Problem 12 - APEX Calculus

Question

Find \(f_{x}, f_{y}, f_{x x}, f_{y y}, f_{x y}\) and \(f_{y x}\). $$ f(x, y)=\frac{4}{x y} $$

Step-by-Step Solution

Answer

\(f_x = -\frac{4}{x^2y}\), \(f_y = -\frac{4}{xy^2}\), \(f_{xx} = \frac{8}{x^3y}\), \(f_{yy} = \frac{8}{xy^3}\), \(f_{xy} = f_{yx} = \frac{4}{x^2y^2}\).

Step 1: 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, \(f_x = \frac{d}{dx}(4x^{-1}y^{-1}) = 4(-1)x^{-2}y^{-1} = -\frac{4}{x^2y}\).

Step 2: 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: \(f_y = \frac{d}{dy}(4x^{-1}y^{-1}) = 4x^{-1}(-1)y^{-2} = -\frac{4}{xy^2}\).

Step 3: 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 \(x\): \(f_{xx} = \frac{d}{dx}(-\frac{4}{x^2y}) = 8x^{-3}y^{-1} = \frac{8}{x^3y}\).

Step 4: 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 \(y\): \(f_{yy} = \frac{d}{dy}(-\frac{4}{xy^2}) = 8x^{-1}y^{-3} = \frac{8}{xy^3}\).

Step 5: 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 \(y\): \(f_{xy} = \frac{d}{dy}(-\frac{4}{x^2y}) = 4x^{-2}y^{-2} = \frac{4}{x^2y^2}\).

Step 6: 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 \(x\): \(f_{yx} = \frac{d}{dx}(-\frac{4}{xy^2}) = 4x^{-2}y^{-2} = \frac{4}{x^2y^2}\).

Key Concepts

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 \( f'(x) = anx^{n-1} \).

• 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.

• 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 \( x \).

By the Clairaut's theorem, for functions that are continuous and have continuous derivatives, \( f_{xy} \) equals \( f_{yx} \). In our function, both mixed derivatives turn out to be \( \frac{4}{x^2y^2} \).This symmetry helps reinforce the consistency of mixed partials in well-behaved functions.

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.

In our example function, where \( f(x, y) = \frac{4}{xy} \), the second partial derivatives are calculated independently:- \( f_{xx} = \frac{8}{x^3y} \)- \( f_{yy} = \frac{8}{xy^3} \)These show how the function bends differently with respect to each individual variable.Pay close attention to these derivatives in optimization problems as they can indicate local maxima, minima, or saddle points.