Partial and total differentiation of multivariate functions

A multivariate function may be differentiated with respect to each variable, which is called partial differentiation. By combining all the partial differentiations, we define total differentiation. The essence of (total) differentiation is a linear approximation. In the case of a univariate function, we approximate the function $y=f(x)$ in the neighbor of a point, say $x=a$, by the tangent line $y=f^{\prime }(a)(x-a)+f(a)$. In the case of a multivariate function, we approximate the function $y=f(x_1,x_2,\cdots ,x_n)$ in the neighbor of a point, say $a=(a_1,a_2,\cdots ,a_n)$, by the tangent hyperplane at the point $a$.

Partial differentiation

Let $f(x,y)$ be a function on an open region $U\subset {\mathbb{R}}^2$ and $(a,b)\in U$. If we fix $y=b$ in $f(x,y)$, we have a univariate function $g(x)=f(x,b)$. Since $U$ is open, there exists $\delta >0$ such that $N_{\delta }(a,b)\subset U$. Therefore $g(x)$ is defined on the open interval $(a-\delta ,a+\delta )$. In other words, the function $g(x)$ is defined in a neighbor of $x=a$.

Remark. We write $N_{\delta }(a,b)$ (rather than $N_{\delta }((a,b))$, to save keystrokes!) to mean the $\delta$-neighbor of the point $(a,b)\in {\mathbb{R}}^2$. □

If $g(x)$ is differentiable at $x=a$, its differential coefficient is called the partial differential coefficient with respect to $x$ (at $(a,b)$) and $\frac{dg}{dx}(a)$ is denoted as $\frac{\partial f}{\partial x}(a,b)$ or $f_x(a,b)$.

Remark. Here is one way to understand the partial differential coefficient. We have a surface $z=f(x,y)$ in ${\mathbb{R}}^3$. Find its cross-section with the plane $y=b$. This cross-section is a curve defined by $z=g(x)=f(x,b)$. The partial differential coefficient $\frac{\partial f}{\partial x}(a,b)=\frac{dg}{dx}(a)$ is the slope of the tangent line of the curve at $x=a$. □

Similarly, if we fix $x=a$ in $f(x,y)$, we have a univariate function $h(y)=f(a,y)$ which is defined in a neighbor of $y=b$. If $\frac{dh}{dy}(b)$ exists, it is called the partial differential coefficient with respect to $y$ (at $b$) and denoted $\frac{\partial f}{\partial y}(a,b)$ or $f_y(a,b)$.

Example. Let $f(x,y)=x^{2}y+2x{y}^2-y^3$. Let us find the partial differential coefficients $f_x(a,b)$ and $f_y(a,b)$. Letting $y=b$, we have $f(x,b)=x^{2}b+2x{b}^2-b^3$. Differentiating the right-hand side with respect to $x$, we have $2xb+2b^2$. Setting $x=a$, we have $$f_x(a,b)=2ab+2b^2.$$ 

Similarly, we have $$f_y(a,b)=a^2+4ab-3b^2.$$ □

Partial derivatives

If the partial differential coefficient $\frac{\partial f}{\partial x}(a,b)$ exists at every $(a,b)\in U$, then it defines a function on $U$. This function is called the partial derivative of $f(x,y)$ with respect to $x$ and is denoted 

$$\frac{\partial f}{\partial x}(x,y)\text{or}{f}_x(x,y).$$

Similarly, we define the partial derivative of $f(x,y)$ with respect to $y$, denoted 

$$\frac{\partial f}{\partial y}(x,y)\text{or}{f}_y(x,y).$$

Example. Let $f(x,y)=5x^3+2x^{2}y-3x{y}^2+y^3$. Then $$\begin{array}{rcl}{f}_x(x,y)& =& 15x^2+4xy-3y^2,\\ f_y(x,y)& =& 2x^2-6xy+3y^2.\end{array}$$ □

Total differentiation

Let us review the notion of differentiation of univariate functions. We defined the differential coefficient of a univariate function $f(x)$ at $x=a$ by

$$f^{\prime }(a)=\frac{df}{dx}(a)=\underset{x\to a}{lim}\frac{f(x)-f(a)}{x-a}.$$

This is equivalent to

$$\underset{x\to a}{lim}\frac{f(x)-f(a)-f^{\prime }(a)(x-a)}{|x-a|}=0,$$

or

$$f(x)=f(a)+f^{\prime }(a)(x-a)+o(|x-a|)$$

where $o$ is Landau's little-o. This equation suggests that the function $y=f(x)$ is approximated by a linear function, namely the tangent of $y=f(x)$ at $x=a$,

$$y=f(a)+f^{\prime }(a)(x-a).$$

Conversely, suppose that the function $y=f(x)$ can be approximated by a linear function in a neighbor of $x=a$:

$$f(x)=f(a)+m(x-a)+o(|x-a|).$$

From this equation, we can see that

$$\underset{x\to a}{lim}\frac{f(x)-f(a)}{x-a}=m.$$

This means that $y=f(x)$ is differentiable at $x=a$ and $f^{\prime }(a)=m$.

In summary, $f(x)$ is approximated by the linear function $f(a)+f^{\prime }(a)(x-a)$ in a neighbor of $x=a$, and its slope is the differential coefficient $f^{\prime }(a)$ itself. Such linear approximation is the essence of differentiation.

The same argument applies to multivariate functions. Differentiating the function $z=f(x,y)$ at the point $P=(a,b)$ is to approximate it by a linear function

$$z=f(a,b)+m(x-a)+n(y-b).$$

That is, for the point $X=(x,y)$ in a neighbor of $P=(a,b)$, we consider the linear approximation

$$\begin{array}{}\text{(Eq:LA)}& f(x,y)=f(a,b)+m(x-a)+n(y-b)+o(\Vert X-P\Vert )\end{array}$$

where $\Vert X-P\Vert =\sqrt{(x-a{)}^2+(y-b{)}^2}=d(X,P)$ is the distance between the points $X$ and $P$. Setting $y=b$ in this equation, we have

$$\underset{x\to a}{lim}\frac{f(x,b)-f(a,b)}{x-a}=m.$$

That is, $f_x(a,b)=m.$ Similarly, we can show that $f_y(a,b)=n$. In summary, if the linear approximation (Eq:LA) holds, it must be

$$f(x,y)=f(a,b)+f_x(a,b)(x-a)+f_y(a,b)(y-b)+o(\Vert X-P\Vert )$$

in a neighbor of $P$.

Definition (Total differentiability)

Let $U$ be an open region in ${\mathbb{R}}^2$ and $P=(a,b)\in U$. The function $f(x,y)$ on $U$ is said to be (totally) differentiable at $(a,b)$ if there exist constants $m$ and $n$ such that

$$f(x,y)=f(a,b)+m(x-a)+n(y-b)+o(\Vert X-P\Vert )\text{ as }X=(x,y)\to P=(a,b)$$

or equivalently,

$$\underset{(x,y)\to (a,b)}{lim}\frac{f(x,y)-f(a,b)-m(x-a)-n(y-b)}{\sqrt{(x-a{)}^2+(y-b{)}^2}}=0.$$

$f(x,y)$ is said to be (totally) differentiable on $U$ if it is (totally) differentiable at every point in $U$.

Remark. The word "totally" in "totally differentiable" is used in contrast to "partially differentiable." However, "totally" may be omitted. If we simply say, "a multivariate function is differentiable," it means the function is totally differentiable. □

From the above discussion, if the function $f(x,y)$ is totally differentiable at $(a,b)$, it is partially differentiable at $(a,b)$, and $m=f_x(a,b)$ and $n=f_y(a,b)$. (The converse is not necessarily true; We will see such an example in a later post.) The linear function

$$z=f(a,b)+f_x(a,b)(x-a)+f_y(a,b)(y-b)$$

is the tanget plane of $z=f(x,y)$ at $(a,b)$. 

Remark. More generally, when the domain is in ${\mathbb{R}}^n$, for the function $y=f(x)=f(x_1,x_2,\cdots ,x_n)$ at the point $a=(a_1,a_2,\cdots ,a_n)$, we have the linear function

$$y=f(a)+f_{x_1}(a)(x_1-a_1)+f_{x_2}(a)(x_2-a_2)+\cdots +f_{x_n}(a)(x_n-a_n)$$

that is the tangent hyperplane of $y=f(x_1,x_2,\cdots ,x_n)$ at $a=(a_1,a_2,\cdots ,a_n)$. □

Example. Let us find the equation of the tangent plane of the surface defined by the function $z=2x^3+y^2$ at $(-1,2,2)$ (make sure this point indeed belongs to the given surface). Let $f(x,y)=2x^3+y^2$. Then

$$\begin{array}{rcl}{f}_x(x,y)& =& 6x^2,\\ f_y(x,y)& =& 2y,\end{array}$$

so that $f_x(-1,2)=6$ and $f_y(-1,2)=4$. Since $f(-1,2)=2$, the tangent plane is given by

$$z=2+6(x+1)+4(y-2),$$

that is,

$$6x+4y-z=0.$$ □