Question

Explain why or why not 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 whether each of the following statements is true or false. If the statement is true, explain why. If the statement is false, provide a counterexample. a) The partial derivative of \(y^{10}\) with respect to x is \(10 y^9\). b) The second-order mixed partial derivative of \(\sqrt{xy}\) is \(\frac{1}{\sqrt{xy}}\). c) If a function 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 not \(10 y^9\), but 0. b) False. The second-order mixed partial derivative of \(\sqrt{xy}\) is not \(\frac{1}{\sqrt{xy}}\), but 0. c) True. According to Clairaut's theorem, if a function has continuous mixed partial derivatives, the order of partial differentiation can be interchanged, so \(f_{xxy}=f_{yxx}\).

Step-by-step solution

Statement a: Partial Derivative

To check whether this statement is true, find the partial derivative of \(y^{10}\) with respect to \(x\): $$\frac{\partial}{\partial x}\left(y^{10}\right) = 0$$ Since the result is not equal to \(10 y^9\), the statement is false.

Statement b: Second-Order Mixed Partial Derivative

To find the second-order mixed partial derivative of \(\sqrt{xy}\), we'll calculate the first-order partial derivative with respect to \(x\): $$\frac{\partial}{\partial x}\left(\sqrt{xy}\right) = \frac{1}{2\sqrt{xy}}\cdot y$$ Now find the partial derivative of the above result with respect to \(y\): $$\frac{\partial^{2}}{\partial x \partial y}\left(\sqrt{xy}\right) = \frac{\partial}{\partial y}\left(\frac{1}{2\sqrt{xy}}\cdot y\right) = 0$$ As the result is not equal to \(\frac{1}{\sqrt{xy}}\), the statement is false.

Statement c: Continuous Partial Derivatives and Equality

Given that \(f\) has continuous partial derivatives of all orders, we want to verify if \(f_{xxy}=f_{yxx}\). According to Clairaut's theorem, if a function has continuous mixed partial derivatives, the order of partial differentiation can be interchanged. Therefore, we have: $$f_{xxy} = f_{yxx}$$ Since the condition is satisfied in the statement, it is true. To sum up: - Statement a is false - Statement b is false - Statement c is true

Key concepts

Second-order Mixed Partial Derivative

In calculus, when we talk about partial derivatives, we are focusing on how a multivariable function changes in response to variations in one variable at a time. The second-order mixed partial derivative is a particular type of partial derivative where the function is differentiated twice, each time with respect to a different variable. For instance, if we have a function of two variables, say, f(x, y), then the second-order mixed partial derivative with respect to x first and y next can be written as \(\frac{\partial^2 f}{\partial y \partial x}\).

When calculating second-order mixed partial derivatives, it's important to remember to handle the operation in steps. We first compute the partial derivative with respect to one variable, and then take the partial derivative of that result with respect to the next variable. This is highlighted in the step-by-step solution to statement b, which shows the intermediate steps and clarifies why the given statement is false. The correct approach to calculation ensures students avoid common mistakes.

Clairaut's Theorem

Clairaut's theorem is a critical theorem in calculus that applies to functions with multiple variables. The formal statement of this theorem says: if a function f(x, y) has continuous second-order mixed partial derivatives, then the mixed partial derivatives can be calculated in any order. In other words, \(\frac{\partial^2 f}{\partial x \partial y} = \frac{\partial^2 f}{\partial y \partial x}\), provided these second-order mixed partial derivatives exist and are continuous on a certain region.

This theorem is demonstrated in the solution for statement c, where the equality of f_xxy and f_yxx depends on the continuity of the partial derivatives. It's essential to ensure that functions meet the necessary conditions for Clairaut's theorem to be applied. Otherwise, the equality cannot be assumed.

Continuous Partial Derivatives

The term 'continuous partial derivatives' implies that a function's partial derivatives do not have any abrupt changes or discontinuities within the region of interest. For any multivariable function like f(x, y), having continuous partial derivatives means that the behavior of the function, as described by the slopes of its tangent planes in any direction, is predictable and smooth.

Continuity of partial derivatives is fundamental when applying theorems such as Clairaut's which relies on this smoothness to guarantee the equality of mixed partial derivatives. In the context of our exercises, statement c assumes the existence of continuous partial derivatives to affirm the truth of the equality f_xxy equals f_yxx. This presupposes a smooth function without abrupt directional changes anywhere in the domain.

Calculus

Calculus is a vast field of mathematics that studies changes. It is divided primarily into two branches: differential calculus, which deals with rates of change and slopes of curves; and integral calculus, which deals with accumulation of quantities and areas under or between curves. Many of the principles of calculus are extended to functions of more than one variable—a field known as multivariable calculus.

In the context of solving problems pertaining to multivariable calculus, such as those in our exercise, students employ partial differentiation, one of the key concepts differential calculus offers. This is used to understand the changes in a function as each variable shifts a little bit, a fundamental skill for tackling more complex problems in physics, engineering, economics, and beyond.