# Partial Derivatives – Definition, Properties, and Example

Knowing how to calculate**partial derivatives**allows one to study and understand the behavior of multivariable functions. This opens a wide range of applications in Calculus such as the tangent planes, Lagrange multipliers, and more. For now, it’s important that we have a strong understanding of partial differentiation.

**In this article, we’ll cover the fundamentals of partial derivatives. This includes the partial derivative’s formal definition, common notations, and the techniques we can apply to calculate first-order, second-order, and even higher-order partial derivatives of different functions!**

*The partial derivatives allow us to understand how a multivariable function changes with respect to a specific variable. Partial differentiation works by treating the rest of the variables as constant.***What Is a Partial Derivative?**

**The partial derivative of a function represents**the derivative of the function

**with respect to one of the function’s variables**. There are instances when functions are defined by more than one independent variable. For multivariable functions, their values will change when one or more of the input values change. \begin{aligned}f(x, y) &= x^2y + 2xy – y^2\\g(x, y) &= \sin xy – \cos xy\\h(x, y, z) &= x^2 – 2xyz + yz + z^2\end{aligned} For us to calculate the rate of change of these functions with respect to one variable, we’ll have to hold the remaining variables constant. We call this

**process partial differentiation**and the rate of change to be the partial derivative of the function with respect to the variable we’re studying. Here’s an example of the curve formed by a function with two variables: $f(x, y) = x^2 + y^2$. If we set $y$ to be equal to $1$, we’ll be focusing on the plane that contains all values of $f(x, y)$ where $y =1$. When this happens, we’re working with $f(x, 1) = x^2 +1$, and it’ll be easier for us to measure the rate of change of $f(x, y)$ with respect to $x$. We’ll be using a similar process to calculate the partial derivatives of multivariable functions. To establish the partial derivative’s formal definition, we need to better understand how variables affect multivariable functions such as $f(x,y) = x^2y + 2xy – y^2$. So, why don’t we construct a table of values highlighting key values of $x$ and $y$?

\begin{aligned}\boldsymbol{x} \backslash^{\displaystyle\boldsymbol{y}}\end{aligned} | \begin{aligned}-2 \end{aligned} | \begin{aligned}-1 \end{aligned} | \begin{aligned}0 \end{aligned} | \begin{aligned}1 \end{aligned} | \begin{aligned}2 \end{aligned} |

\begin{aligned}-2 \end{aligned} | \begin{aligned}-4 \end{aligned} | \begin{aligned}\color{Teal}-1 \end{aligned} | \begin{aligned}0 \end{aligned} | \begin{aligned}1 \end{aligned} | \begin{aligned}4 \end{aligned} |

\begin{aligned}-1 \end{aligned} | \begin{aligned}-2 \end{aligned} | \begin{aligned}\color{Teal} 0\end{aligned} | \begin{aligned}0 \end{aligned} | \begin{aligned}-2\end{aligned} | \begin{aligned}-6\end{aligned} |

\begin{aligned}0 \end{aligned} | \begin{aligned}-4 \end{aligned} | \begin{aligned}\color{Teal} -1 \end{aligned} | \begin{aligned}0 \end{aligned} | \begin{aligned}-1 \end{aligned} | \begin{aligned}-4 \end{aligned} |

\begin{aligned}\color{Teal}1 \end{aligned} | \begin{aligned}\color{Teal}-10 \end{aligned} | \begin{aligned}\color{Teal}-4 \end{aligned} | \begin{aligned}\color{Teal}0 \end{aligned} | \begin{aligned}\color{Teal}2 \end{aligned} | \begin{aligned}\color{Teal}2 \end{aligned} |

\begin{aligned}2 \end{aligned} | \begin{aligned}-20 \end{aligned} | \begin{aligned}\color{Teal}-9 \end{aligned} | \begin{aligned}0 \end{aligned} | \begin{aligned}7 \end{aligned} | \begin{aligned}12\end{aligned} |

\begin{aligned}\boldsymbol{h = -1}\end{aligned} | \begin{aligned}\boldsymbol{h = 1}\end{aligned} |

\begin{aligned}f\prime(1) &=\dfrac{f(1 + -1) – f(1)}{-1}\\&= \dfrac{f(1 + -1, -1) – f(1, -1)}{-1}\\ &=\dfrac{f(0, -1) – f(1, -1)}{-1}\\&= \dfrac{1 – -4}{-1}\\&= -5\end{aligned} | \begin{aligned}f\prime(1) &=\dfrac{f(1 + 1) – f(1)}{1}\\&= \dfrac{f(1 + 1, -1) – f(1, -1)}{1}\\&=\dfrac{f(2, -1) – f(1, -1)}{1}\\&= \dfrac{-9 – -4}{1}\\&= -5\end{aligned} |

**Partial Derivative Definition**

To understand the underlying principles behind partial derivatives, let’s first understand the formal definition of partial derivatives through limits. Let’s say we have a continuous function, $z = f(x, y)$, we can define the following:
i. The **partial derivative of**$\boldsymbol{f(x, y)}$

**with respect to**$\boldsymbol{x}$ can be defined as: \begin{aligned}f_x(x, y) &= \lim_{h \rightarrow 0} \dfrac{f(x + h, y) – f(x, y)}{h}\end{aligned} ii. The

**partial derivative of**$\boldsymbol{f(x, y)}$

**with respect to**$\boldsymbol{y}$ can be defined as: \begin{aligned}f_y(x, y) &= \lim_{h \rightarrow 0} \dfrac{f(x , y + h) – f(x, y)}{h}\end{aligned} For example, we can show that when $f(x, y) = x^2 + y^2$, $f_x(x, y) = 2x$ by using the limit-based definition of $f(x, y)$ as shown below. \begin{aligned}f_x(x, y) &= \lim_{h \rightarrow 0} \dfrac{f(x + h, y) – f(x, y)}{h}\\&= \lim_{h \rightarrow 0} \dfrac{(x + h)^2 + y^2 – (x^2 + y^2)}{h}\\&= \lim_{h \rightarrow 0} \dfrac{x^2 + 2xh + h^2 +y^2 – x^2 – y^2}{h}\\ &= \lim_{h \rightarrow 0} \dfrac{2xh + h^2}{h}\\& = \lim_{h \rightarrow 0} \dfrac{h(2x + h)}{h}\\&= \lim_{h \rightarrow 0} (2x + h)\\&= 2x\end{aligned} This confirms that the partial derivative of $f(x, y)$ with respect to $x$ is equal to $2x$. As with our derivative lessons in the past, calculating the derivative of a function is tedious if we use the limit-based definition. This is why it’s important that we establish easier notations and processes for partial differentiation.

**Partial Derivative Notation**

There are different notations we can use to represent partial derivatives. One of the most common notations is $\dfrac{\partial f}{\partial x}$ or $\dfrac{\partial f}{\partial y}$. We call the symbol $\partial$, “del”, and this is one of the surest ways to know that we’re evaluating the partial derivative of the function.
Let’s say we have $f(x, y) = 2x^2 – 4xy + y^2$, we can represent the partial derivatives:
- with respect to $x$ as $\dfrac{\partial f}{\partial x}$
- with respect to $y$ as $\dfrac{\partial f}{\partial y}$

\begin{aligned}\boldsymbol{\dfrac{\partial f}{\partial x}}\end{aligned} | \begin{aligned}\boldsymbol{\dfrac{\partial f}{\partial y}}\end{aligned} |

\begin{aligned}\dfrac{\partial f}{\color{Teal}\partial x} &=\underbrace{\dfrac{\partial }{\color{Teal}\partial x} (2{\color{Teal}x^2} – 4{\color{Teal}x}y + y^2)}_{\displaystyle{\text{Take the derivative of }x}}\\&\phantom{xxx}\text{Treat} y \text{ as a constant}\\&= 2({\color{Teal}2x}) – 4({\color{Teal}1})y + 0\\&= 4x – 4y \end{aligned} | \begin{aligned}\dfrac{\partial f}{\color{DarkOrange}\partial y} &=\underbrace{\dfrac{\partial }{\color{DarkOrange}\partial y} (2{x^2} – 4x{\color{DarkOrange}y} + {\color{DarkOrange}y^2})}_{\displaystyle{\text{Take the derivative of }y}}\\&\phantom{xxx}\text{Treat} x \text{ as a constant}\\&= 0- 4x({\color{DarkOrange}1}) + {\color{DarkOrange}2y}\\&= -4x + 2y\\&= 2y – 4x \end{aligned} |

**There are other ways for us to represent partial derivatives such as $f_x$ or $f_1$. For $f_1$, the number represents the order of the variable, so for this case, we want to take the partial derivative of $f$ with respect to the first variable.**

Suppose that $z = f(x, y)$ is a continuous function, we can write their partial derivatives using any of the following notations: \begin{aligned}f_x(x, y) = f_x = \dfrac{\partial}{\partial x}f(x, y) = \dfrac{\partial z}{\partial x} = f_1 = D_1f = D_xf\\f_y(x, y) = f_y = \dfrac{\partial}{\partial y}f(x, y) = \dfrac{\partial z}{\partial y} = f_2 = D_2f = D_yf\end{aligned} |

- “Del $f$ over del $x$.”
- “Partial of $f$ with respect to $x$.”
- “The partial derivative of $f$ with respect to $x$.”
- “Partial derivative of $f$ with respect to the first variable.”

**second**

**partial derivative of**$\boldsymbol{f(x, y)}$

**with respect to**$\boldsymbol{x}$

**then**$\boldsymbol{x}$ can be represented as: \begin{aligned}\dfrac{\partial}{\partial x}\left( \dfrac{\partial f}{\partial x}\right ) = \dfrac{\partial^2 f}{\partial x^2} = f_{xx}\end{aligned} ii. The

**second**

**partial derivative of**$\boldsymbol{f(x, y)}$

**with respect to**$\boldsymbol{y}$

**then**$\boldsymbol{y}$ can be represented as: \begin{aligned}\dfrac{\partial}{\partial y}\left( \dfrac{\partial f}{\partial y}\right ) = \dfrac{\partial^2 f}{\partial y^2} = f_{yy}\end{aligned} iii. The

**second**

**partial derivative of**$\boldsymbol{f(x, y)}$

**with respect to**$\boldsymbol{x}$

**then**$\boldsymbol{y}$ can be represented as: \begin{aligned}\dfrac{\partial}{\partial y}\left( \dfrac{\partial f}{\partial x}\right ) = \dfrac{\partial^2 f}{\partial y \partial x} = f_{xy}\end{aligned} iv. The

**second**

**partial derivative of**$\boldsymbol{f(x, y)}$

**with respect to**$\boldsymbol{y}$

**then**$\boldsymbol{x}$ can be represented as: \begin{aligned}\dfrac{\partial}{\partial x}\left( \dfrac{\partial f}{\partial y}\right ) = \dfrac{\partial^2 f}{\partial x \partial y} = f_{yx}\end{aligned} Keep in mind that these notations will only be valid when the limit exists for each condition. We can also apply a similar process to represent higher-order partial derivatives, but for now, let’s understand how we can calculate partial derivatives and second partial derivatives.

**How To Do Partial Derivatives?**

We can calculate partial derivatives by applying the definition of partial differentiation. Keep in mind that we only need to find the derivative of functions with respect to one variable by keeping the rest of the variables constant.
Here are some pointers to remember when calculating first-order partial derivatives:
- Identify the variable we’re differentiating. For example, when working with $\dfrac{\partial f}{\partial x}$, we different $f(x)$ with respect to $x$.
- Treat the rest of the variables as constants.
- Apply fundamental derivative rules to different $f(x)$ with respect to our variable.

- $\dfrac{\partial f}{\partial x} = f_x = 4xy – 4$
- $\dfrac{\partial f}{\partial y} = f_y = 2x^2 – 4x$

- Start differentiating $f(x)$ with respect to the first variable, $v_1$.
- Differentiate the resulting expression with respect to $v_2$ and treating $v_1$ as constant.

**Use the formal definition of partial derivative to find $f_x(x, y)$ when we have the function, $f(x, y) = xy^2 + x^2y$.**

*Example 1*

__Solution__As we have discussed earlier, we can define the partial derivative of a function in terms of limits as shown below. \begin{aligned}f_x(x, y) &= \lim_{h \rightarrow 0} \dfrac{f(x + h, y) – f(x, y)}{h}\end{aligned} Use this definition then simplify the numerator as shown below. \begin{aligned}f_x(x, y) &= \lim_{h \rightarrow 0} \dfrac{f(x + h, y)- f(x, y)}{h}\\ &= \lim_{h \rightarrow 0} \dfrac{(x + h)y^2 + (x + h)^2y- (xy^2 + x^2y)}{h}\\&= \lim_{h \rightarrow 0} \dfrac{(xy^2 + y^2h) + (x^2 + 2xh + h^2)y – xy^2 – x^2y}{h}\\&= \lim_{h \rightarrow 0} \dfrac{xy^2 + y^2h + x^2y + 2xhy + h^2y- xy^2 – x^2y}{h}\end{aligned} Cancel the common factor, $h$, shared by the numerator and denominator then evaluate the limit of the resulting expression as $h \rightarrow 0$. \begin{aligned}f_x(x, y) &= \lim_{h \rightarrow 0} y^2 + 2xy + hy\\&= y^2 + 2xy\end{aligned} This means that the partial derivative of $f(x, y)$ with respect to $x$ is equal to $y^2 + 2xy$.

**Determine all of the first-order partial derivatives of the following functions. a. $f(x, y) = x^3 – 2\sqrt{y} + \dfrac{5}{x} – 6$ b. $f(x, y, z) = x^2y + 6yz^2 – 6\sin x \cos y + 8xz$**

*Example 2*__Solution__We’re being asked to write down all of the first-order partial derivatives of each of the functions, so for our first function, we need to determine $\dfrac{\partial f}{\partial x}$ and $\dfrac{\partial f}{\partial y}$.

- To calculate $\dfrac{\partial f}{\partial x}$ or $f_x(x, y)$, we’ll take the derivative of the function with respect to $x$ by treating $y$ as a constant.
- Similarly, we can calculate $\dfrac{\partial f}{\partial y}$ or $f_y(x, y)$ by treating $x$ as a constant.

\begin{aligned}\boldsymbol{\dfrac{\partial f}{\partial x} = f_x(x, y)}\end{aligned} | \begin{aligned}\dfrac{\partial }{\partial x} f(x, y) &= \dfrac{\partial f}{\partial {\color{Teal}x}}\left({\color{Teal}x^3} – 2\sqrt{y} + {\color{Teal}\dfrac{5}{x} }- 6 \right )\\&= ({\color{Teal}3x^2}) – 0 + ({\color{Teal}-5x^{-1 -1}}) – 0\\&= 3x^2 – 5x^{-2}\\&= 3x^2 – \dfrac{5}{x^2}\end{aligned} |

\begin{aligned}\boldsymbol{\dfrac{\partial f}{\partial y} = f_y(x, y)}\end{aligned} | \begin{aligned}\dfrac{\partial }{\partial x} f(x, y) &= \dfrac{\partial f}{\partial {\color{DarkOrange}y}}\left(0 – {\color{DarkOrange} 2\sqrt{y} } +\dfrac{5}{x} {\color{DarkOrange} – 6} \right )\\&= x^3 – \left({\color{DarkOrange}2 \cdot \dfrac{1}{2}y^{\frac{1}{2}-1}}\right) + 0 + ({\color{DarkOrange} 0})\\&=- y^{-\frac{1}{2}}\\&= -\dfrac{1}{\sqrt{y}}\end{aligned} |

\begin{aligned}\boldsymbol{\dfrac{\partial f}{\partial x} = f_x(x, y, z)}\end{aligned} | \begin{aligned}\dfrac{\partial }{\partial x} f(x, y, z) &= \dfrac{\partial f}{\partial {\color{Teal}x}} ({\color{Teal}x^2}y + 6yz^2 – 6{\color{Teal}\sin x} \cos y + {\color{Teal}8x}z)\\&= ({\color{Teal} 2x})y + 0 – 6({\color{Teal}}\cos x)\cos y + ({\color{Teal}8 \cdot 1})z\\&= 2xy – 6 \cos x\cos y + 8z\end{aligned} |

\begin{aligned}\boldsymbol{\dfrac{\partial f}{\partial y} = f_y(x, y, z)}\end{aligned} | \begin{aligned}\dfrac{\partial }{\partial x} f(x, y, z) &= \dfrac{\partial f}{\partial {\color{DarkOrange}y}} (x^2{\color{DarkOrange}y} + {\color{DarkOrange}6y}z^2 – 6\sin x{\color{DarkOrange}\cos y} + 8xz)\\&= x^2({\color{DarkOrange} 1}) + 6({\color{DarkOrange}1})z^2 – 6\sin x \cdot{\color{DarkOrange}-\sin y} + 0\\&= x^2 + 6z^2 + 6\sin x \sin y\end{aligned} |

\begin{aligned}\boldsymbol{\dfrac{\partial f}{\partial y} = f_z(x, y, z)}\end{aligned} | \begin{aligned}\dfrac{\partial }{\partial x} f(x, y, z) &= \dfrac{\partial f}{\partial {\color{Purple}z}} (x^2y + 6y{\color{Purple}z^2 }- 6\sin x\cos y + 8x{\color{Purple}z})\\&= 0 + 6y({\color{Purple}2z}) – 0 + 8x({\color{Purple}1})\\&= 12yz + 8x\end{aligned} |

**Suppose that we have $f(x,y, z) = x^2y^4z – 2x^2y + 12xy^2z^2$, determine the following higher order partial derivatives: a. $\dfrac{\partial^2 f}{\partial y^2}$ b. $\dfrac{\partial^2 f}{\partial x \partial z}$ c. $\dfrac{\partial^3 f}{\partial y \partial x \partial z}$**

*Example 3*__Solution__To evaluate $\dfrac{\partial^2 f}{\partial y^2}$, we’ll need to differentiate $f(x, y, z)$ with respect to $y$ twice in a row. Treat both $x$ and $z$ as constant values as shown below. \begin{aligned}\dfrac{\partial^2 f}{\partial y^2}&= \dfrac{\partial }{\partial y}\left(\dfrac{\partial f}{\partial y} \right ) \\&= \dfrac{\partial }{\partial y}\left[\dfrac{\partial }{\partial y} (x^2y^4z – 2x^2y + 12xy^2z^2) \right]\\&= \dfrac{\partial}{\partial y}\left[ x^2(4y^3)z – 2x^2(1) + 12xz^2(2y)\right]\\&= \dfrac{\partial}{\partial y}\left(4x^2y^3z – 2x^2 + 24xyz^2 \right ) \end{aligned} Repeat the process to find the expression for $\dfrac{\partial^2 f}{\partial y^2}$. \begin{aligned}\dfrac{\partial^2 f}{\partial y^2} &= \dfrac{\partial}{\partial y}\left(4x^2y^3z – 2x^2 + 24xyz^2 \right ) \\&= 4x^2(3y^2)z – 0 + 24xz^2(1)\\&= 12x^2y^2z + 24xz^2 \end{aligned} a. This means that $\dfrac{\partial^2 f}{\partial y^2}$ is equal to $12x^2y^2z + 24xz^2$. For the next time, we’ll take the partial derivative of $f(x, y, z)$ with respect to $z$ first. \begin{aligned}\dfrac{\partial^2 f}{\partial x \partial z} &= \dfrac{\partial}{\partial x}\left[\dfrac{\partial}{\partial z} (x^2y^4z – 2x^2y + 12xy^2z^2) \right] \\&= \dfrac{\partial}{\partial x} \left[x^2y^4 (1) – 0 + 12xy^2(2z) \right ]\\&= \dfrac{\partial}{\partial x}\left(x^2y^4 + 24xy^2z \right ) \end{aligned} Now, differentiate the resulting expression with respect to $x$. \begin{aligned}\dfrac{\partial^2 f}{\partial x \partial z} &= \dfrac{\partial}{\partial x}\left(x^2y^4 + 24xy^2z \right )\\&= y^4(2x) + 24yz(1)\\&= 2y^4x + 24y^2z \end{aligned} b. Hence, $\dfrac{\partial^2 f}{\partial x \partial z}$ is equivalent to $2y^4x + 24y^2z$. We can rewrite $\dfrac{\partial^3 f}{\partial y \partial x \partial z}$ as $\dfrac{\partial}{\partial y} \left(\dfrac{\partial^2 f}{\partial x \partial z} \right)$, so let’s use our result from the Q3b then differentiate it with respect to $y$. \begin{aligned}\dfrac{\partial^3 f}{\partial y \partial x \partial z} &= \dfrac{\partial}{\partial y}\left( \dfrac{\partial^2 f}{\partial x \partial z}\right ) \\ &= \dfrac{\partial}{\partial y} (2y^4x + 24y^2z ) \\&= 2x(4y^3) + 24z(2y)\\&= 8xy^3 + 48yz\end{aligned} c. This means that $\dfrac{\partial^3 f}{\partial y \partial x \partial z}$ is equal to $8xy^3 + 48yz$.

**Practice Question**

1. Use the formal definition of partial derivative to find $f_y(x, y)$ when we have the function, $f(x, y) = x^2 – 2xy + y^2$.
2. Determine all of the first-order partial derivatives of the following functions.
a. $f(x, y) = 2y^2 – 3\sqrt{x} + 4xy^2 – \dfrac{x}{y}$
b. $f(x, y, z) = x – 6xyz + y\sin x\cos z – \sqrt{yz}$
c. $f(x, y,z) = \sin(x^3y + z) – \cos(y^2 – xz)$
3. Suppose that we have $f(x,y, z) = -3x^2y + 3xyz – 4y^2z^2 + 6\sin x \cos z$, determine the following higher order partial derivatives:
a. $\dfrac{\partial^2 f}{\partial x^2}$
b. $\dfrac{\partial^2 f}{\partial y \partial z}$
c. $\dfrac{\partial^3 f}{\partial x \partial y \partial z}$
**Answer Key**

1.
$ \begin{aligned}f_y(x, y) &= \lim_{h \rightarrow 0} \dfrac{[x^2 – 2x(y + h) + (y + h)^2] – (x^2 – 2xy + y^2)}{h}\\&= \lim_{h \rightarrow 0} \dfrac{(x^2 – 2xy – 2hx + y^2 + 2yh + h^2 – x^2 + 2xy – y^2)}{h}\\&=\lim_{h \rightarrow 0} \dfrac{(- 2hx + 2yh + h^2)}{h}\\&=\lim_{h \rightarrow 0} -2x + 2y + h\\&= -2x + 2y\end{aligned}$
2.
a. $f_x = 4y^2- \dfrac{1}{y}- \dfrac{3}{2\sqrt{x}}$, $f_y = y+ 8yx+ \dfrac{x}{y^2}$
b. $f_x = y\cos x \cos z – 6yz +1$, $f_y = -6xz+\sin x \cos z- \dfrac{\sqrt{z}}{2\sqrt{y}}$, $f_z = -6xy- y\sin x \sin z- \dfrac{\sqrt{y}}{2\sqrt{z}}$
c. $f_x = \cos \left(x^3y +z\right) \cdot \:3yx^2 – z\sin \left(y^2 – zx\right)$, $f_y = x^3 \cos \left(x^3y + z\right) + 2y\sin \left(y^2 – xz\right)$, $f_z = \cos \left(x^3y + z\right) – x\sin \left(y^2 – xz\right)$
3.
a. $\dfrac{\partial^2 f}{\partial x^2} = -6y – 6\sin x\cos z$
b. $\dfrac{\partial^2 f}{\partial y \partial z} = 3x – 16yz$
c. $\dfrac{\partial^3 f}{\partial x \partial y \partial z} = 3$
*3D images/mathematical drawings are created with GeoGebra.*

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