Question

Verify that \(f_{x y}=f_{y x}\) for the following functions. $$f(x, y)=\sqrt{x y}$$

Short answer

Question: Verify that the mixed partial derivatives of the function \(f(x, y) = \sqrt{xy}\) are equal. Answer: For the given function \(f(x, y) = \sqrt{xy}\), we have computed the mixed partial derivatives \(f_{xy}\) and \(f_{yx}\) and found that they are equal: \(f_{xy} = f_{yx} = \frac{1}{2\sqrt{xy}}-\frac{x}{2\sqrt{xy}}\). Therefore, we have verified that the mixed partial derivatives are equal.

Step-by-step solution

Compute the first partial derivatives \(f_x\) and \(f_y\).

To compute the second-order mixed partial derivatives, we need to find \(f_x = \frac{\partial f}{\partial x}\) and \(f_y = \frac{\partial f}{\partial y}\) first. $$f_x = \frac{\partial}{\partial x} (\sqrt{xy}) = \frac{1}{2\sqrt{xy}}\frac{\partial}{\partial x}(xy) = \frac{y}{2\sqrt{xy}}$$ $$f_y = \frac{\partial}{\partial y} (\sqrt{xy}) = \frac{1}{2\sqrt{xy}}\frac{\partial}{\partial y}(xy) = \frac{x}{2\sqrt{xy}}$$

Compute the second-order mixed partial derivatives \(f_{xy}\) and \(f_{yx}\).

Now that we have \(f_x\) and \(f_y\), we can compute the mixed partial derivatives \(f_{xy}\) and \(f_{yx}\). $$f_{xy} = \frac{\partial}{\partial y} \left(\frac{y}{2\sqrt{xy}}\right)$$ $$f_{yx} = \frac{\partial}{\partial x} \left(\frac{x}{2\sqrt{xy}}\right)$$

Compute the derivatives and show that they're equal.

We can compute the derivatives and then compare them to see if they are equal. $$f_{xy} = \frac{1}{2\sqrt{xy}}\frac{\partial}{\partial y}(y) - \frac{y}{2(xy)^{\frac{3}{2}}}\frac{\partial}{\partial y}(xy) = \frac{1}{2\sqrt{xy}} - \frac{xy}{2(xy)^{\frac{3}{2}}} = \frac{1}{2\sqrt{xy}}-\frac{x}{2\sqrt{xy}}$$ $$f_{yx} = \frac{1}{2\sqrt{xy}}\frac{\partial}{\partial x}(x) - \frac{x}{2(xy)^{\frac{3}{2}}}\frac{\partial}{\partial x}(xy) = \frac{1}{2\sqrt{xy}} - \frac{xy}{2(xy)^{\frac{3}{2}}} = \frac{1}{2\sqrt{xy}}-\frac{y}{2\sqrt{xy}}$$ Since \(f_{xy} = f_{yx}\), we have verified that the mixed partial derivatives are equal for the given function \(f(x, y) = \sqrt{xy}\).

Key concepts

Understanding Partial Derivatives

When studying functions with multiple variables, understanding how changing one variable affects the function while keeping the others constant is essential. This concept is precisely what partial derivatives are about. To illustrate, consider a surface that represents a two-variable function, such as height over a geographical area. If you want to know how steep the hill is in the north-south direction at a particular point, you look at the rate at which the height changes as you move north or south, regardless of any changes in the east-west direction.

Mathematically, if you have a function like \( f(x, y) = \sqrt{xy} \), the partial derivative of \( f \) with respect to \( x \) (noted as \( f_x \)) measures how \( f \) changes as \( x \) varies, while \( y \) stays fixed. Similarly, \( f_y \) would measure change along \( y \) while \( x \) is constant. The calculation of \( f_x \) and \( f_y \) involves treating the non-differentiated variable as a constant and differentiating accordingly, hence why \( f_x \) and \( f_y \) for \( \sqrt{xy} \) are \( \frac{y}{2\sqrt{xy}} \) and \( \frac{x}{2\sqrt{xy}} \) respectively.

Exploring Second-Order Derivatives

Once you understand first-order partial derivatives, the next step is to look at how these derivatives change—that's where second-order derivatives come into play. These derivatives are essentially the 'derivatives of derivatives' and give us information about the curvature of the function's graph.

For our function \( f(x, y) = \sqrt{xy} \) from the previous example, the second-order partial derivatives include \( f_{xx} \), \( f_{yy} \) (the 'pure' second-order derivatives), and \( f_{xy} \) or \( f_{yx} \) (the 'mixed' second-order derivatives). Mixed partial derivatives like \( f_{xy} \) and \( f_{yx} \) tell us how the function's rate of change in one variable changes as we change another variable.

In layman's terms, it is like evaluating how the steepness of our hill in the north-south direction changes as we move east or west. The calculations for mixed second-order derivatives can get complex, but they provide profound insights into the function's behavior, such as indicating points of inflection or concavity.

Clairaut's theorem and Mixed Partial Derivatives

A pivotal piece in understanding mixed partial derivatives is Clairaut's theorem, a rule that under certain conditions, ensures the equality of mixed partial derivatives, regardless of the order of differentiation. To comply with Clairaut's theorem, the mixed partial derivatives must be continuous at the point of interest.

In the context of our function \( f(x, y) = \sqrt{xy} \), we had to prove that \( f_{xy} = f_{yx} \) for any point on the surface. As computed in the exercise, we indeed find that both mixed partial derivatives are equal, in this case to \( \frac{1}{2\sqrt{xy}} - \frac{x}{2\sqrt{xy}} \) and \( \frac{1}{2\sqrt{xy}} - \frac{y}{2\sqrt{xy}} \) respectively.

Clairaut's theorem is not just a mere mathematical curiosity; it has real-world implications. For instance, in engineering and physics, the theorem implies that the order in which forces or effects are applied doesn't change the final state of a system within certain conditions. This profound result is fundamental in fields like thermodynamics, elasticity, and fluid dynamics, where understanding and predicting system behavior is crucial.