Problem 20 - APEX Calculus

Question

Find \(f_{x}, f_{y}, f_{x x}, f_{y y}, f_{x y}\) and \(f_{y x}\). $$ f(x, y)=(2 x+5 y) \sqrt{y} $$

Step-by-Step Solution

Answer

\(f_x = 2\sqrt{y}, f_y = \frac{x}{\sqrt{y}} + \frac{15}{2}\sqrt{y}, f_{xx} = 0, f_{yy} = -\frac{x}{2y^{3/2}} + \frac{15}{4y^{1/2}}, f_{xy} = f_{yx} = \frac{1}{\sqrt{y}}.\)

Step 1: Find the First Partial Derivative with respect to x

To find \( f_x \), we need to differentiate the function \( f(x, y) = (2x + 5y)\sqrt{y} \) with respect to \( x \), treating \( y \) as constant. Since \( 5y \sqrt{y} \) is constant with respect to \( x \), the derivative simplifies to:\[ f_x = \frac{\partial}{\partial x} \left((2x + 5y)\sqrt{y} \right) = 2\sqrt{y}\]

Step 2: Find the First Partial Derivative with respect to y

To find \( f_y \), differentiate \( f(x, y) = (2x + 5y)\sqrt{y} \) with respect to \( y \). We use the product rule because both terms are functions of \( y \):\[f_y = \frac{\partial}{\partial y} \left( (2x + 5y)\sqrt{y} \right) = 2x \cdot \frac{1}{2\sqrt{y}} + 5\sqrt{y} + 5y \cdot \frac{1}{2\sqrt{y}} \]Simplifying:\[ f_y = \frac{x}{\sqrt{y}} + \frac{5y}{2\sqrt{y}} + 5\sqrt{y} = \frac{x}{\sqrt{y}} + \frac{5}{2}\sqrt{y} + 5\sqrt{y} \]Combine like terms:\[ f_y = \frac{x}{\sqrt{y}} + \frac{15}{2}\sqrt{y} \]

Step 3: Find the Second Partial Derivative with Respect to x Twice

Since \( f_x = 2\sqrt{y} \), differentiate \( f_x \) again with respect to \( x \) to find \( f_{xx} \):\[f_{xx} = \frac{\partial}{\partial x} (2\sqrt{y}) = 0\]This is because \( 2\sqrt{y} \) does not depend on \( x \).

Step 4: Find the Second Partial Derivative with Respect to y Twice

Differentiate \( f_y = \frac{x}{\sqrt{y}} + \frac{15}{2}\sqrt{y} \) with respect to \( y \) to find \( f_{yy} \). Apply the power rule:\[f_{yy} = \frac{\partial}{\partial y} \left( \frac{x}{\sqrt{y}} + \frac{15}{2}\sqrt{y} \right)\]Calculating:\[f_{yy} = \frac{\partial}{\partial y} \left( x y^{-1/2} + \frac{15}{2} y^{1/2} \right)\]\[f_{yy} = x \left(-\frac{1}{2} y^{-3/2} \right) + \frac{15}{2} \cdot \frac{1}{2} y^{-1/2}\]Simplify:\[f_{yy} = -\frac{x}{2y^{3/2}} + \frac{15}{4y^{1/2}} \]

Step 5: Find the Mixed Partial Derivative with Respect to x then y

To find \( f_{xy} \), differentiate \( f_x = 2\sqrt{y} \) with respect to \( y \).\[f_{xy} = \frac{\partial}{\partial y} (2\sqrt{y}) = 2 \cdot \frac{1}{2\sqrt{y}} = \frac{1}{\sqrt{y}}\]

Step 6: Find the Mixed Partial Derivative with Respect to y then x

To find \( f_{yx} \), differentiate \( f_y = \frac{x}{\sqrt{y}} + \frac{15}{2}\sqrt{y} \) with respect to \( x \).\[f_{yx} = \frac{\partial}{\partial x} \left( \frac{x}{\sqrt{y}} + \frac{15}{2}\sqrt{y} \right) = \frac{1}{\sqrt{y}}\]Since the mixed partial derivatives \( f_{xy} \) and \( f_{yx} \) are equal, this is consistent with Clairaut's theorem for continuous functions.

Key Concepts

First Partial Derivative

The first partial derivative is the process of differentiating a multivariable function with respect to one variable while keeping the others constant. It measures how the function changes as one variable changes, similar to finding the slope of the function along that variable's axis. In the exercise, we aim to find first partial derivatives of the function \( f(x, y) = (2x + 5y)\sqrt{y} \).

• For \( f_x \): We differentiate with respect to \( x \). Here, \( y \) is treated as a constant. This means we only derive the term \( 2x \) since everything else is a constant with respect to \( x \). The result is \( f_x = 2\sqrt{y} \).
• For \( f_y \): We differentiate with respect to \( y \). This case involves the product rule because \( (2x + 5y)\sqrt{y} \) is a product of two y-dependent components: \( 2x + 5y \) and \( \sqrt{y} \). Using the product rule, we further simplify to get \( f_y = \frac{x}{\sqrt{y}} + \frac{15}{2}\sqrt{y} \).

These derivatives give us insight into the function’s behavior along each variable's directions.

Second Partial Derivative

The second partial derivative involves differentiating a first partial derivative. This can be with respect to the same variable twice or two different variables.

• \( f_{xx} \): Differentiate \( f_x = 2\sqrt{y} \) again with respect to \( x \). Since \( y \) acts as a constant and \( 2\sqrt{y} \) is independent of \( x \), \( f_{xx} = 0 \). This indicates that \( f_x \) doesn't change with \( x \).
• \( f_{yy} \): Differentiate \( f_y = \frac{x}{\sqrt{y}} + \frac{15}{2}\sqrt{y} \) with respect to \( y \). Apply the power rule for simplification: \( f_{yy} = -\frac{x}{2y^{3/2}} + \frac{15}{4y^{1/2}} \). This provides the acceleration of change in \( y \) direction.

Understanding the second partial derivatives highlights concavity or curvature of the function related to each variable.

Mixed Partial Derivative

Mixed partial derivatives involve differentiating a function with respect to two different variables in sequence. Our objective is to find \( f_{xy} \) and \( f_{yx} \), which cross-checks the partial derivatives of \( x \) first then \( y \), and \( y \) first then \( x \).

• \( f_{xy} \): Differentiate \( f_x = 2\sqrt{y} \) with respect to \( y \). After applying the rules, we find \( f_{xy} = \frac{1}{\sqrt{y}} \).
• \( f_{yx} \): Differentiate \( f_y = \frac{x}{\sqrt{y}} + \frac{15}{2}\sqrt{y} \) with respect to \( x \). The result is the same: \( f_{yx} = \frac{1}{\sqrt{y}} \).

These results being equal confirms Clairaut's theorem, assuring us that for continuous functions, the mixed derivatives are consistent irrespective of differentiation order. This is useful in confirming the reliability of calculations in complex functions.