# Partial Derivatives of Multivariable Functions

Examples and exercises on the calculations of partial derivatives are presented.

While ordinary derivatives deal with functions of a single variable, partial derivatives are a type of derivative that generalize the concept of ordinary derivatives to multivariable functions.

Formally, the partial derivative of a function $$f(x_1, x_2, ..., x_n)$$ with respect to one of its variables, say $$x_i$$, is denoted by $$\dfrac{\partial f}{\partial x_i}$$. It represents the rate of change of the function $$f$$ with respect to the variable $$x_i$$, while holding all other variables constant.

Mathematically, the partial derivative of $$f$$ with respect to $$x_i$$ is defined as: $\dfrac{\partial f}{\partial x_i} = \lim_{h \to 0} \dfrac{f(x_1, x_2, ..., x_i + h, ..., x_n) - f(x_1, x_2, ..., x_i, ..., x_n)}{h}$ In words, this definition says that the partial derivative of $$f$$ with respect to $$x_i$$ is the limit of the difference quotient as $$h$$ approaches zero, where $$h$$ represents a small change in the variable $$x_i$$, while keeping all other variables constant.

Partial derivatives allow us to analyze how a function changes with respect to one of its variables while keeping the others fixed.
When calculating the partial derivative of a function $$f(x, y)$$ with respect to $$x$$, denoted as $$\dfrac{\partial f}{\partial x}$$, we treat $$y$$ as a constant.
Similarly, when calculating the partial derivative of $$f$$ with respect to $$y$$, denoted as $$\dfrac{\partial f}{\partial y}$$, we treat $$x$$ as a constant. We only consider how $$f$$ changes with respect to variations in $$y$$, while keeping $$x$$ constant.
This is the fundamental concept behind partial derivatives, allowing us to analyze how a function changes with respect to one variable while holding others constant.

Partial derivatives are used extensively in calculus, differential equations, optimization, and various fields of science and engineering, including physics, economics, and engineering. They play a crucial role in the study of multivariable calculus and the analysis of systems with multiple independent variables. A partial derivative calculator is included and may be used to check calculkations.

## Examples with Solutions

### Example 1

Calculate the partial derivatives of $$f$$ with respect to $$x$$ denoted by $$\dfrac{\partial f}{\partial x}$$ and the partial derivatives of $$f$$ with respect to $$y$$ denoted by $$\dfrac{\partial f}{\partial y}$$ where $$f$$ is given by $f(x, y) = 3x^2 + 4xy - y^2$.

### Solution to Example 1

#### 1. Derivative of $$3x^2 + 4xy - y^2$$ with respect to $$x$$:

We calculate the partial derivative of $$f$$ with respect to $$x$$, denoted as $$\dfrac{\partial f}{\partial x}$$. $\dfrac{\partial f}{\partial x} = \dfrac{\partial}{\partial x} (3x^2 + 4xy - y^2)$ Using the rule of the sum, we write $\dfrac{\partial f}{\partial x} = \dfrac{\partial}{\partial x} (3x^2) + \dfrac{\partial}{\partial x} (4xy) - \dfrac{\partial}{\partial x} (y^2)$ NOTE that when calculating the partial derivative of a function $$f(x, y)$$ with respect to $$x$$, we treat $$y$$ as a constant. Using the power rule for differentiation, we have: $\dfrac{\partial}{\partial x} (3x^2) = 6x$ $\dfrac{\partial}{\partial x} (4xy) = 4y$ $\dfrac{\partial}{\partial x} (y^2) = 0$ Hence $\dfrac{\partial f}{\partial x} = 6x + 4y - 0$ $\dfrac{\partial f}{\partial x} = 6x + 4y$.

#### 2. Derivative of $$3x^2 + 4xy - y^2$$ with respect to $$y$$:

We now calculate the partial derivative of $$f$$ with respect to $$y$$, denoted as $$\dfrac{\partial f}{\partial y}$$. $\dfrac{\partial f}{\partial x} = \dfrac{\partial}{\partial y} (3x^2 + 4xy - y^2)$ Using the rule of the sum, we write $\dfrac{\partial f}{\partial x} = \dfrac{\partial}{\partial y} (3x^2) + \dfrac{\partial}{\partial y} (4xy) - \dfrac{\partial}{\partial y} (y^2)$ NOTE that when calculating the partial derivative of a function $$f(x, y)$$ with respect to $$y$$, we treat $$x$$ as a constant. Using different rules for differentiation, we have: $\dfrac{\partial}{\partial y} (3x^2) = 0$ $\dfrac{\partial}{\partial y} (4xy)= 4x$ $\dfrac{\partial}{\partial y} (y^2) = 2y$ Hence $\dfrac{\partial f}{\partial y} = 0 + 4x - 2y$ $\dfrac{\partial f}{\partial y} = 4x - 2y$.

### Example 2

Calculate the partial derivatives of $$g$$ with respect to $$x$$, $$y$$ and $$z$$, where $$g$$ is given by $g(x, y, z) = e^{xy} \cos(z)$.

### Solution to Example 2

To calculate the partial derivatives of $$g(x, y, z) = e^{xy} \cos(z)$$ with respect to $$x$$, $$y$$, and $$z$$, we treat each variable as independent and differentiate each term of $$g$$ with respect to the respective variable while holding the other variables constant. Let's calculate each partial derivative step by step:

#### 1. Partial derivative with respect to $$x$$:

$\dfrac{\partial g}{\partial x} = \dfrac{\partial}{\partial x} \left( e^{xy} \cos(z) \right)$ Using the product rule for differentiation: $\dfrac{\partial}{\partial x} \left( e^{xy} \cos(z) \right) = \dfrac{\partial}{\partial x} \left( e^{xy} \right) \cos(z) + e^{xy} \dfrac{\partial}{\partial x} \left( \cos(z) \right)$ Now, we calculate each term separately: $\dfrac{\partial}{\partial x} \left( e^{xy} \right) = y e^{xy}$ $\dfrac{\partial}{\partial x} \left( \cos(z) \right) = 0$ So, putting it all together: $\dfrac{\partial g}{\partial x} = y e^{xy} \cos(z)$

#### 2. Partial derivative with respect to $$y$$:

$\dfrac{\partial g}{\partial y} = \dfrac{\partial}{\partial y} \left( e^{xy} \cos(z) \right)$ Using the product rule for differentiation: $\dfrac{\partial}{\partial y} \left( e^{xy} \cos(z) \right) = \dfrac{\partial}{\partial y} \left( e^{xy} \right) \cos(z) + e^{xy} \dfrac{\partial}{\partial y} \left( \cos(z) \right)$ Now, we calculate each term separately: $\dfrac{\partial}{\partial y} \left( e^{xy} \right) = x e^{xy}$ Since $$\cos(z)$$ does not depend on $$y$$, its derivative with respect to $$y$$ is zero: $\dfrac{\partial}{\partial y} \left( \cos(z) \right) = 0$ So, putting it all together: $\dfrac{\partial g}{\partial y} = x e^{xy} \cos(z)$

#### 3. Partial derivative with respect to $$z$$:

$\dfrac{\partial g}{\partial z} = \dfrac{\partial}{\partial z} \left( e^{xy} \cos(z) \right)$ Using the product rule for differentiation: $\dfrac{\partial}{\partial z} \left( e^{xy} \cos(z) \right) = \dfrac{\partial}{\partial z} \left( e^{xy} \right) \cos(z) + e^{xy} \dfrac{\partial}{\partial z} \left( \cos(z) \right)$ Now, we calculate each term separately: a. Derivative of $$e^{xy}$$ with respect to $$z$$: Since $$e^{xy}$$ does not depend on $$z$$, its derivative with respect to $$z$$ is zero: $\dfrac{\partial}{\partial z} \left( e^{xy} \right) = 0$ b. Derivative of $$\cos(z)$$ with respect to $$z$$: $\dfrac{\partial}{\partial z} \left( \cos(z) \right) = -\sin(z)$ So, putting it all together: $\dfrac{\partial g}{\partial z} = - e^{xy} \sin(z)$ Therefore, the partial derivatives of $$g$$ with respect to $$x$$, $$y$$, and $$z$$ are: $\dfrac{\partial g}{\partial x} = y e^{xy} \cos(z)$ $\dfrac{\partial g}{\partial y} = x e^{xy} \cos(z)$ $\dfrac{\partial g}{\partial z} = - e^{xy} \sin(z)$

## Exercises with Solutions

Find the partial derivatives of the functions
1. $$g(u,v) = u^2 \; v^2 + e^{u^2+v^2}$$
2. $$f(x,y,z) = \sin (xy )\;\ln (xyz )$$
3. $$h(x,y,z) = \dfrac{z}{x \;y \;z +1}$$

### Solutions to the Above Exercises

1. $$\dfrac{\partial g}{\partial u} = 2 \;u \;v^2 + 2u \; e^{u^2+v^2}$$
$$\dfrac{\partial g}{\partial v} = 2 \;v \;u^2 + 2v \; e^{u^2+v^2}$$

2. $$\dfrac{\partial f}{\partial x} = y \;\cos(xy)\;\ln (xyz )+\dfrac{\sin (xy )}{x}$$
$$\dfrac{\partial f}{\partial y} = x\;\cos (xy )\;\ln (xyz )+\dfrac{\sin (xy )}{y}$$
$$\dfrac{\partial f}{\partial z} = \dfrac{\sin (xy)}{z}$$

3. $$\dfrac{\partial h}{\partial x} = -\dfrac{y z^2 }{\left(zxy+1\right)^2}$$
$$\dfrac{\partial h}{\partial y} = -\dfrac{x z^2}{\left(zxy+1\right)^2}$$
$$\dfrac{\partial h}{\partial z} = \dfrac{1}{\left(zxy+1\right)^2}$$