Directional Derivative Of F(x, Y) = X E^(xy) + Sin(y) Explained

Introduction

In multivariable calculus, the directional derivative is a concept that extends the idea of a derivative to more than one dimension. Specifically, it measures the rate of change of a function along a given direction. This article will walk you through the process of finding the directional derivative of the function $f(x, y) = x e^{xy} + \sin(y)$ in the direction of the unit vector $u = \left\langle-\frac{\sqrt{2}}{2}, \frac{\sqrt{2}}{2}\right\rangle$. This problem combines concepts from calculus and vector analysis, making it a valuable exercise for students and professionals alike. Let's delve into the steps required to solve this problem effectively.

Understanding the Directional Derivative

The directional derivative, denoted as $D_u f(x, y)$, quantifies how much a function $f$ changes at a specific point $(x, y)$ when moving in the direction of a unit vector $u$. Essentially, it's the slope of the function in that particular direction. To calculate it, we use the gradient of the function and the unit vector. The gradient, represented as $\nabla f(x, y)$, is a vector composed of the partial derivatives of $f$ with respect to each variable. In a two-dimensional space, the gradient is given by $\nabla f(x, y) = \left\langle \frac{\partial f}{\partial x}, \frac{\partial f}{\partial y} \right\rangle$. The directional derivative is then found by taking the dot product of the gradient and the unit vector, expressed as $D_u f(x, y) = \nabla f(x, y) \cdot u$. Understanding this foundational concept is crucial for tackling problems involving rates of change in multiple dimensions, with applications ranging from physics to economics. By mastering the directional derivative, one gains insights into how functions behave across various directions, enabling a more nuanced analysis of complex systems and models.

Step-by-Step Calculation

To find the directional derivative $D_u f(x, y)$ of the function $f(x, y) = x e^{xy} + \sin(y)$ in the direction of the unit vector $u = \left\langle-\frac{\sqrt{2}}{2}, \frac{\sqrt{2}}{2}\right\rangle$, we'll proceed through a series of steps. First, we need to compute the partial derivatives of $f$ with respect to $x$ and $y$. The partial derivative with respect to $x$, denoted as $\frac{\partial f}{\partial x}$, is found by differentiating $f$ with respect to $x$ while treating $y$ as a constant. Applying this to our function, we get $\frac{\partial f}{\partial x} = e^{xy} + xye^{xy}$. Similarly, the partial derivative with respect to $y$, denoted as $\frac{\partial f}{\partial y}$, is found by differentiating $f$ with respect to $y$ while treating $x$ as a constant. This yields $\frac{\partial f}{\partial y} = x^2 e^{xy} + \cos(y)$. These partial derivatives form the components of the gradient vector, which is a critical element in determining the directional derivative. Once we have the gradient, we can then compute the dot product with the unit vector to find the rate of change in the specified direction. The methodical approach ensures that we accurately capture the function's behavior along the given vector, providing valuable insights into its directional properties.

1. Compute Partial Derivatives

First, we need to compute the partial derivatives of $f(x, y) = x e^{xy} + \sin(y)$ with respect to $x$ and $y$. These partial derivatives will form the components of the gradient vector, which is essential for calculating the directional derivative. The process involves treating one variable as a constant while differentiating with respect to the other. This technique allows us to isolate the rate of change along each axis, providing a clear picture of how the function behaves in different directions. Mastering this step is crucial for any multivariable calculus problem, as it lays the groundwork for more advanced concepts such as optimization and tangent planes. Let's start by finding the partial derivative with respect to $x$ and then move on to the partial derivative with respect to $y$, ensuring we have a complete understanding of the function's behavior in the $xy$-plane.

Partial Derivative with Respect to x

To find the partial derivative of $f(x, y) = x e^{xy} + \sin(y)$ with respect to $x$, denoted as $\frac{\partial f}{\partial x}$, we treat $y$ as a constant and differentiate the function with respect to $x$. Applying the product rule to the term $x e^{xy}$, we get:

$$\frac{\partial}{\partial x}(x e^{xy}) = e^{xy} + x(y e^{xy}) = e^{xy} + xy e^{xy}$$

The derivative of $\sin(y)$ with respect to $x$ is 0 because $y$ is treated as a constant. Therefore,

$$\frac{\partial f}{\partial x} = e^{xy} + xy e^{xy}$$

This result indicates how the function $f$ changes as $x$ varies, keeping $y$ constant. The expression $e^{xy} + xy e^{xy}$ provides a measure of the function's sensitivity to changes in the $x$-direction at a specific point $(x, y)$. Understanding this partial derivative is crucial for computing the gradient vector and, ultimately, the directional derivative. The combination of the exponential term and the product of $x$ and $y$ in the derivative highlights the function's complex behavior in the $x$-direction, making this calculation a fundamental step in our analysis.

Partial Derivative with Respect to y

Next, we compute the partial derivative of $f(x, y) = x e^{xy} + \sin(y)$ with respect to $y$, denoted as $\frac{\partial f}{\partial y}$. In this case, we treat $x$ as a constant and differentiate the function with respect to $y$. The derivative of $x e^{xy}$ with respect to $y$ is found using the chain rule:

$$\frac{\partial}{\partial y}(x e^{xy}) = x(x e^{xy}) = x^2 e^{xy}$$

The derivative of $\sin(y)$ with respect to $y$ is $\cos(y)$. Therefore,

$$\frac{\partial f}{\partial y} = x^2 e^{xy} + \cos(y)$$

This partial derivative tells us how the function $f$ changes as $y$ varies, holding $x$ constant. The expression $x^2 e^{xy} + \cos(y)$ quantifies the function's sensitivity to changes in the $y$-direction at a specific point $(x, y)$. This calculation, along with the partial derivative with respect to $x$, provides a complete picture of the function's instantaneous rates of change along the two primary axes. The combination of exponential and trigonometric terms in the derivative highlights the interplay between different types of functions in determining the overall behavior of $f$. Understanding this partial derivative is essential for computing the gradient vector and, consequently, the directional derivative.

2. Form the Gradient Vector

Now that we have computed the partial derivatives, we can form the gradient vector $\nabla f(x, y)$. The gradient vector is a vector whose components are the partial derivatives of the function. It points in the direction of the greatest rate of increase of the function. In this case, the gradient vector is given by:

$$\nabla f(x, y) = \left\langle \frac{\partial f}{\partial x}, \frac{\partial f}{\partial y} \right\rangle = \left\langle e^{xy} + xy e^{xy}, x^2 e^{xy} + \cos(y) \right\rangle$$

The gradient vector is a crucial concept in multivariable calculus, providing a powerful tool for analyzing the behavior of functions in multiple dimensions. It not only indicates the direction of steepest ascent but also its magnitude represents the maximum rate of change at a given point. By understanding the gradient vector, we can gain insights into the function's topography, identifying critical points, saddle points, and other features that define its overall shape. The components of the gradient, which we calculated in the previous steps, are now combined to form this vector, making it a tangible representation of the function's directional tendencies. This vector will be used in the next step to compute the directional derivative, which measures the rate of change along a specific direction.

3. Compute the Dot Product

To find the directional derivative $D_u f(x, y)$, we compute the dot product of the gradient vector $\nabla f(x, y)$ and the unit vector $u$. The dot product, also known as the scalar product, is an operation that takes two vectors and returns a scalar. In this context, it projects the gradient vector onto the direction of the unit vector, giving us the rate of change of the function in that specific direction. The formula for the dot product of two vectors $\mathbf{a} = \langle a_1, a_2 \rangle$ and $\mathbf{b} = \langle b_1, b_2 \rangle$ is $\mathbf{a} \cdot \mathbf{b} = a_1b_1 + a_2b_2$. Applying this to our problem, we will multiply the corresponding components of the gradient vector and the unit vector and then sum the results. This process will yield a scalar value that represents the directional derivative, providing a precise measure of how the function changes along the given direction. Mastering the dot product is essential for understanding directional derivatives and their applications in various fields, such as physics and engineering.

Given the gradient vector $\nabla f(x, y) = \left\langle e^{xy} + xy e^{xy}, x^2 e^{xy} + \cos(y) \right\rangle$ and the unit vector $u = \left\langle-\frac{\sqrt{2}}{2}, \frac{\sqrt{2}}{2}\right\rangle$, we compute their dot product:

$$\begin{aligned} D_u f(x, y) &= \nabla f(x, y) \cdot u \\ &= \left\langle e^{xy} + xy e^{xy}, x^2 e^{xy} + \cos(y) \right\rangle \cdot \left\langle-\frac{\sqrt{2}}{2}, \frac{\sqrt{2}}{2}\right\rangle \\ &= -\frac{\sqrt{2}}{2}(e^{xy} + xy e^{xy}) + \frac{\sqrt{2}}{2}(x^2 e^{xy} + \cos(y)) \\ &= \frac{\sqrt{2}}{2}(-e^{xy} - xy e^{xy} + x^2 e^{xy} + \cos(y)) \end{aligned}$$

4. Simplify the Result

Finally, we simplify the expression obtained from the dot product to get the directional derivative $D_u f(x, y)$. Simplification involves combining like terms and expressing the result in its most concise form. This step is crucial for making the result more understandable and easier to use in further calculations or analyses. In our case, we have the expression:

$$D_u f(x, y) = \frac{\sqrt{2}}{2}(-e^{xy} - xy e^{xy} + x^2 e^{xy} + \cos(y))$$

There are no further simplifications possible in terms of combining like terms, so this is the final expression for the directional derivative. This result tells us the rate of change of the function $f(x, y)$ at a given point $(x, y)$ in the direction of the unit vector $u$. The simplified form allows us to easily evaluate the directional derivative at specific points and understand how the function's behavior varies across different locations and directions. This final step ensures that our answer is both accurate and presentable, completing the process of finding the directional derivative.

Final Answer

The directional derivative of $f(x, y) = x e^{xy} + \sin(y)$ in the direction of the unit vector $u = \left\langle-\frac{\sqrt{2}}{2}, \frac{\sqrt{2}}{2}\right\rangle$ is:

$$D_u f(x, y) = \frac{\sqrt{2}}{2}(-e^{xy} - xy e^{xy} + x^2 e^{xy} + \cos(y))$$

This final answer represents the rate of change of the function $f(x, y)$ at any point $(x, y)$ in the direction specified by the unit vector $u$. It combines the effects of the partial derivatives and the direction vector to give a precise measure of how the function's value changes as we move along that direction. The expression includes exponential and trigonometric terms, reflecting the original function's complexity and the interplay between different types of functions. This result is valuable for various applications, such as optimization problems, where we seek to find the direction of the steepest ascent or descent, and in physics, where it can describe the rate of change of a field along a particular path. Understanding and calculating directional derivatives is a fundamental skill in multivariable calculus, providing insights into the behavior of functions in multiple dimensions.

Conclusion

In conclusion, we have successfully found the directional derivative of the function $f(x, y) = x e^{xy} + \sin(y)$ in the direction of the unit vector $u = \left\langle-\frac{\sqrt{2}}{2}, \frac{\sqrt{2}}{2}\right\rangle$. The process involved computing partial derivatives, forming the gradient vector, and then taking the dot product of the gradient vector with the unit vector. The final result is:

$$D_u f(x, y) = \frac{\sqrt{2}}{2}(-e^{xy} - xy e^{xy} + x^2 e^{xy} + \cos(y))$$

This directional derivative represents the rate of change of the function in the specified direction, providing valuable insights into the function's behavior in the $xy$-plane. The step-by-step approach we followed ensures a clear and methodical solution, which is essential for understanding and applying multivariable calculus concepts. Mastering these techniques allows for a deeper analysis of functions in multiple dimensions, with applications ranging from optimization problems to physical simulations. The directional derivative is a powerful tool for understanding how functions change along specific paths, and this example demonstrates its practical application in a complex scenario. This exercise reinforces the importance of partial derivatives, gradient vectors, and the dot product in the broader context of multivariable calculus.