Question

Determine whether the following statements are true and give an explanation or counterexample. a. \(\frac{\partial}{\partial x}\left(y^{10}\right)=10 y^{9}\) b. \(\frac{\partial^{2}}{\partial x \partial y}(\sqrt{x y})=\frac{1}{\sqrt{x y}}\) c. If \(f\) has continuous partial derivatives of all orders, then \(f_{x x y}=f_{y x x}\)

Short answer

Question: Determine the truth of the following statements involving partial derivatives: a. \(\frac{\partial}{\partial x}(y^{10}) = 10y^9\) b. \(\frac{\partial^2}{\partial x \partial y}(\sqrt{xy}) = \frac{1}{\sqrt{xy}}\) c. If \(f\) has continuous partial derivatives of all orders, then \(f_{xxy} = f_{yxx}\). Answer: a. False. The partial derivative of \(y^{10}\) with respect to \(x\) is 0, not \(10y^9\). We provided a counterexample where \(y=1\), showing that \(\frac{\partial}{\partial x}(y^{10}) = 0\) and \(10y^9 = 10\). b. False. The second-order mixed partial derivative of \(\sqrt{xy}\) is \(\frac{1}{4} \cdot \frac{1}{x^{\frac{3}{2}}y^{\frac{1}{2}}} \cdot x\), not \(\frac{1}{\sqrt{xy}}\). We provided a counterexample where \(x = y = 1\), showing that \(\frac{\partial^2}{\partial x \partial y}(\sqrt{xy}) = \frac{1}{4}\) and \(\frac{1}{\sqrt{xy}} = 1\). c. True. According to Clairaut's theorem, if a function has continuous second-order mixed partial derivatives, the order of differentiation does not matter. Thus, \(f_{xxy} = f_{yxx}\).

Step-by-step solution

Find the partial derivative of \(y^{10}\) with respect to \(x\) (Statement a)

Since the function \(y^{10}\) does not depend on \(x\), the partial derivative of \(y^{10}\) with respect to \(x\) is 0.

Show that the partial derivative in Step 1 is not equal to \(10 y^9\) and provide a counterexample (Statement a)

According to our calculation, \(\frac{\partial}{\partial x}(y^{10}) = 0\), which is not equal to \(10 y^9\). To provide a counterexample, consider \(y = 1\). The partial derivative is still 0, while \(10 y^9 = 10\).

Determine the second-order mixed partial derivative of \(\sqrt{xy}\) (Statement b)

First, we find the partial derivative with respect to \(x\): \(\frac{\partial}{\partial x}(\sqrt{xy}) = \frac{1}{2} \cdot \frac{1}{\sqrt{xy}} \cdot y\). Next, we find the partial derivative with respect to \(y\): \(\frac{\partial}{\partial y}\left(\frac{1}{2} \cdot \frac{1}{\sqrt{xy}} \cdot y\right) = \frac{1}{4} \cdot \frac{1}{x^{\frac{3}{2}}y^{\frac{1}{2}}} \cdot x\).

Show that the second-order mixed partial derivative in Step 3 is not equal to \(\frac{1}{\sqrt{xy}}\) and provide a counterexample (Statement b)

According to our calculation, \(\frac{\partial^2}{\partial x \partial y}(\sqrt{xy}) = \frac{1}{4} \cdot \frac{1}{x^{\frac{3}{2}}y^{\frac{1}{2}}} \cdot x\), which is not equal to \(\frac{1}{\sqrt{xy}}\). To provide a counterexample, consider \(x = y = 1\). The second-order mixed partial derivative is \(\frac{1}{4}\), while \(\frac{1}{\sqrt{xy}} = 1\).

Argue the truth of Statement c based on Clairaut's theorem

Statement c states that if \(f\) has continuous partial derivatives of all orders, then \(f_{xxy} = f_{yxx}\). This statement is true according to Clairaut's theorem, which states that if a function has continuous second-order mixed partial derivatives, then the order of differentiation does not matter, and we can write: \(f_{xxy} = f_{xyx} = f_{yxx} = f_{yx^n}\). So, based on our analysis: Statement a is false, and we provided a counterexample. Statement b is false, and we provided a counterexample. Statement c is true, and we used Clairaut's theorem to justify it.

Key concepts

Mixed Partial Derivatives

Mixed partial derivatives are derivatives of a function taken with respect to different variables. For a function of two variables, say \(f(x, y)\), you might first differentiate with respect to \(x\) and then \(y\), or the opposite order. These are called mixed because they involve more than one variable.

Mixed partial derivatives are important because they provide insight into how a change in one variable affects the change brought about by another. Understanding their order and behavior can lead to deeper comprehension of multidimensional functions.

A typical question about mixed partial derivatives is whether changing the order of differentiation yields the same result. This is often linked to Clairaut's Theorem, which provides conditions under which swapping the order does not change the outcome. However, if a function does not meet these conditions, mixed partial derivatives can indeed differ.

Clairaut's Theorem

Clairaut's Theorem is a pivotal concept when discussing mixed partial derivatives. It states that for a function with continuous second-order partial derivatives, the order of differentiation does not matter.

This means if we have a function \(f\), and it has continuous second-order partial derivatives, then \(\frac{\partial^2 f}{\partial x \partial y} = \frac{\partial^2 f}{\partial y \partial x}\).

The importance of Clairaut's Theorem lies in simplifying calculations in multivariable calculus. It assures us that under certain conditions, the sequence of differentiating does not influence the result, making problem-solving more intuitive. When solving problems or verifying the equality of mixed derivatives, one can rely on this theorem for functions with the necessary continuity conditions.

Continuous Partial Derivatives

Continuous partial derivatives ensure stability and predictability in a function's behavior. A function has continuous partial derivatives if its partial derivatives do not jump abruptly and are smooth in their changes.

Continuity of partial derivatives is crucial when applying Clairaut's Theorem. This continuous nature allows for the interchangeability of differentiation orders in mixed partial derivatives.

This concept is not just limited to second-order derivatives. It extends to higher orders, ensuring that complexities do not arise when approaching functions analytically. Continuous partial derivatives allow for consistent results across computations, aiding in reliable mathematical modeling.