
# 2.2: Partial Derivatives

Now that we have an idea of what functions of several variables are, and what a limit of such a function is, we can start to develop an idea of a derivative of a function of two or more variables. We will start with the notion of a partial derivative.

Definition 2.3

Let $$f (x, y)$$ be a real-valued function with domain $$D$$ in $$\mathbb{R}^2$$, and let $$(a,b)$$ be a point in $$D$$. Then the partial derivative of $$\textbf{f}$$ at $$(a,b)$$ with respect to $$x$$, denoted by $$\dfrac{∂f}{∂x}(a,b)$$, is defined as

$\dfrac{∂f}{∂x}(a,b)=\lim \limits_{h \to 0}\dfrac{f(a+h,b)-f(a,b)}{h} \label{Eq2.2}$

and the partial derivative of $$f$$ at $$(a,b)$$ with respect to $$y$$, denoted by $$\dfrac{∂f}{∂y}(a,b)$$, is defined as

$\dfrac{∂f}{∂x}(a,b) = \lim \limits_{h \to 0}\dfrac{f(a+h,b)-f(a,b)}{h}\label{Eq2.3}$

Note: The symbol $$∂$$ is pronounced “del”.

Recall that the derivative of a function $$f (x)$$ can be interpreted as the rate of change of that function in the (positive) $$x$$ direction. From the definitions above, we can see that the partial derivative of a function $$f (x, y)$$ with respect to $$x$$ or $$y$$ is the rate of change of $$f (x, y)$$ in the (positive) $$x$$ or $$y$$ direction, respectively. What this means is that the partial derivative of a function $$f (x, y)$$ with respect to $$x$$ can be calculated by treating the $$y$$ variable as a constant, and then simply differentiating $$f (x, y)$$ as if it were a function of $$x$$ alone, using the usual rules from single-variable calculus. Likewise, the partial derivative of $$f (x, y)$$ with respect to $$y$$ is obtained by treating the $$x$$ variable as a constant and then differentiating $$f (x, y)$$ as if it were a function of $$y$$ alone.

Example 2.10

Find $$\dfrac{∂f}{∂x} (x, y)$$ and $$\dfrac{∂f}{∂y} (x, y)$$ for the function $$f (x, y) = x^2y+ y^3$$ .

Solution

Treating $$y$$ as a constant and differentiating $$f (x, y)$$ with respect to $$x$$ gives

$\nonumber \dfrac{∂f}{∂x}(x,y)=2xy$

and treating $$x$$ as a constant and differentiating $$f (x, y)$$ with respect to $$y$$ gives

$\nonumber \dfrac{∂f}{∂y}(x,y)=x^2+3y^2$

We will often simply write $$\dfrac{∂f}{∂x}$$ and $$\dfrac{∂f}{∂y}$$ instead of $$\dfrac{∂f}{∂x} (x, y)$$ and $$\dfrac{∂f}{∂y} (x, y)$$.

Example 2.11

Find $$\dfrac{∂f}{∂x}$$ and $$\dfrac{∂f}{∂y}$$ for the function $$f (x, y) = \dfrac{\sin{(xy^2)}}{ x^2 +1}$$ .

Solution

Treating $$y$$ as a constant and differentiating $$f (x, y)$$ with respect to $$x$$ gives

$\nonumber \dfrac{∂f}{∂x}=\dfrac{(x^2+1)(y^2\cos{(xy^2)})-(2x)\sin{(xy^2)}}{(x^2+1)^2}$

and treating $$x$$ as a constant and differentiating $$f (x, y)$$ with respect to $$y$$ gives

$\nonumber \dfrac{∂f}{∂y}=\dfrac{2xy\cos{(xy^2)}}{x^2+1}$

Since both $$\dfrac{∂f}{∂x}$$ and $$\dfrac{∂f}{∂y}$$ are themselves functions of $$x$$ and $$y$$, we can take their partial derivatives with respect to $$x$$ and $$y$$. This yields the higher-order partial derivatives

$\nonumber \dfrac{∂^2f}{∂x^2}=\dfrac{∂}{∂x}\left ( \dfrac{∂f}{∂x}\right) \quad \dfrac{∂^2f}{∂y^2}=\dfrac{∂}{∂y}\left ( \dfrac{∂f}{∂y}\right )$

$\nonumber \dfrac{∂^2f}{∂y∂x}=\dfrac{∂}{∂y}\left ( \dfrac{∂f}{∂x}\right) \quad \dfrac{∂^2f}{∂x∂y}=\dfrac{∂}{∂x}\left ( \dfrac{∂f}{∂y}\right )$

$\nonumber \dfrac{∂^3f}{∂x^3}=\dfrac{∂}{∂x}\left ( \dfrac{∂^2f}{∂x^2}\right) \quad \dfrac{∂^3f}{∂y^3}=\dfrac{∂}{∂y}\left ( \dfrac{∂^2f}{∂y^2}\right )$

$\nonumber \dfrac{∂^3f}{∂y∂x^2}=\dfrac{∂}{∂y}\left ( \dfrac{∂^2f}{∂x^2}\right) \quad \dfrac{∂^3f}{∂x∂y^2}=\dfrac{∂}{∂x}\left ( \dfrac{∂^2f}{∂y^2}\right )$

$\nonumber \dfrac{∂^3f}{∂y^2∂x}=\dfrac{∂}{∂y}\left ( \dfrac{∂^2f}{∂y∂x}\right) \quad \dfrac{∂^3f}{∂x^2∂y}=\dfrac{∂}{∂x}\left ( \dfrac{∂^2f}{∂x∂y}\right )$

$\nonumber \dfrac{∂^3f}{∂x∂y∂x}=\dfrac{∂}{∂x}\left ( \dfrac{∂^2f}{∂y∂x}\right) \quad \dfrac{∂^3f}{∂y∂x∂y}=\dfrac{∂}{∂y}\left ( \dfrac{∂^2f}{∂x∂y}\right )$

$\nonumber \vdots$

Example 2.12

Find the partial derivatives $$\dfrac{∂f}{∂x}$$ , $$\dfrac{∂f}{∂y}$$ , $$\dfrac{∂^2f}{∂x^2}$$ , $$\dfrac{∂^2f}{∂y^2}$$ , $$\dfrac{∂^2f}{∂y∂x}$$ and $$\dfrac{∂^2f}{∂x∂y}$$ for the function $$f (x, y) = e^{x^2y} + xy^3$$ .

Solution

Proceeding as before, we have

$\nonumber \begin{split}\dfrac{∂f}{∂x}&=2xye^{x^2y}+y^3 \\ \nonumber \dfrac{∂^2f}{∂x^2}&=\dfrac{∂}{∂x}(2xye^{x^2y}+y^3) \\ \nonumber &=2ye^{x^2y}+4x^2y^2e^{x^2y}\\ \nonumber \dfrac{∂^2f}{∂y∂x}&=\dfrac{∂}{∂y}(2xye^{x^2y}+y^3) \\ \nonumber &=2xe^{x^2y}+2x^3ye^{x^2y}+3y^2 \\ \nonumber \end{split} \qquad \begin{split}\dfrac{∂f}{∂y}&=x^2e^{x^2y}+3xy^2 \\ \nonumber \dfrac{∂^2f}{ ∂y^2}&=\dfrac{∂}{∂y}(x^2e^{x^2y}+3xy^2) \\ \nonumber &=x^4e^{x^2y}+6xy \\ \nonumber \dfrac{∂^2f}{∂x∂y}&=\dfrac{∂}{∂x}(x^2e^{x^2y}+3xy^2) \\ \nonumber &= 2xe^{x^2y}+2x^3ye^{x^2y}+3y^2 \\ \end{split}$

Higher-order partial derivatives that are taken with respect to different variables, such as $$\dfrac{∂^2f}{∂y∂x}$$ and $$\dfrac{∂^2f}{∂x∂y}$$ , are called mixed partial derivatives. Notice in the above example that $$\dfrac{∂^2f}{∂y∂x} = \dfrac{∂^2f}{∂x∂y}$$. It turns that this will usually be the case. Specifically, whenever both $$\dfrac{∂^2f}{∂y∂x}$$ and $$\dfrac{∂^2f}{∂x∂y}$$ are continuous at a point $$(a,b)$$, then they are equal at that point. All the functions we will deal with will have continuous partial derivatives of all orders, so you can assume in the remainder of the text that

$\nonumber \dfrac{∂^2f}{∂y∂x}=\dfrac{∂^2f}{∂x∂y}\text{ for all }(x,y)\text{ in the domain of }f$

In other words, it doesn’t matter in which order you take partial derivatives. This applies even to mixed partial derivatives of order 3 or higher.

The notation for partial derivatives varies. All of the following are equivalent:

$\nonumber \dfrac{∂f}{∂x} : f_x(x,y),\quad f_1(x,y),\quad D_x(x,y),\quad D_1(x,y)$

$\nonumber \dfrac{∂f}{∂y} : f_y(x,y),\quad f_2(x,y),\quad D_y(x,y),\quad D_2(x,y)$

$\nonumber \dfrac{∂^2f}{∂x^2} : f_{xx}(x,y),\quad f_{11}(x,y),\quad D_{xx}(x,y),\quad D_{11}(x,y)$

$\nonumber \dfrac{∂^2f}{∂y^2} : f_{yy}(x,y),\quad f_{22}(x,y),\quad D_{yy}(x,y),\quad D_{22}(x,y)$

$\nonumber \dfrac{∂^2f}{∂y∂x} : f_{xy}(x,y),\quad f_{12}(x,y),\quad D_{xy}(x,y),\quad D_{12}(x,y)$

$\nonumber \dfrac{∂^2f}{∂x∂y} : f_{yx}(x,y),\quad f_{21}(x,y),\quad D_{yx}(x,y),\quad D_{21}(x,y)$