Problem 51 - Calculus Early Transcendentals

Question

Find all the second partial derivatives. $$f(x, y)=x^{3} y^{5}+2 x^{4} y$$

Step-by-Step Solution

Answer

The second partial derivatives are \(f_{xx} = 6x y^5 + 24x^2 y\), \(f_{yy} = 20x^3 y^3\), and \(f_{xy} = f_{yx} = 15x^2 y^4 + 8x^3\).

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

We start by differentiating the function \(f(x,y) = x^3 y^5 + 2x^4y\) with respect to \(x\). The first term, \(x^3 y^5\), becomes \(3x^2 y^5\) applying the power rule while treating \(y^5\) as a constant. The second term, \(2x^4y\), becomes \(8x^3y\). Thus, the first partial derivative with respect to \(x\) is:\[ f_x = 3x^2 y^5 + 8x^3 y \]

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

Next, differentiate the function \(f(x,y) = x^3 y^5 + 2x^4y\) with respect to \(y\). For the first term \(x^3 y^5\), the derivative is \(5x^3 y^4\) applying the power rule, treating \(x^3\) as a constant. For the second term \(2x^4y\), the derivative simplifies to \(2x^4\). Thus, the first partial derivative with respect to \(y\) is:\[ f_y = 5x^3 y^4 + 2x^4 \]

Step 3: Compute the Second Partial Derivative with respect to x (\(f_{xx}\))

To find \(f_{xx}\), we take the partial derivative of \(f_x\) with respect to \(x\). This derivative is calculated from \(f_x = 3x^2 y^5 + 8x^3 y\). Differentiating term by term, the first term gives \(6x y^5\), and the second term gives \(24x^2 y\). Thus, the second partial derivative with respect to \(x\) is:\[ f_{xx} = 6x y^5 + 24x^2 y \]

Step 4: Compute the Second Partial Derivative with respect to y (\(f_{yy}\))

To find \(f_{yy}\), we take the partial derivative of \(f_y\) with respect to \(y\). This derivative is calculated from \(f_y = 5x^3 y^4 + 2x^4\). The first term becomes \(20x^3 y^3\), following the power rule, and the second term, \(2x^4\), is considered constant with respect to \(y\) and results in zero. Hence, the second partial derivative with respect to \(y\) is:\[ f_{yy} = 20x^3 y^3 \]

Step 5: Compute Mixed Partial Derivative \(f_{xy}\)

For \(f_{xy}\), compute the partial derivative of \(f_x = 3x^2 y^5 + 8x^3 y\) with respect to \(y\). The first term \(3x^2 y^5\) differentiates to \(15x^2 y^4\), and the second term \(8x^3 y\) differentiates to \(8x^3\). Thus, the mixed partial derivative \(f_{xy}\) is:\[ f_{xy} = 15x^2 y^4 + 8x^3 \]

Step 6: Compute Mixed Partial Derivative \(f_{yx}\)

For \(f_{yx}\), which should be the same as \(f_{xy}\) because mixed partial derivatives are generally equal, we find it by differentiating \(f_y = 5x^3 y^4 + 2x^4\) with respect to \(x\). The first term \(5x^3 y^4\) becomes \(15x^2 y^4\), and the second term \(2x^4\) becomes \(8x^3\). Thus, \(f_{yx}\) is:\[ f_{yx} = 15x^2 y^4 + 8x^3 \]

Key Concepts

Partial Differentiation

Partial differentiation is a fundamental concept in calculus where you focus on changing one variable while keeping others constant. This technique allows us to analyze functions with multiple variables, giving insights into how the function behaves if one variable changes. In this exercise, we explore a function of two variables, \(f(x, y) = x^3 y^5 + 2x^4y\). By taking partial derivatives, we derive the effect of small changes in each variable on the function's value.

To find the first partial derivative with respect to \(x\), treat \(y\) as constant. Use the power rule to differentiate each term. Do the same for \(y\), treating \(x\) as constant. This way, the understanding of how each variable influences the function is enriched. Partial differentiation provides essential insights in fields ranging from physics to economics.

Multivariable Calculus

Multivariable calculus extends calculus concepts to functions with more than one variable. Unlike single-variable calculus, you assess interactions between multiple variables. In the exercise, this includes examining the function \(f(x, y)\) by computing partial derivatives with respect to both \(x\) and \(y\).

Here, you start by considering how each variable independently affects the function by finding the first partial derivatives \(f_x\) and \(f_y\). Next, enrich this analysis by computing second derivatives, like \(f_{xx}\), \(f_{yy}\), and the mixed second partial derivatives \(f_{xy}\) and \(f_{yx}\).

Multivariable calculus provides a framework for understanding complex systems like fluid dynamics, and in this function, it allows analysis of more intricate surfaces represented by equations.

Mixed Partial Derivatives

Mixed partial derivatives help in understanding how simultaneous changes in two variables influence a function. They show changes across different dimensions. For example, in the exercise, \(f_{xy}\) and \(f_{yx}\) are mixed partial derivatives of the function \(f(x, y) = x^3 y^5 + 2x^4y\).

The intriguing part of mixed partial derivatives is their symmetry, often expressed through Clairaut's theorem. The theorem states that if the mixed derivatives are continuous, then \(f_{xy} = f_{yx}\).

• This property is used to ease the process of finding derivatives and can verify computation accuracy.
• Mixed partials help assess how changes in one direction impact changes in another, applicable in designing circuits or optimization problems.

Overall, they serve crucial roles in both theoretical and practical mathematics.