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Find all four second-order partial derivatives of the given function \(f\). $$ f(x, y)=\sin x \cos y $$

Short Answer

Expert verified
The second-order partial derivatives are: \(f_{xx} = -\sin x \cos y\), \(f_{yy} = -\sin x \cos y\), and \(f_{xy} = f_{yx} = -\cos x \sin y\).

Step by step solution

01

Find the first-order partial derivative with respect to x

To find the partial derivative of the function with respect to \(x\), treat \(y\) as a constant. Differentiate \(f(x, y) = \sin x \cos y\) with respect to \(x\):\[f_x(x, y) = \cos x \cdot \cos y\]
02

Find the first-order partial derivative with respect to y

Now find the partial derivative of the function with respect to \(y\), treating \(x\) as a constant. Differentiate \(f(x, y) = \sin x \cos y\) with respect to \(y\):\[f_y(x, y) = -\sin x \cdot \sin y\]
03

Find the second-order partial derivative with respect to x twice

Differentiate \(f_x(x, y) = \cos x \cdot \cos y\) with respect to \(x\) again to find \(f_{xx}\):\[f_{xx}(x, y) = -\sin x \cdot \cos y\]
04

Find the second-order partial derivative with respect to y twice

Differentiate \(f_y(x, y) = -\sin x \cdot \sin y\) with respect to \(y\) again to find \(f_{yy}\):\[f_{yy}(x, y) = -\sin x \cdot \cos y\]
05

Find the second-order mixed partial derivative f_{xy}

Differentiate \(f_x(x, y) = \cos x \cdot \cos y\) with respect to \(y\) to find \(f_{xy}\):\[f_{xy}(x, y) = -\cos x \cdot \sin y\]
06

Confirm the equality of mixed partial derivatives f_{xy} and f_{yx}

Differentiate \(f_y(x, y) = -\sin x \cdot \sin y\) with respect to \(x\) to find \(f_{yx}\):\[f_{yx}(x, y) = -\cos x \cdot \sin y\] The mixed partial derivatives \(f_{xy}\) and \(f_{yx}\) are equal, confirming they are legitimate derivatives.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

First-Order Partial Derivatives
Understanding first-order partial derivatives is the first step towards grasping more complex derivative concepts like second-order and mixed partial derivatives. A first-order partial derivative measures how a function changes as one variable changes while keeping the other variables constant. This is particularly useful when dealing with functions of multiple variables, as it helps us understand the behavior in a specific direction.

For the function \( f(x, y) = \sin x \cos y \), we calculated the partial derivatives with respect to both \( x \) and \( y \).
  • The partial derivative with respect to \( x \), \( f_x(x, y) = \cos x \cdot \cos y \), tells us how the function changes as \( x \) changes, while keeping \( y \) constant.
  • The partial derivative with respect to \( y \), \( f_y(x, y) = -\sin x \cdot \sin y \), shows how the function changes as \( y \) changes, with \( x \) remaining constant.
This concept plays a vital role in fields such as physics, engineering, and economics to model changes in multi-variable scenarios.
Mixed Partial Derivatives
Mixed partial derivatives emerge when a function of multiple variables is differentiated with respect to two different variables. It's an important concept indicating interactions between the variables. For a function \( f(x, y) \), mixed partial derivatives are represented as \( f_{xy} \) or \( f_{yx} \). These denote the partial derivative of \( f_x \) with respect to \( y \), and \( f_y \) with respect to \( x \), respectively.

In calculating the mixed partial derivatives of our function:
  • We found \( f_{xy}(x, y) = -\cos x \cdot \sin y \) by taking the derivative of \( f_x(x, y) \) with respect to \( y \).
  • We ensured consistency by also calculating \( f_{yx}(x, y) = -\cos x \cdot \sin y \) by differentiating \( f_y(x, y) \) with respect to \( x \).
In most cases, \( f_{xy} \) equals \( f_{yx} \), confirming the equality of mixed partial derivatives and reflecting a symmetry property important in mathematical analysis.
Function Differentiation
Function differentiation is the cornerstone of calculus, extending to multiple variables with partial derivatives. When dealing with functions involving several variables, differentiation provides insights into how small changes in one or more variables impact the entire function.

As illustrated by differentiating \( f(x, y) = \sin x \cos y \), we extract vital information about the rate of change individually for each variable through first-order partial derivatives. Moving from first-order to second-order derivatives allows us to explore deeper levels of curvature and interaction between variables.

For instance:
  • Second-order derivatives \( f_{xx}(x, y) = -\sin x \cdot \cos y \) and \( f_{yy}(x, y) = -\sin x \cdot \cos y \) provide insights into how the graph of \( f \) bends or curves, similar to how second derivatives work for functions of a single variable.
  • Mixed partial derivatives extend this further, revealing interaction effects, such as symmetry (equality) between them.
This comprehensive toolset is essential for understanding complex systems and continuously varying phenomena.

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Most popular questions from this chapter

Complete the description of the behavior of the second-order approximation near a critical point a by considering the case that \(\operatorname{det} H(\mathbf{a}) \neq 0\) and \(A=0\), where \(H(\mathbf{a})=\left[\begin{array}{ll}A & B \\ B & C\end{array}\right]\). (a) Show that \(\operatorname{det} H(\mathbf{a})<0\). (b) Show that the quadratic term \(A h^{2}+2 B h k+C k^{2}=2 B h k+C k^{2}\) of the second-order approximation can be written as a difference of two perfect squares.

Consider the problem of finding the maximum and minimum values of the function \(f(x, y)=\) \(x y\) subject to the constraint \(x+2 y=1\) (a) Explain why a minimum value does not exist. (b) On the other hand, you may assume that a maximum does exist. Use Lagrange multipliers to find the maximum value and the point at which it is attained.

Let a be a point of \(\mathbb{R}^{n}\), and let \(B=B(\mathbf{a}, r)\) be an open ball centered at a. Let \(f: B \rightarrow \mathbb{R}\) be a differentiable function. If \(\mathrm{b}\) is any point of \(B,\) show that there exists a point \(\mathbf{c}\) on the line segment connecting a and \(\mathbf{b}\) such that: $$ f(\mathbf{b})-f(\mathbf{a})=\nabla f(\mathbf{c}) \cdot(\mathbf{b}-\mathbf{a}) $$ Note that this is a generalization of the mean value theorem to real-valued functions of more than one variable.

Find all values of \(c\) such that, at every point of intersection of the spheres $$ (x-c)^{2}+y^{2}+z^{2}=3 \quad \text { and } \quad x^{2}+(y-1)^{2}+z^{2}=1 $$ the respective tangent planes are perpendicular to one another.

Consider the function \(f: \mathbb{R}^{2} \rightarrow \mathbb{R}\) defined by: $$ f(x, y)=\left\\{\begin{array}{ll} \frac{x^{3}+2 y^{3}}{x^{2}+y^{2}} & \text { if }(x, y) \neq(0,0), \\ 0 & \text { if }(x, y)=(0,0) . \end{array}\right. $$ (a) Find the values of the partial derivatives \(\frac{\partial f}{\partial x}(0,0)\) and \(\frac{\partial f}{\partial y}(0,0)\). (This shouldn't require much calculation.) (b) Use the definition of differentiability to determine whether \(f\) is differentiable at (0,0) .

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