Question

Find the first partial derivatives with respect to \(x\) and with respect to \(y\). $$ z=x e^{x+y} $$

Short answer

The first partial derivative of \(z = xe^{x+y}\) with respect to \(x\) is \(e^{x+y} + x(e^{x+y})\) and with respect to \(y\) is \(x(e^{x+y})\).

Step-by-step solution

Find the First Partial Derivative with respect to \(x\)

We have \(z = xe^{x+y}\) and need to find \(\frac{\partial z}{\partial x}\). We have a product of two functions of \(x\), \(x\) and \(e^{x + y}\). Using the product rule, we get \(\frac{\partial z}{\partial x} = e^{x+y} + x \frac{\partial}{\partial x} (e^{x+y})\). Now, applying the chain rule to \(e^{x+y}\), we get\(\frac{\partial z}{\partial x} = e^{x+y} + x(e^{x+y})\)

Find the First Partial Derivative with respect to \(y\)

Next, we need to find \(\frac{\partial z}{\partial y}\). Here, we treat \(x\) as a constant. So, \(\frac{\partial z}{\partial y} = x \frac{\partial}{\partial y}( e^{x+y})\). Using the chain rule, we find that \(\frac{\partial z}{\partial y} = x(e^{x+y})\)

Key concepts

Product Rule

The product rule is an essential tool in calculus for finding derivatives when dealing with functions that are products of two or more functions. In essence, it helps us differentiate expressions like the one found in this exercise, where the function \(z\) is composed of two parts: \(x\) and \(e^{x+y}\). To apply the product rule, you use the formula:

• If \(u(x)\) and \(v(x)\) are functions of \(x\), then the derivative of their product is given by \(\frac{d}{dx}[u(x)v(x)] = u'(x)v(x) + u(x)v'(x)\).

In the context of partial derivatives, this rule is adjusted slightly. When calculating \(\frac{\partial z}{\partial x}\), consider \(x\) as a function and \(e^{x+y}\) as another function of \(x\). Hence, the partial derivative of \(z\) with respect to \(x\) involves differentiation of both parts:

• Differentiate \(x\) while keeping \(e^{x+y}\) unchanged.
• Keep \(x\) unchanged and differentiate \(e^{x+y}\) with respect to \(x\).

By applying this method to the problem, you correctly obtained the partial derivative of the function \(z = x e^{x+y}\) with respect to \(x\).

Chain Rule

The chain rule is another cornerstone of calculus, particularly useful when differentiating compositions of functions. It allows us to find the derivative of a function relative to an inner dependent variable. In this exercise, the function \(e^{x+y}\) acts as a composition where \(y\) is inside the exponential function. The chain rule states that:

• If \(g(u)\) is a function and \(u = u(x)\) is another function, then the derivative of \(g(u(x))\) is \(\frac{dg}{du} \cdot \frac{du}{dx}\).

In the partial derivative context, when finding \(\frac{\partial}{\partial y}( e^{x+y})\), treat \(x+y\) as a composition of functions. Here:

• Differentiating \(e^{x+y}\) with respect to \(y\) involves applying the chain rule: the derivative of \(e^{x+y}\) with respect to \(x+y\) is \(e^{x+y}\), and then multiply by the derivative of \(x+y\) with respect to \(y\), which is \(1\).

This explains how the derivative simplifies to \(x(e^{x+y})\), as shown in the solution.

Multivariable Calculus

Multivariable calculus extends the concepts of calculus to functions with more than one variable. The importance of multivariable calculus is profound as it deals with functions that naturally depend on several inputs, such as the function \(z = x e^{x+y}\) used in this exercise. In this realm, partial derivatives come into play as they allow us to understand how changes in each variable individually affect the function's outcome.When finding a partial derivative, it's crucial to hold all other variables constant, similar to analyzing the function's behavior in a single-variable calculus context. This approach gives rise to differentiation with respect to one variable at a time. For example:

• When computing \(\frac{\partial z}{\partial x}\), consider \(y\) as a constant.
• Conversely, for \(\frac{\partial z}{\partial y}\), take \(x\) as a constant.

Multivariable calculus not only includes differentiation but also integration over spaces of higher dimensions, which enables solving more complex, real-world problems that involve multiple variables. This exercise illustrates the differentiation aspect, laying the foundation for continuous exploration of multivariable functions.