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Magazine Chapter 14: Problem 43
Find all the second-order partial derivatives of the functions in Exercises \(41-50 .\) $$g(x, y)=x^{2} y+\cos y+y \sin x$$
Short Answer
Expert verified
The second-order partial derivatives are: \( \frac{\partial^2 g}{\partial x^2} = 2y - y\sin(x) \), \( \frac{\partial^2 g}{\partial y^2} = -\cos(y) \), and mixed derivatives \( \frac{\partial^2 g}{\partial y \partial x} = \frac{\partial^2 g}{\partial x \partial y} = 2x + \cos(x) \).
Step by step solution
01
Find the First-Order Partial Derivative with respect to x
To find the first partial derivative of \( g(x, y) \) with respect to \( x \), apply the rules of differentiation to each term of the function where \( x \) appears:- For \( x^2y \), the derivative with respect to \( x \) is \( 2xy \), because \( y \) is treated as a constant.- \( \cos(y) \) does not depend on \( x \), so its derivative is \( 0 \).- For \( y \sin(x) \), the derivative with respect to \( x \) is \( y \cos(x) \), treating \( y \) as a constant.Thus, the partial derivative \( \frac{\partial g}{\partial x} = 2xy + y\cos(x) \).
02
Find the First-Order Partial Derivative with respect to y
Next, find the partial derivative with respect to \( y \):- For \( x^2y \), the derivative with respect to \( y \) is \( x^2 \), treating \( x \) as a constant.- For \( \cos(y) \), the derivative is \( -\sin(y) \).- For \( y \sin(x) \), the derivative with respect to \( y \) is \( \sin(x) \), treating \( x \) as a constant.Thus, the partial derivative \( \frac{\partial g}{\partial y} = x^2 - \sin(y) + \sin(x) \).
03
Find the Second-Order Partial Derivative with respect to x twice
Differentiating \( \frac{\partial g}{\partial x} = 2xy + y\cos(x) \) with respect to \( x \):- For \( 2xy \), the derivative with respect to \( x \) is \( 2y \).- For \( y\cos(x) \), the derivative with respect to \( x \) is \( -y\sin(x) \).Thus, the second-order partial derivative \( \frac{\partial^2 g}{\partial x^2} = 2y - y\sin(x) \).
04
Find the Second-Order Partial Derivative with respect to y twice
Differentiating \( \frac{\partial g}{\partial y} = x^2 - \sin(y) + \sin(x) \) with respect to \( y \):- For \( x^2 \), the derivative with respect to \( y \) is \( 0 \).- For \( -\sin(y) \), the derivative is \( -\cos(y) \).- For \( \sin(x) \), the derivative with respect to \( y \) is \( 0 \).Thus, the second-order partial derivative \( \frac{\partial^2 g}{\partial y^2} = -\cos(y) \).
05
Find the Mixed Partial Derivative with respect to x then y
Differentiating \( \frac{\partial g}{\partial x} = 2xy + y\cos(x) \) with respect to \( y \):- For \( 2xy \), the derivative with respect to \( y \) is \( 2x \).- For \( y\cos(x) \), the derivative with respect to \( y \) is \( \cos(x) \), treating \( x \) as a constant.Thus, the mixed partial derivative \( \frac{\partial^2 g}{\partial y \partial x} = 2x + \cos(x) \).
06
Find the Mixed Partial Derivative with respect to y then x
Now, differentiate \( \frac{\partial g}{\partial y} = x^2 - \sin(y) + \sin(x) \) with respect to \( x \):- For \( x^2 \), the derivative with respect to \( x \) is \( 2x \).- \( -\sin(y) \) does not depend on \( x \), so its derivative is \( 0 \).- For \( \sin(x) \), the derivative with respect to \( x \) is \( \cos(x) \).Thus, the mixed partial derivative \( \frac{\partial^2 g}{\partial x \partial y} = 2x + \cos(x) \).
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
First-Order Partial Derivatives
First-order partial derivatives help us understand how a function changes as we vary one of its variables while holding the others constant. For a function of two variables, like \( g(x, y) = x^2 y + \cos y + y \sin x \), these derivatives tell us the rate at which the function changes in the direction of a single variable.
- To find the partial derivative of \( g \) with respect to \( x \), we differentiate each term of the function considering \( y \) as a constant. Here, the derivative becomes \( \frac{\partial g}{\partial x} = 2xy + y\cos(x) \). Note that terms that do not include \( x \) disappear because their derivative is zero with respect to \( x \).
- Similarly, for the derivative of \( g \) with respect to \( y \), treat \( x \) as a constant. The resulting expression is \( \frac{\partial g}{\partial y} = x^2 - \sin(y) + \sin(x) \).
Second-Order Partial Derivatives
Second-order partial derivatives are simply derivatives of the first-order partial derivatives. They provide insight into the curvature or concavity of the function in a given direction.
- For instance, when we differentiate \( \frac{\partial g}{\partial x} \) with respect to \( x \) again, which means taking the second derivative of \( g \) with respect to \( x \), we calculate \( \frac{\partial^2 g}{\partial x^2} = 2y - y\sin(x) \). This helps us understand how quickly the rate of change in the \( x \) direction is itself changing.
- Likewise, differentiating \( \frac{\partial g}{\partial y} \) with respect to \( y \) gives \( \frac{\partial^2 g}{\partial y^2} = -\cos(y) \). This tells us about the acceleration or concavity of the function along the \( y \) axis.
Mixed Partial Derivatives
Mixed partial derivatives involve differentiating with respect to more than one variable, taken in different orders. They reveal interactions between different variables of a function.
- For \( g(x, y) = x^2 y + \cos y + y \sin x \), finding \( \frac{\partial^2 g}{\partial y \partial x} \) involves taking \( \frac{\partial g}{\partial x} \) and differentiating it with respect to \( y \). This yields \( \frac{\partial^2 g}{\partial y \partial x} = 2x + \cos(x) \).
- Similarly, \( \frac{\partial^2 g}{\partial x \partial y} \) means differentiating \( \frac{\partial g}{\partial y} \) with respect to \( x \), also resulting in \( 2x + \cos(x) \).
