Chapter 12: Problem 12
Find \(f_{x}, f_{y}, f_{x x}, f_{y y}, f_{x y}\) and \(f_{y x}\). $$ f(x, y)=\frac{4}{x y} $$
Short Answer
Expert verified
\(f_x = -\frac{4}{x^2y}\), \(f_y = -\frac{4}{xy^2}\), \(f_{xx} = \frac{8}{x^3y}\), \(f_{yy} = \frac{8}{xy^3}\), \(f_{xy} = f_{yx} = \frac{4}{x^2y^2}\).
Step by step solution
01
Differentiate with respect to x
First, we differentiate the function with respect to \(x\). Given \(f(x, y) = \frac{4}{xy}\), rewrite it as \(f(x, y) = 4x^{-1}y^{-1}\). Applying the power rule, \(f_x = \frac{d}{dx}(4x^{-1}y^{-1}) = 4(-1)x^{-2}y^{-1} = -\frac{4}{x^2y}\).
02
Differentiate with respect to y
Next, differentiate the function with respect to \(y\). From the rewritten function \(f(x, y) = 4x^{-1}y^{-1}\), apply the power rule: \(f_y = \frac{d}{dy}(4x^{-1}y^{-1}) = 4x^{-1}(-1)y^{-2} = -\frac{4}{xy^2}\).
03
Second derivative with respect to x
Now, take the second derivative with respect to \(x\). From \(f_x = -\frac{4}{x^2y}\), differentiate again with respect to \(x\): \(f_{xx} = \frac{d}{dx}(-\frac{4}{x^2y}) = 8x^{-3}y^{-1} = \frac{8}{x^3y}\).
04
Second derivative with respect to y
Take the second derivative with respect to \(y\). From \(f_y = -\frac{4}{xy^2}\), differentiate again with respect to \(y\): \(f_{yy} = \frac{d}{dy}(-\frac{4}{xy^2}) = 8x^{-1}y^{-3} = \frac{8}{xy^3}\).
05
Mixed partial derivative \(f_{xy}\)
Now, differentiate \(f_x\) with respect to \(y\). We have \(f_x = -\frac{4}{x^2y}\), so take the derivative with respect to \(y\): \(f_{xy} = \frac{d}{dy}(-\frac{4}{x^2y}) = 4x^{-2}y^{-2} = \frac{4}{x^2y^2}\).
06
Mixed partial derivative \(f_{yx}\)
Differentiate \(f_y\) with respect to \(x\). Starting from \(f_y = -\frac{4}{xy^2}\), take the derivative with respect to \(x\): \(f_{yx} = \frac{d}{dx}(-\frac{4}{xy^2}) = 4x^{-2}y^{-2} = \frac{4}{x^2y^2}\).
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Power Rule
The power rule is a fundamental tool in calculus, especially when you differentiate functions.Its simplicity and versatility make it a helpful shortcut for finding derivatives of functions with exponents.When you have a function of the form \( f(x) = ax^n \), where \( a \) and \( n \) are constants, the derivative is \( f'(x) = anx^{n-1} \).
For the given function \( f(x, y) = \frac{4}{xy} = 4x^{-1}y^{-1} \), we apply the power rule to both \( x \) and \( y \).This allows us to find the partial derivatives \( f_x \) and \( f_y \), by treating one variable at a time.Remember:- Differentiate with respect to one variable while considering the others constant.- Power rule helps simplify terms like \( x^{-1} \) into more manageable integers for differentiation.
- This rule is applicable to each variable individually.
- When differentiating with respect to a specific variable, treat the other variables as constants.
For the given function \( f(x, y) = \frac{4}{xy} = 4x^{-1}y^{-1} \), we apply the power rule to both \( x \) and \( y \).This allows us to find the partial derivatives \( f_x \) and \( f_y \), by treating one variable at a time.Remember:- Differentiate with respect to one variable while considering the others constant.- Power rule helps simplify terms like \( x^{-1} \) into more manageable integers for differentiation.
Mixed Partial Derivatives
Mixed partial derivatives involve taking derivatives with respect to different variables successively.They help us understand how a function changes when two different variables change simultaneously.Partial derivatives are taken "mixedly" when you alternate differentiation between variables.
In the original exercise, we calculated \( f_{xy} \) and \( f_{yx} \) as mixed partials.
In the original exercise, we calculated \( f_{xy} \) and \( f_{yx} \) as mixed partials.
- First, compute \( f_x \) and then differentiate it with respect to \( y \).
- Start with \( f_y \) and differentiate again, but this time with respect to \( x \).
Second Derivative
The second derivative provides insight into the curvature or concavity of a function.It represents how the rate of change itself is changing.For single-variable calculus, the second derivative can indicate points of inflection, concavity up, or down.In multivariable functions, they explain the behavior along different axes.
- Analogous to the first derivative, you take the second derivative by differentiating the first derivative again.
- For the original function, the second derivatives \( f_{xx} \) and \( f_{yy} \) were derived by differentiating \( f_x \) and \( f_y \) respectively.