Question

The purpose of this exercise is to show that if \(f, f_{x}\) and \(f_{y}\) exist on a neighborhood \(N\) of \(\left(x_{0}, y_{0}\right)\) and \(f_{x}\) and \(f_{y}\) are differentiable at \(\left(x_{0}, y_{0}\right),\) then \(f_{x y}\left(x_{0}, y_{0}\right)=\) \(f_{y x}\left(x_{0}, y_{0}\right) .\) Suppose that the open square $$\left\\{(x, y)|| x-x_{0}|<| h|,| y-y_{0}|<| h \mid\right\\}$$ is in \(N\). Consider $$B(h)=f\left(x_{0}+h, y_{0}+h\right)-f\left(x_{0}+h, y_{0}\right)-f\left(x_{0}, y_{0}+h\right)+f\left(x_{0}, y_{0}\right)$$ (a) Use the mean value theorem as we did in the proof of Theorem 5.3 .3 to write $$B(h)=\left[f_{x}\left(\widehat{x}, y_{0}+k\right)-f_{x}\left(\hat{x}, y_{0}\right)\right] h$$ where \(\widehat{x}\) is between \(x_{0}\) and \(x_{0}+h\). Then use the differentiability of \(f_{x}\) at \(\left(x_{0}, y_{0}\right)\) to infer that $$B(h)=h^{2} f_{x y}\left(x_{0}, y_{0}\right)+h E_{1}(h), \quad \text { where } \quad \lim _{h \rightarrow 0} \frac{E_{1}(h)}{h}=0$$ (b) Use the mean value theorem to write $$B(h)=\left[f_{y}\left(x_{0}+h, \widehat{y}\right)-f_{y}\left(x_{0}, \widehat{y}\right)\right] h$$ where \(\widehat{y}\) is between \(y_{0}\) and \(y_{0}+h\). Then use the differentiability of \(f_{y}\) at \(\left(x_{0}, y_{0}\right)\) to infer that $$B(h)=h^{2} f_{y x}\left(x_{0}, y_{0}\right)+h E_{2}(h), \quad \text { where } \quad \lim _{h \rightarrow 0} \frac{E_{2}(h)}{h}=0$$ (c) Infer from (a) and (b) that \(f_{x y}\left(x_{0}, y_{0}\right)=f_{y x}\left(x_{0}, y_{0}\right)\).

Short answer

Question: Prove that if \(f\), \(f_{x}\), and \(f_{y}\) exist on a neighborhood \(N\) of \((x_{0}, y_{0})\) and \(f_{x}\) and \(f_{y}\) are differentiable at \((x_{0}, y_{0})\), then \(f_{xy}(x_0, y_0) = f_{yx}(x_0, y_0)\) using the Mean Value Theorem.

Step-by-step solution

Initial Definitions

Let's state the given function \(B(h)\): $$B(h) = f(x_0+h, y_0+h) - f(x_0+h, y_0) - f(x_0, y_0+h) + f(x_0, y_0)$$ We need to prove that \(f_{xy}(x_0, y_0) = f_{yx}(x_0, y_0)\).

- Applying Mean Value Theorem on \(x\)

Applying the Mean Value theorem in terms of \(x\), we have: $$B(h) = \left[f_x(\hat{x}, y_0+h) - f_x(\hat{x}, y_0)\right]h$$ with \(\hat{x}\) between \(x_0\) and \(x_0+h\). Now we know that \(f_{x}\) is differentiable, so by using the definition of differentiability we can state: $$B(h) = h^2 f_{xy}(x_0, y_0) + h E_1(h), \quad \text { where } \quad \lim _{h \rightarrow 0} \frac{E_{1}(h)}{h}=0$$

- Applying Mean Value Theorem on \(y\)

Applying the Mean Value theorem in terms of \(y\), we have: $$B(h) = \left[f_y(x_0+h, \hat{y}) - f_y(x_0, \hat{y})\right]h$$ with \(\hat{y}\) between \(y_0\) and \(y_0+h\). Similarly, we use the differentiability of \(f_{y}\) : $$B(h) = h^2 f_{yx}(x_0, y_0) + h E_2(h), \quad \text { where } \quad \lim _{h \rightarrow 0} \frac{E_{2}(h)}{h}=0$$

- Equating both expressions of \(B(h)\)

Now, let's equate the two expressions for \(B(h)\) : $$h^2 f_{xy}(x_0, y_0) + h E_1(h) = h^2 f_{yx}(x_0, y_0) + h E_2(h)$$ If we divide both sides by \(h^2\) we have: $$f_{xy}(x_0, y_0) + \frac{E_1(h)}{h} = f_{yx}(x_0, y_0) + \frac{E_2(h)}{h}$$ Now, we know that \(\lim _{h \rightarrow 0} \frac{E_{1}(h)}{h}=0\) and \(\lim _{h \rightarrow 0} \frac{E_{2}(h)}{h}=0\), so we can take the limit as \(h\) approaches \(0\) on both sides of the equation: $$\lim_{h \rightarrow 0} \left(f_{xy}(x_0, y_0) + \frac{E_1(h)}{h}\right) = \lim_{h \rightarrow 0} \left(f_{yx}(x_0, y_0) + \frac{E_2(h)}{h}\right)$$ Therefore, we arrive at the conclusion that: $$f_{xy}(x_0, y_0) = f_{yx}(x_0, y_0)$$ This proves that if \(f\), \(f_{x}\), and \(f_{y}\) exist on a neighborhood \(N\) of \((x_{0}, y_{0})\) and \(f_{x}\) and \(f_{y}\) are differentiable at \((x_{0}, y_{0})\), then \(f_{xy}(x_0, y_0) = f_{yx}(x_0, y_0)\).

Key concepts

mean value theorem

The Mean Value Theorem (MVT) is a fundamental concept in calculus that acts as a bridge between values of a function at certain points and the behavior of its difference quotient. The theorem typically applies to continuous functions and their derivatives on an interval. For a differentiable function, the MVT guarantees at least one point (**c**) in the interval where the derivative equals the average rate of change over that interval.
This theorem is crucial when dealing with partial derivatives.

Here's how it applies in the context of multivariable functions and, specifically, to the exercise at hand:

• By applying the MVT separately to the function's behavior in the direction of each coordinate axis (first for the **x** direction and then for the **y** direction), we can approximate how the function changes.
• In the given exercise, MVT aids in expressing the change in the function as a derivative multiplied by the distance **h** between points, along paths parallel to the axes.
  Thus, \[B(h) = \left[f_x(\hat{x}, y_0+k) - f_x(\widehat{x}, y_0)\right] h\],
  and similarly for \[f_y\] on path **y**.
• This usage allows us to express complex changes in functions with multiple variables in a simpler, more manageable form, building towards showing the equality of mixed partial derivatives.

differentiability

Differentiability is more than just having a derivative; it's about how smoothly a function behaves at a point. For single-variable functions, differentiability at a point implies that the function can be well-approximated by a linear function in the immediate vicinity of that point.
Now, when we extend this idea to functions of several variables, we require not just partial derivatives, but also their continuous behavior—thus contributing to the consistency of these partial derivatives.

In the context of this exercise:

• The exercise hinges on \(f_x\) and \(f_y\) being differentiable. This is significant because that leads to errors **E1(h)** and **E2(h)** vanishing faster than **h** as **h** approaches zero.
• This behavior is captured by the equations:
  \[\lim_{h \to 0} \frac{E_1(h)}{h} = 0\] and \[\lim_{h \to 0} \frac{E_2(h)}{h} = 0\]
  which ensure the smoothness of function **f** at point \((x_0, y_0)\).
• The continuity and differentiability of partial derivatives are fundamental in confirming that the mixed partial derivatives, \(f_{xy}\) and \(f_{yx}\), are equal at that point, as they express the rate and manner of change of the function over small intervals.

mixed partial derivatives

In calculus, mixed partial derivatives signify differentiating a multivariable function with respect to two or more variables. This concept extends the idea of repeated differentiation into higher dimensions and reveals insights about the shape and curvature of surfaces described by the functions.
For most continuous and well-behaved functions, mixed partial derivatives are commutative, meaning
\(f_{xy}(x, y) = f_{yx}(x, y)\).
This is known as Clairaut's Theorem or Schwarz's Theorem, named after the mathematicians who made notable contributions to this area.

In the given exercise, the concept plays a pivotal role:

• The task is to prove that the mixed partial derivatives \(f_{xy}(x_0, y_0)\) and \(f_{yx}(x_0, y_0)\) are equal assuming differentiability conditions for function **f** in a neighborhood around \((x_0, y_0)\).
• The sequence of applying mean value theorem, followed by ensuring differentiability, knots these concepts together. It shows that, when functions and their first partial derivatives are smooth enough (differentiable and continuous), the paths to take these derivatives imply no difference in the limits.
• Equations like
  \((B(h)=h^2 f_{xy}(x_0, y_0)+h E_1(h))\)
  and
  \((B(h)=h^2 f_{yx}(x_0, y_0)+h E_2(h)) \) emphasize that the change in the function by traversing in different variable orders results in the same incremental adjustments, hence proving the equality of these mixed partials.

By equating limits, we thus deduce the symmetry of mixed partial derivatives under the given conditions.