Partial Derivatives

Now let $f(x,y)$ be a function of two variables. We consider the problem of finding derivatives of $f(x,y)$. For the present, we shall consider two kinds of derivatives: if $(x_0,y_0)$ is a point, we shall call the derivative at $x_0$ of the function $g(x)=f(x,y_0)$ the partial derivative of $f(x,y)$ with respect to $x$ at $(x_0,y_0)$ and write it as

$$f_x(x_0,y_0),\text{or }\frac{\partial f}{\partial x}(x_0,y_0)\cdot$$

Likewise, the partial derivative of $f(x,y)$ with respect to $y$ at $(x_0,y_0)$, written as

$$f_y(x_0,y_0),\text{or }\frac{\partial f}{\partial y}(x_0,y_0)$$

is the derivative of the function $h(y)=f(x_0,y)$ at $y=y_0$. Thus we have

$$f_x(x_0,y_0)=\underset{h\to 0}{lim}\frac{f(x_0+h,y_0)-f(x_0,y_0)}{h},$$$$f_y(x_0,y_0)=\underset{k\to 0}{lim}\frac{f(x_0,y_0+k)-f(x_0,y_0)}{k}\cdot$$

Then it should be evident what we mean by $f_{xx},f_{xy},f_{yx},f_{yy}$; for instance, $f_{xy}(x_0,y_0)=(f_x{)}_y(x_0,y_0)$, or in words, the partial derivative with respect to $y$ of the partial derivative with respect to $x$ of $f(x,y)$. To find it, we must use

\begin{aligned} f_{xy}(x_0, y_0) &= \lim_{k \to 0} \frac{f_x(x_0, y_0+k) - f_x(x_0, y_0)}{k} \\ &= \lim_{k \to 0} \lim_{h \to 0} \frac{\frac{f(x_0+h, y_0+k) - f(x_0, y_0+k)}{h} - \frac{f(x_0+h, y_0) - f(x_0, y_0)}{h}}{k} \\ &= \lim_{k \to 0} \lim_{h \to 0} \frac{f(x_0+h, y_0+k) - f(x_0, y_0+k) - f(x_0+h, y_0) + f(x_0, y_0)}{hk} \cdot \end{aligned}

Our numerator is thus a combination of the values of the function at the four corners of the rectangle shown below, with the corresponding signs:

Likewise,

$$f_{yx}(x_0,y_0)=\underset{h\to 0}{lim}\underset{k\to 0}{lim}\frac{f(x_0+h,y_0+k)-f(x_0+h,y_0)-f(x_0,y_0+k)+f(x_0,y_0)}{hk}\cdot$$

The fact that the two terms which are involved in taking limits are equal does not mean that the limits are equal; for we are not taking the same limit in both cases. For example, consider

$$\underset{h\to 0}{lim}\underset{k\to 0}{lim}\frac{h^2-k^2}{h^2+k^2}=\underset{h\to 0}{lim}\frac{h^2}{h^2}=\underset{h\to 0}{lim}1=1\cdot$$

On the other hand,

$$\underset{k\to 0}{lim}\underset{h\to 0}{lim}\frac{h^2-k^2}{h^2+k^2}=\underset{k\to 0}{lim}\frac{-k^2}{k^2}=\underset{k\to 0}{lim}-1=-1\cdot$$

We do know, by the existence of $f_x$, that we can apply the mean value theorem to $f(x,y)$ when considered as a function of $x$ for finding $y$. Thus we have

$$f(x_0+h,y_0+k)-f(x_0,y_0+k)=h{f}_x(\xi ,y_0+k),$$

where $\xi$ is between $x_0$ and $x_0+h$, and

f(x_0 + h, y_0) - f(x_0, y_0) = h f_x(\xi', y_0),

where \xi' is between $x_0$ and $x_0+h$. Suppose for the moment that \xi = \xi'; then the numerator in the expressions whose limits are the mixed second partial derivatives is

$$h(f_x(\xi ,y_0+k)-f_x(\xi ,y_0))=hk{f}_{xy}(\xi ,\eta )$$

for some $\eta$ between $y_0$ and $y_0+k$. Thus we would have, for instance,

$$f_{yx}(x_0,y_0)=\underset{h\to 0}{lim}\underset{k\to 0}{lim}{f}_{xy}(\xi ,\eta ),$$

where $(\xi ,\eta )$ is in our rectangle. Then if we knew that

$$\underset{h\to 0}{lim}\underset{k\to 0}{lim}{f}_{xy}(\xi ,\eta )=f_{xy}(x_0,y_0),$$

we would have the result that

$$f_{xy}(x_0,y_0)=f_{yx}(x_0,y_0)\cdot$$

However, this is not always true. It is left as an exercise to show that for the function

f(x, y) =\left\{\begin{aligned} &\frac{4xy(x^2 - y^2)}{x^2 + y^2}, && (x, y) \neq (0, 0) \\ \\ &0, && (x, y) = (0, 0) \end{aligned}\right\} ,

we have $f_{xy}(0,0)=+4,f_{yx}(0,0)=-4$. But in this example, neither $f_{xy}$ nor $f_{yx}$ is continuous at $(0,0)$. If $f_{xy}$ is continuous at $(x_0,y_0)$, then we do have

$$\underset{h\to 0}{lim}\underset{k\to 0}{lim}{f}_{xy}(\xi ,\eta )=f_{xy}(x_0,y_0),$$

and we would have shown $f_{xy}(x_0,y_0)=f_{yx}(x_0,y_0)$ if we could be sure that \xi = \xi'. But we can arrange this:

Let

$$g(x)=f(x,y_0+k)-f(x,y_0)\cdot$$

Then

$$g(x_0+h)-g(x_0)=f(x_0+h,y_0+k)-f(x_0+h,y_0)-f(x_0,y_0+k)+f(x_0,y_0)$$

is exactly the numerator of the expression we are considering. Since $g(x)$ is differentiable, with g'(x) = f_x(x, y_0 + k) - f_x(x, y_0), we have

g(x_0 + h) - g(x_0) = h g'(\xi),

where $\xi$ is between $x_0$ and $x_0+h$. Thus

\begin{aligned} f(x_0 + h, y_0 + k) - f(x_0 + h, y_0) - &f(x_0, y_0 + k) + f(x_0, y_0) \\[6pt] &= h\bigg(f_x(\xi, y_0+k) - f_x(\xi, y_0)\bigg). \end{aligned}

Now this is equal to $hk{f}_{xy}(\xi ,\eta )$ for some $\eta$ between $y_0$ and $y_0+k$, and we can finish the proof as when we assumed \xi = \xi'. What we have proved is the following:

In practically all applications, we do have $f_{yx}=f_{xy}$, so you will see this property used freely in most books.

EXERCISES