Chapter 13: Problem 85
Confirm that the mixed second-order partial derivatives of \(f\) are the same. $$ f(x, y)=4 x^{2}-8 x y^{4}+7 y^{5}-3 $$
Short Answer
Expert verified
The mixed partial derivatives \( f_{xy} \) and \( f_{yx} \) are both \(-32y^3\), therefore they are equal.
Step by step solution
01
Find the first partial derivative with respect to x
To find the first partial derivative of \( f(x, y) = 4x^2 - 8xy^4 + 7y^5 - 3 \) with respect to \( x \), treat \( y \) as a constant and differentiate with respect to \( x \). The derivative is obtained by applying the power rule and terms involving only \( y \) become zero because they become constants. Therefore, \[ f_x(x, y) = \frac{\partial}{\partial x}(4x^2 - 8xy^4 + 7y^5 - 3) = 8x - 8y^4 \]
02
Find the first partial derivative with respect to y
Now, find the first partial derivative of \( f(x, y) = 4x^2 - 8xy^4 + 7y^5 - 3 \) with respect to \( y \), treating \( x \) as a constant. Differentiate accordingly:\[ f_y(x, y) = \frac{\partial}{\partial y}(4x^2 - 8xy^4 + 7y^5 - 3) = -32xy^3 + 35y^4 \]
03
Find the second mixed partial derivative \(f_{xy}\)
The next step is to find the second mixed partial derivative by first differentiating \( f_x(x, y) = 8x - 8y^4 \) with respect to \( y \):\[ f_{xy} = \frac{\partial}{\partial y}(8x - 8y^4) = -32y^3 \]
04
Find the second mixed partial derivative \(f_{yx}\)
Now, find the second mixed partial derivative by differentiating \( f_y(x, y) = -32xy^3 + 35y^4 \) with respect to \( x \):\[ f_{yx} = \frac{\partial}{\partial x}(-32xy^3 + 35y^4) = -32y^3 \]
05
Compare the mixed partial derivatives
Finally, compare the mixed partial derivatives found in Steps 3 and 4. We have \[ f_{xy} = -32y^3 \] and \[ f_{yx} = -32y^3 \]. Since both derivatives are equal, this confirms that the mixed second-order partial derivatives of \( f \) are the same.
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Mixed Second-Order Derivatives
When dealing with functions of multiple variables, mixed second-order derivatives play a crucial role. They involve taking the derivative of a function first with respect to one variable, and then with the second. For instance, to find \( f_{xy} \), we first differentiate with respect to \( x \) and then with respect to \( y \). The beauty of mixed second-order derivatives lies in Clairaut's Theorem, which states that if the mixed partial derivatives are continuous, they will be equal: \( f_{xy} = f_{yx} \). This theorem simplifies analysis, verifying symmetry and ensuring consistency in mathematical models. Thus, confirming that these derivatives are equal provides assurance in the calculations.
Mixed partial derivatives are a fundamental concept in calculus, particularly in multivariable calculus, as they help in understanding how functions change when two variables vary simultaneously.
Mixed partial derivatives are a fundamental concept in calculus, particularly in multivariable calculus, as they help in understanding how functions change when two variables vary simultaneously.
Power Rule
The power rule is a basic principle for differentiation that simplifies the process of finding derivatives. It states that the derivative of a term \( x^n \) is \( nx^{n-1} \). This principle is utilized to derive both single-variable and partial derivatives. For instance, when differentiating \( 4x^2 \) with respect to \( x \), applying the power rule results in \( 8x \). To correctly use the power rule, identify the exponent of the variable in each term:
- If \( n = 2 \), as in \( x^2 \), then the derivative is \( 2x^{2-1} = 2x \).
- For a negative or fractional exponent, apply the same rule.
Differentiation with Respect to Variables
In multivariable calculus, differentiating a function with respect to one of its variables involves treating other variables as constants. This process highlights the unique role each variable plays. For example, when calculating \( \frac{\partial}{\partial x} \) for \( f(x, y) = 4x^2 - 8xy^4 + 7y^5 - 3 \), the variable \( y \) is treated as a constant:
- The term \( 4x^2 \) becomes \( 8x \) after differentiation with respect to \( x \).
- The term \( -8xy^4 \) simplifies to \( -8y^4 \), treating \( y^4 \) as a constant.