Problem 11 - APEX Calculus

Question

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

Step-by-Step Solution

Answer

\(f_x = \frac{1}{y}\), \(f_y = -\frac{x}{y^2}\), \(f_{xx} = 0\), \(f_{yy} = \frac{2x}{y^3}\), \(f_{xy}= f_{yx} = -\frac{1}{y^2}\).

Step 1: Differentiate with respect to x

We need to find the partial derivative of the function with respect to \(x\). The derivative of \(x/y\) with respect to \(x\) is \(1/y\). Thus, \( f_x = \frac{1}{y} \).

Step 2: Differentiate with respect to y

Next, find the partial derivative with respect to \(y\). The function \(x/y\) can be seen as \(x \cdot y^{-1}\). Using the chain rule, \( f_y = -\frac{x}{y^2} \).

Step 3: Second partial derivative with respect to x, twice

To find \( f_{xx} \), we differentiate \(f_x = \frac{1}{y}\) with respect to \(x\). Since \(\frac{1}{y}\) is constant with respect to \(x\), \( f_{xx} = 0 \).

Step 4: Second partial derivative with respect to y, twice

To find \( f_{yy} \), we differentiate \(f_y = -\frac{x}{y^2}\) with respect to \(y\). Using the quotient rule again, \( f_{yy} = \frac{2x}{y^3} \).

Step 5: Differentiate first with respect to x, then with y

To find \( f_{xy} \), we differentiate \(f_x = \frac{1}{y}\) with respect to \(y\). Differentiating \(\frac{1}{y}\) with respect to \(y\) gives \(-\frac{1}{y^2}\), thus \( f_{xy} = -\frac{1}{y^2} \).

Step 6: Differentiate first with respect to y, then with x

To find \( f_{yx} \), we differentiate \(f_y = -\frac{x}{y^2}\) with respect to \(x\). Since \(x\) differentiates to \(1\), \( f_{yx} = -\frac{1}{y^2} \).

Key Concepts

Multivariable Calculus

In multivariable calculus, we explore functions involving more than one variable. Imagine a surface in space, which can be described by a function of two variables, say, \( f(x, y) \). Such functions lead us to consider how changes in each individual variable affect the output or the value of the function.
Partial derivatives are central in this area. They allow us to examine the rate of change of a multivariable function with respect to one variable, while keeping the others constant. This insight is not only crucial in mathematics but also finds applications in physics, engineering, and economics.
In our example, \( f(x, y) = \frac{x}{y} \), partial derivatives \( f_x \) and \( f_y \) give us the rate of change of the function with respect to \( x \) and \( y \), respectively.
This understanding sets the stage for calculating more complex derivatives and analyzing the behavior of multivariable functions in various contexts.

Chain Rule

The chain rule is a fundamental concept in calculus, playing a key role when we need to differentiate composite functions. In the context of partial derivatives, it helps us handle situations where one of the variables itself is a function of yet another variable.
When differentiating \( f(x, y) = \frac{x}{y} \) with respect to \( y \), the function can be rewritten in a form that involves an exponent: \( x \cdot y^{-1} \).

• The chain rule tells us that the derivative of \( y^{-1} \) with respect to \( y \) is \(-y^{-2} \).
• Hence, when we multiply by the constant \( x \), we arrive at the partial derivative \( f_y = -\frac{x}{y^2} \).

The chain rule thus simplifies the process of finding derivatives, especially when dealing with functions of multiple variables and complicated expressions.

Quotient Rule

The quotient rule is another vital calculus technique that's employed whenever we differentiate functions expressed as a quotient. It is particularly useful when the numerator and the denominator are both differentiable functions.
Here, the function \( f(x, y) = \frac{x}{y} \) requires the quotient rule when calculating the second partial derivative \( f_{yy} \).
If using this rule, remember:

• The rule states that if \( u \) and \( v \) are functions of \( x \), then the derivative of \( \frac{u}{v} \) is given by \( \frac{v\, u' - u\, v'}{v^2} \).
• When we apply the quotient rule to find \( f_{yy} \), we compute \( f_y \) and again differentiate with respect to \( y \), resulting in \( \frac{2x}{y^3} \).

The quotient rule can be intricate, but breaking it down into simple steps ensures accurate differentiation.

Second Partial Derivatives

Second partial derivatives represent the derivative of a derivative, providing deeper insights into the curvature and shape of a multivariable function. These derivatives measure how the rate of change itself is changing.
In the given exercise, we calculated several second partial derivatives: \( f_{xx} \), \( f_{yy} \), \( f_{xy} \), and \( f_{yx} \).

• \( f_{xx} \) results in \( 0 \) because \( f_x \) is constant with respect to \( x \).
• The derivative \( f_{yy} \) is shown as \( \frac{2x}{y^3} \), revealing how \( f_y \) changes with respect to \( y \).
• For \( f_{xy} \) and \( f_{yx} \), the cross partial derivatives equal \(-\frac{1}{y^2} \), indicating that mixed partial derivatives are often equal due to Clairaut's theorem.

These derivatives help elucidate the behavior of the function's surface, helping us understand phenomena like concavity, saddle points, and more in multivariable calculus.