Understanding Partial Derivatives of Functions

A partial derivative quantifies the instantaneous rate of change of a function with several variables when only one variable varies, with all others held constant. Partial differentiation forms a fundamental procedure in multivariable calculus.

Formal Definition and Notation of the Partial Derivative

Definition: Let $f$ be a function of $n$ variables, i.e., $f(x_1, x_2, ..., x_n)$. The partial derivative of $f$ with respect to the variable $x_i$ at the point $(a_1, ..., a_n)$ is defined as:

$\displaystyle \left.\frac{\partial f}{\partial x_i}\right|_{(a_1, ..., a_n)} = \lim_{h\to 0} \frac{f(a_1,...,a_i+h,...,a_n) - f(a_1,...,a_i,...,a_n)}{h}$

Common notations for the partial derivative of $f$ with respect to $x$ (assuming $f$ of two variables $x$ and $y$):

• $\dfrac{\partial f}{\partial x}$
• $f_x(x, y)$
• $\partial_x f(x, y)$

Partial derivatives exist for functions of more than two variables by following the same limiting process, treating others as constants.

Evaluation of Partial Derivatives for Multivariable Functions

To compute the partial derivative of $f(x, y, ...)$ with respect to one variable (say $x$), treat every other variable as a constant, and differentiate as in single-variable calculus.

If $f(x, y) = 2x^2 y^3$, then:

Example: Find $\dfrac{\partial f}{\partial x}$ and $\dfrac{\partial f}{\partial y}$.

Given $f(x, y) = 2x^2 y^3$.

Holding $y$ constant:

$\dfrac{\partial f}{\partial x} = \dfrac{\partial}{\partial x} (2x^2 y^3) = 4x y^3$

Holding $x$ constant:

$\dfrac{\partial f}{\partial y} = \dfrac{\partial}{\partial y} (2x^2 y^3) = 6x^2 y^2$

Properties and Notational Aspects of Partial Differentiation

Partial differentiation distinguishes between the variables, with explicit notation to indicate the variable differentiated. Derivatives of functions in one variable use $d$, while partial derivatives use the symbol $\partial$ for multivariable cases. The notation always indicates the variable of differentiation in a subscript or denominator.

Second and higher-order partial derivatives can be formed by applying partial differentiation multiple times, potentially in different orders:

$\displaystyle f_{xx} = \frac{\partial^2 f}{\partial x^2}$, $\qquad f_{xy} = \frac{\partial^2 f}{\partial y \partial x}$

The mixed partial derivatives $f_{xy}$ and $f_{yx}$ may be equal if the function is sufficiently smooth (Clairaut’s theorem for continuous second derivatives).

Evaluation for Standard Function Types and Rules

To compute partial derivatives in common settings:

• Polynomial functions: Differentiate as usual, treating other variables as constants.
• Product, quotient, and chain rules: Use analogously as for single-variable, but hold non-differentiated variables constant throughout.
• Logarithmic, exponential, and trigonometric functions: Differentiate with respect to the variable of interest, with all others constant.

For product $f(x, y) = x^2 \sin y$, $\displaystyle \frac{\partial f}{\partial x} = 2x \sin y$ (treated $y$ as constant).

For quotient $f(x, y) = \dfrac{x^2 + y}{y}$, $\displaystyle \frac{\partial f}{\partial x} = \dfrac{2x}{y}$ (treated $y$ as constant).

See Differential Calculus for analogous single-variable differentiation rules.

Partial Derivatives for Implicitly Defined and Composite Functions

If a function is defined implicitly, such as $F(x, y, z) = 0$ where $z = z(x, y)$, use implicit differentiation: when differentiating with respect to $x$, apply the chain rule to $z$-terms.

Example: Let $F(x, y, z) = x^2 + y^2 + z^2 - 1 = 0$ with $z = z(x, y)$. Find $\displaystyle \frac{\partial z}{\partial x}$.

Differentiate both sides with respect to $x$:

$2x + 2z \dfrac{\partial z}{\partial x} = 0$

$\implies \dfrac{\partial z}{\partial x} = -\dfrac{x}{z}$

For composite (chain rule) scenarios, suppose $f(x, y)$ where $x$ and $y$ both depend on $t$, then:

$\dfrac{df}{dt} = \dfrac{\partial f}{\partial x} \dfrac{dx}{dt} + \dfrac{\partial f}{\partial y} \dfrac{dy}{dt}$

For more on such relations, see Functional Equations and Differential Equations.

Worked Problems Involving Partial Differentiation

Example: If $f(x, y) = e^{x^2 y}$, compute $\dfrac{\partial f}{\partial x}$ and $\dfrac{\partial f}{\partial y}$.

$\dfrac{\partial f}{\partial x} = \dfrac{\partial}{\partial x}(e^{x^2 y}) = e^{x^2 y} \cdot 2x y$

$\dfrac{\partial f}{\partial y} = \dfrac{\partial}{\partial y}(e^{x^2 y}) = e^{x^2 y} \cdot x^2$

Example: If $f(x, y) = \ln(xy)$, find $\dfrac{\partial f}{\partial x}$ and $\dfrac{\partial f}{\partial y}$.

$\dfrac{\partial f}{\partial x} = \dfrac{1}{x}$ (holding $y$ constant).

$\dfrac{\partial f}{\partial y} = \dfrac{1}{y}$ (holding $x$ constant).

Example: For $f(x, y) = \dfrac{x}{y^2 + 1}$, calculate $\dfrac{\partial f}{\partial x}$ and $\dfrac{\partial f}{\partial y}$.

$\dfrac{\partial f}{\partial x} = \dfrac{1}{y^2 + 1}$ (since $y$ is constant).

$\dfrac{\partial f}{\partial y} = x \cdot \dfrac{-2y}{(y^2 + 1)^2}$ (by chain rule for $y$).

Example: If $z = x^2 y + y^2 \sin x$, compute $\displaystyle \frac{\partial^2 z}{\partial y \partial x}$.

First, $\dfrac{\partial z}{\partial x} = 2x y + y^2 \cos x$.

Now, $\dfrac{\partial}{\partial y}(2x y + y^2 \cos x) = 2x + 2y \cos x$.

Distinctions and Common Errors in Partial Differentiation

Partial derivatives differ from ordinary derivatives by requiring all variables except the one being differentiated to be constant in each step. Misidentifying which variable is held constant is a frequent student error, particularly in product or chain rule scenarios.

For additional practice on mixed partials, higher-order derivatives, and multi-step reasoning, consult Partial Derivative Of Functions and related concept pages.