) or \(y\). For \(-xy\), this means finding the derivatives \(-y\) and \(-x\) respectively.
• Finally, multiply these two derivatives together. For example, when finding \(\frac{\partial g}{\partial x}\), multiply \(e^{-xy}\) by \(-y\), resulting in \(-y e^{-xy}\).

Understanding the chain rule allows us to find these partial derivatives efficiently and is pivotal when working with more complex or nested functions.

Multivariable Calculus

Multivariable calculus extends calculus concepts to functions of multiple variables, rather than just a single variable. This branch of calculus deals with functions like \( g(x, y) = e^{-xy} \) that depend on two or more variables. Partial derivatives are a key concept here; they help us understand how changes in one variable affect the function while keeping other variables constant.

• Partial derivatives provide insight into the rate of change in multivariable functions. In our exercise, \(\frac{\partial g}{\partial x}\) shows how the function changes with \(x\) while \(y\) is constant.
• Understanding these derivatives is crucial for applications in fields such as physics, engineering, and economics, where systems often depend on several parameters.
• In practical terms, this means solving problems that involve surfaces rather than lines or curves, adding a new depth to understanding changes across dimensions.

Knowing how to work in multiple variables is the gateway to more advanced topics like gradient, divergence, and curl, which are more geometric in nature.

Differentiation

At its core, differentiation is the process of finding how a function changes as its input changes. In our problem, we deal with partial differentiation which focuses on functions with more than one variable. This is slightly different from ordinary differentiation with just one variable, as here we differentiate with respect to one variable while considering others as constants.

• To differentiate \( g(x, y) = e^{-xy} \) with respect to \(x\), treat \(y\) as constant and apply the chain rule to get \( \frac{\partial g}{\partial x} = -y e^{-xy} \).
• Similarly, when differentiating with respect to \(y\), treat \(x\) as constant. This gives us \( \frac{\partial g}{\partial y} = -x e^{-xy} \).
• This method of maintaining one variable constant is what allows us to study the "slice" or "profile" view of a complex multi-dimensional surface.

Understanding partial differentiation is fundamental, as it introduces how changes in specific directions affect the function, allowing us to analyze and predict outcomes in multi-dimensional environments effectively.