Problem 8 - Active Calculus Multivariable

Question

Let \(f(x, y)=(-(2 x+y))^{6}\). Then \(\frac{\partial^{2} f}{\partial x \partial y}\) __________. \(\frac{\partial^{3} f}{\partial x \partial y \partial x}=\) __________. \(\frac{\partial^{3} f}{\partial x^{2} \partial y}=\) __________.

Step-by-Step Solution

Answer

\(\frac{\partial^{2} f}{\partial x \partial y} = -60( -(2x+y)^4)\) \(\frac{\partial^{3} f}{\partial x \partial y \partial x} = 480( -(2x+y)^3)\) \(\frac{\partial^{3} f}{\partial x^{2} \partial y} = 240( -(2x+y)^2)\)

Step 1: Find the given function, f(x, y)

We are provided with the function: \(f(x, y) = (-(2x+y))^6\)

Step 2: Find the first-order partial derivatives

Find the first-order partial derivatives of \(f(x, y)\) with respect to x and y. \(\frac{\partial f}{\partial x} = \frac{d}{dx}( -(2x+y)^6) = 6( -(2x+y)^5)(-2)\) \(\frac{\partial f}{\partial y} = \frac{d}{dy}( -(2x+y)^6) = 6( -(2x+y)^5)(-1)\)

Step 3: Find the second mixed partial derivative

Compute the second mixed partial derivative, \(\frac{\partial^{2} f}{\partial x \partial y}\): \(\frac{\partial^{2} f}{\partial x \partial y} = \frac{\partial}{\partial y} ( \frac{\partial f}{\partial x}) = \frac{\partial}{\partial y}(6( -(2x+y)^5)(-2)) = -60( -(2x+y)^4)\)

Step 4: Find the first two third-order partial derivatives

Compute the first two third-order partial derivatives, \(\frac{\partial^{3} f}{\partial x \partial y \partial x}\) and \(\frac{\partial^{3} f}{\partial x^{2} \partial y}\): 1. \(\frac{\partial^{3} f}{\partial x \partial y \partial x} = \frac{\partial}{\partial x}( \frac{\partial^{2} f}{\partial x \partial y}) = \frac{\partial}{\partial x}( -60( -(2x+y)^4)) = 480( -(2x+y)^3)\) 2. \(\frac{\partial^{3} f}{\partial x^{2} \partial y} = \frac{\partial^{2}}{\partial x^2}( \frac{\partial f}{\partial y}) = \frac{\partial^{2}}{\partial x^2}(6( -(2x+y)^5)(-1)) = 240( -(2x+y)^2)\) So we have our final answers: \(\frac{\partial^{2} f}{\partial x \partial y} = -60( -(2x+y)^4)\) \(\frac{\partial^{3} f}{\partial x \partial y \partial x} = 480( -(2x+y)^3)\) \(\frac{\partial^{3} f}{\partial x^{2} \partial y} = 240( -(2x+y)^2)\)

Key Concepts

Multivariable Calculus

Multivariable calculus is an extension of single-variable calculus to more than one variable. In courses and real-world applications, we often come across functions that depend on two or more variables. For example, the volume of a gas may depend on its temperature and pressure, making it a function of two variables. This branch of mathematics allows us to analyze such functions to find rates of change, optimizations, and more.

When dealing with functions of multiple variables, we use partial derivatives to explore how the function changes as we tweak one variable at a time, holding the others constant. Understanding how to compute partial derivatives is crucial as it lays the groundwork for more complex concepts like optimization and integration in multivariable contexts. In practice, knowing how to find such derivatives equips students with the ability to tackle problems in physics, engineering, economics, and beyond...

Mixed Partial Derivatives

The concept of mixed partial derivatives comes into play when we're dealing with functions of more than one variable. A mixed partial derivative involves taking the derivative with respect to one variable, and then taking a derivative of the result with respect to another, different variable.

In our example with the function f(x, y), after finding the first partial derivatives, we calculated the mixed partial derivative \(\frac{\partial^{2} f}{\partial x \partial y}\) by differentiating \(\frac{\partial f}{\partial x}\) with respect to y. This process can uncover how a function's rate of change in one direction is affected by a change in another direction. It's important to note that for functions that are class C² (twice differentiable and continuous), the mixed partial derivatives \(\frac{\partial^{2} f}{\partial x \partial y}\) and \(\frac{\partial^{2} f}{\partial y \partial x}\) are equal, which is known as Clairaut's Theorem.

The ability to compute mixed partial derivatives is immensely useful in optimization problems and when dealing with the second derivative test in multivariable calculus.

Third-Order Partial Derivatives

Going a step further, we encounter third-order partial derivatives when we continue the process of differentiation. These are derived by applying the operation of taking partial derivatives three times, and they can be all with respect to the same variable or involve two or more variables.

In the context of the provided exercise, we considered two instances of third-order partial derivatives: \(\frac{\partial^{3} f}{\partial x \partial y \partial x}\) and \(\frac{\partial^{3} f}{\partial x^{2} \partial y}\). The first one involves taking the partial derivative of the mixed second-order derivative \(\frac{\partial^{2} f}{\partial x \partial y}\) with respect to x. The second one involves differentiating the mixed second-order derivative \(\frac{\partial f}{\partial y}\) with respect to x twice. The computation of these higher-order derivatives helps in understanding the behavior of a multivariable function more deeply and can be particularly helpful in Taylor series expansions in multiple dimensions as well as in solving certain types of differential equations.