Partial Derivatives | Mathematics

Let $f : D \subseteq \mathbb{R}^n \to \mathbb{R}$. The partial derivative of $f$ with respect to $x_i$ at $\mathbf{a} = (a_1, \ldots, a_n)$ is

$\frac{\partial f}{\partial x_i}(\mathbf{a}) = \lim_{h \to 0} \frac{f(a_1, \ldots, a_i + h, \ldots, a_n) - f(a_1, \ldots, a_n)}{h}$

Provided the limit exists. This is the rate of change of $f$ in the direction of the $x_i$-axis, Holding all other variables fixed.

Notation. Common notations for the partial derivative with respect to $x_i$ include $f_{x_i}$, $\partial_i f$ And $\frac{\partial f}{\partial x_i}$. We use these interchangeably.

Theorem 1.1 (Clairaut’s Theorem / Schwarz’s Theorem). If $f_{xy}$ and $f_{yx}$ are continuous on an Open set containing $(a, b)$ Then

$\frac{\partial^2 f}{\partial x \partial y}(a,b) = \frac{\partial^2 f}{\partial y \partial x}(a,b)$

Proof. Define the second-order difference function

$\Delta(h, k) = f(a+h,\, b+k) - f(a+h,\, b) - f(a,\, b+k) + f(a, b)$

For $h, k \neq 0$. Define $\phi(s) = f(s, b+k) - f(s, b)$. Then $\Delta(h,k) = \phi(a+h) - \phi(a)$. By the Mean Value Theorem, there exists $\theta_1 \in (0, 1)$ such that

$\Delta(h, k) = h \cdot \phi'(a + \theta_1 h) = h \left[f_x(a + \theta_1 h,\, b+k) - f_x(a + \theta_1 h,\, b)\right]$

Apply the Mean Value Theorem again to the function $g(t) = f_x(a + \theta_1 h,\, t)$ on $[b, b+k]$. There exists $\theta_2 \in (0, 1)$ such that

$\Delta(h, k) = hk \cdot f_{xy}(a + \theta_1 h,\, b + \theta_2 k)$

Similarly, by reversing the order of application, there exist $\theta_3, \theta_4 \in (0,1)$ such That

$\Delta(h, k) = hk \cdot f_{yx}(a + \theta_3 h,\, b + \theta_4 k)$

For $h, k \neq 0$ we have

$f_{xy}(a + \theta_1 h,\, b + \theta_2 k) = f_{yx}(a + \theta_3 h,\, b + \theta_4 k)$

Taking the limit as $(h, k) \to (0, 0)$ and using continuity of $f_{xy}$ and $f_{yx}$We obtain $f_{xy}(a, b) = f_{yx}(a, b)$. $\blacksquare$

Intuition. Clairaut’s theorem tells us that, under a mild regularity condition (continuity of the Mixed second partials), the order in which we differentiate does not matter. Without this Condition, the mixed partials may differ.

Definition. $f : D \subseteq \mathbb{R}^n \to \mathbb{R}$ is differentiable at $\mathbf{a}$ if There exists a linear map $L : \mathbb{R}^n \to \mathbb{R}$ such that

$\lim_{\mathbf{h} \to \mathbf{0}} \frac{f(\mathbf{a} + \mathbf{h}) - f(\mathbf{a}) - L(\mathbf{h})}{\lVert \mathbf{h} \rVert} = 0$

When $f$ is differentiable at $\mathbf{a}$The linear map $L$ is given by the gradient.

Remark. Existence of all partial derivatives at a point does not imply differentiability at That point. The canonical counterexample is

$f(x,y) = \begin{cases} \dfrac{xy}{x^2 + y^2} & \mathrm{if\ }(x,y) \neq (0,0), \\ 0 & \mathrm{if\ }(x,y) = (0,0). \end{cases}$

Both $f_x(0,0)$ and $f_y(0,0)$ exist (and equal $0$), yet $f$ is not even continuous at the origin, Hence not differentiable.

The gradient of $f$ at $\mathbf{a}$ is

$\nabla f(\mathbf{a}) = \left(\frac{\partial f}{\partial x_1}(\mathbf{a}), \ldots, \frac{\partial f}{\partial x_n}(\mathbf{a})\right)$

The linear approximation of $f$ near $\mathbf{a}$ is

$f(\mathbf{a} + \mathbf{h}) \approx f(\mathbf{a}) + \nabla f(\mathbf{a}) \cdot \mathbf{h}$

Theorem 1.2. If all partial derivatives of $f$ exist and are continuous in a neighbourhood of $\mathbf{a}$ Then $f$ is differentiable at $\mathbf{a}$.

Remark. Functions whose partial derivatives exist and are continuous on an open set $U$ are called $C^1(U)$. Theorem 1.2 says $C^1 \implies$ differentiable. The converse is false: there exist Differentiable functions whose partial derivatives are not continuous.

Proposition. If $f$ is differentiable at $\mathbf{a}$ Then $f$ is continuous at $\mathbf{a}$.

Proof. From the definition of differentiability:

$f(\mathbf{a} + \mathbf{h}) - f(\mathbf{a}) = L(\mathbf{h}) + \varepsilon(\mathbf{h})\lVert \mathbf{h} \rVert$

Where $L$ is linear and $\varepsilon(\mathbf{h}) \to 0$ as $\mathbf{h} \to \mathbf{0}$. As $\mathbf{h} \to \mathbf{0}$ Both terms on the right vanish, so $f(\mathbf{a} + \mathbf{h}) \to f(\mathbf{a})$. $\blacksquare$

The directional derivative of $f$ at $\mathbf{a}$ in the direction of a unit vector $\mathbf{u}$ is

$D_{\mathbf{u}} f(\mathbf{a}) = \lim_{h \to 0} \frac{f(\mathbf{a} + h\mathbf{u}) - f(\mathbf{a})}{h}$

Theorem 1.3. If $f$ is differentiable at $\mathbf{a}$ Then

$D_{\mathbf{u}} f(\mathbf{a}) = \nabla f(\mathbf{a}) \cdot \mathbf{u}$

Proof. Since $f$ is differentiable at $\mathbf{a}$

$\frac{f(\mathbf{a} + h\mathbf{u}) - f(\mathbf{a})}{h} = \frac{\nabla f(\mathbf{a}) \cdot (h\mathbf{u}) + \varepsilon(h\mathbf{u}) \lVert h\mathbf{u} \rVert}{h}$

$= \nabla f(\mathbf{a}) \cdot \mathbf{u} + \varepsilon(h\mathbf{u}) \lVert \mathbf{u} \rVert$

Where $\varepsilon(\mathbf{h}) \to 0$ as $\mathbf{h} \to \mathbf{0}$. Taking $h \to 0$ gives the result. $\blacksquare$

Corollary 1.4. The gradient points in the direction of steepest ascent, and $\lVert \nabla f \rVert$ Is the rate of steepest ascent.

Proof. By the Cauchy—Schwarz inequality, $\lvert \nabla f \cdot \mathbf{u} \rvert \leq \lVert \nabla f \rVert \cdot \lVert \mathbf{u} \rVert = \lVert \nabla f \rVert$ With equality when $\mathbf{u}$ is parallel to $\nabla f$. $\blacksquare$

Theorem 1.5 (Multivariable Chain Rule). If $\mathbf{g} : \mathbb{R}^m \to \mathbb{R}^n$ is Differentiable at $\mathbf{a}$ and $f : \mathbb{R}^n \to \mathbb{R}$ is differentiable at $\mathbf{g}(\mathbf{a})$ Then

$\nabla (f \circ \mathbf{g})(\mathbf{a}) = J\mathbf{g}(\mathbf{a})^T \nabla f(\mathbf{g}(\mathbf{a}))$

Where $J\mathbf{g}$ is the Jacobian matrix of $\mathbf{g}$.

Proof. Write $h(t) = f(\mathbf{g}(\mathbf{a} + t\mathbf{v}))$ for a fixed direction $\mathbf{v}$. Then

$\frac{h(t) - h(0)}{t} = \frac{f(\mathbf{g}(\mathbf{a} + t\mathbf{v})) - f(\mathbf{g}(\mathbf{a}))}{t}$

Let $\mathbf{k} = \mathbf{g}(\mathbf{a} + t\mathbf{v}) - \mathbf{g}(\mathbf{a})$. By differentiability of $\mathbf{g}$ $\mathbf{k} = J\mathbf{g}(\mathbf{a})(t\mathbf{v}) + o(t)$ And $\mathbf{k} \to \mathbf{0}$ as $t \to 0$. By Differentiability of $f$:

$f(\mathbf{g}(\mathbf{a}) + \mathbf{k}) - f(\mathbf{g}(\mathbf{a})) = \nabla f(\mathbf{g}(\mathbf{a})) \cdot \mathbf{k} + o(\lVert \mathbf{k} \rVert)$

$= \nabla f(\mathbf{g}(\mathbf{a})) \cdot [J\mathbf{g}(\mathbf{a})(t\mathbf{v}) + o(t)] + o(t)$

Dividing by $t$ and taking $t \to 0$:

$h'(0) = \nabla f(\mathbf{g}(\mathbf{a})) \cdot J\mathbf{g}(\mathbf{a})\mathbf{v} = [J\mathbf{g}(\mathbf{a})^T \nabla f(\mathbf{g}(\mathbf{a}))] \cdot \mathbf{v}$

Since $\mathbf{v}$ was arbitrary, $\nabla h(0) = J\mathbf{g}(\mathbf{a})^T \nabla f(\mathbf{g}(\mathbf{a}))$. $\blacksquare$

Problem. Let $f(x, y) = x^2 y$ and let $x = \cos t$, $y = \sin t$. Find $\frac{d}{dt} f(\cos t, \sin t)$ Using the chain rule, and verify by direct substitution.

Problem. Let $f(x, y) = x^2 y + \sin(xy)$. Compute $\nabla f$ and find the directional derivative At $(1, \pi)$ in the direction $\mathbf{u} = (1/\sqrt{2}, 1/\sqrt{2})$.

Solution.

$\frac{\partial f}{\partial x} = 2xy + y\cos(xy)$

$\frac{\partial f}{\partial y} = x^2 + x\cos(xy)$

$\nabla f(1, \pi) = (2\pi + \pi\cos(\pi), 1 + \cos(\pi)) = (2\pi - \pi, 1 - 1) = (\pi, 0)$

$D_{\mathbf{u}} f(1, \pi) = \nabla f(1, \pi) \cdot \mathbf{u} = \pi \cdot \frac{1}{\sqrt{2}} + 0 = \frac{\pi}{\sqrt{2}}$ $\blacksquare$

Problem. Let $f(x, y, z) = x^2 y\, e^z + \sin(xz)$. Compute $\nabla f$ and evaluate it at $(1, 0, \pi)$.

Problem. Find the directional derivative of $f(x,y) = x^2 y^3$ at $(1, -1)$ in the direction of $\mathbf{v} = (3, -4)$.

Suppose $F(x, y, z) = 0$ defines $z$ implicitly as a function of $x$ and $y$ near a point $(a, b, c)$ with $F_z(a, b, c) \neq 0$. By the Implicit Function Theorem, there exists a $C^1$ function $\varphi$ defined on a neighbourhood of $(a, b)$ such that $\varphi(a, b) = c$ and $F(x, y, \varphi(x, y)) = 0$.

Differentiating $F(x, y, \varphi(x, y)) = 0$ with respect to $x$:

$F_x + F_z \cdot \frac{\partial z}{\partial x} = 0 \implies \frac{\partial z}{\partial x} = -\frac{F_x}{F_z}$

Similarly, $\frac{\partial z}{\partial y} = -\frac{F_y}{F_z}$.

Proposition 1.6 (Implicit Function Theorem, special case). If $F : \mathbb{R}^3 \to \mathbb{R}$ is $C^1$ and $F(a,b,c) = 0$ with $F_z(a,b,c) \neq 0$ Then there exist neighbourhoods $U$ of $(a,b)$ and $V$ of $c$ and a unique $C^1$ function $\varphi : U \to V$ with $\varphi(a,b) = c$ and $F(x, y, \varphi(x,y)) = 0$ for all $(x,y) \in U$.

Problem. If $x^2 y + y^2 z + z^2 x = 3$Find $\frac{\partial z}{\partial x}$ and $\frac{\partial z}{\partial y}$ at the point $(1, 1, 1)$.

Theorem 1.7 (Taylor’s Theorem). Let $f : U \subseteq \mathbb{R}^n \to \mathbb{R}$ be of class $C^{k+1}$ On an open convex set $U$ And let $\mathbf{a} \in U$. Then for all $\mathbf{x} \in U$:

$f(\mathbf{x}) = f(\mathbf{a}) + \nabla f(\mathbf{a}) \cdot (\mathbf{x} - \mathbf{a}) + \frac{1}{2!}(\mathbf{x} - \mathbf{a})^T H_f(\mathbf{a})(\mathbf{x} - \mathbf{a}) + \cdots + R_k$

Where $H_f$ is the Hessian matrix and the remainder $R_k$ can be written in Lagrange form:

$R_k = \frac{1}{(k+1)!} \sum_{\lvert \alpha \rvert = k+1} \frac{(k+1)!}{\alpha!} D^{\alpha} f(\mathbf{c})\, (\mathbf{x} - \mathbf{a})^{\alpha}$

For some $\mathbf{c}$ on the line segment joining $\mathbf{a}$ and $\mathbf{x}$.

For $n = 2$ and $k = 2$The second-order Taylor expansion is:

$f(a+h, b+k) = f(a,b) + f_x h + f_y k + \frac{1}{2}\left(f_{xx} h^2 + 2f_{xy} hk + f_{yy} k^2\right) + R_2$

Where all partial derivatives are evaluated at $(a, b)$ and the remainder is

$R_2 = \frac{1}{6}\left(f_{xxx} h^3 + 3f_{xxy} h^2 k + 3f_{xyy} hk^2 + f_{yyy} k^3\right)\Big|_{\mathbf{c}}$

Proof (sketch). Define $\phi(t) = f(\mathbf{a} + t(\mathbf{x} - \mathbf{a}))$ for $t \in [0, 1]$. Apply the single-variable Taylor theorem to $\phi$ at $t = 0$:

$\phi(1) = \phi(0) + \phi'(0) + \frac{1}{2!}\phi''(0) + \cdots + \frac{1}{k!}\phi^{(k)}(0) + \frac{1}{(k+1)!}\phi^{(k+1)}(\tau)$

For some $\tau \in (0, 1)$. By the multivariable chain rule, $\phi'(t) = \nabla f(\mathbf{a} + t(\mathbf{x}-\mathbf{a})) \cdot (\mathbf{x}-\mathbf{a})$ And higher Derivatives involve higher-order partial derivatives of $f$. Substituting $\mathbf{c} = \mathbf{a} + \tau(\mathbf{x}-\mathbf{a})$ yields the result. $\blacksquare$

A partial derivative measures the rate of change of a function when you move in one coordinate direction while holding all others fixed. If $f(x,y)$ represents the temperature at position $(x,y)$, then $f_x$ tells you how fast the temperature changes as you walk east, ignoring north-south motion. The gradient vector $\nabla f$ collects all partial derivatives into a single object that points in the direction of steepest ascent, and its magnitude tells you the slope in that direction.

The key subtlety in multivariable calculus is that knowing all partial derivatives does not guarantee the function is well-approximated by a linear map. A function can have all partial derivatives at a point yet oscillate wildly, making it non-differentiable. The remedy is to require that the partial derivatives be continuous ($C^1$), which ensures the function is genuinely differentiable. The multivariable chain rule extends the single-variable version: the rate of change of $f(\mathbf{g}(t))$ depends on how $f$ changes at each coordinate (the gradient) times how each coordinate changes with $t$ (the Jacobian). The Hessian matrix, containing all second-order partial derivatives, captures the curvature of the surface and is essential for optimisation.

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    A[1_Partial Derivatives] --> B[Key Concepts]
    A --> C[Core Principles]
    A --> D[Practical Applications]
    B --> E[Fundamental definitions]
    C --> F[Design patterns]
    D --> G[Real-world usage]

• Multiple Integrals: Partial derivatives and multiple integrals are linked by the fundamental theorem of calculus in higher dimensions.
• Vector Calculus: The gradient, divergence, and curl are built from partial derivatives and form the core operators of vector calculus.
• Optimization: Critical points are found by setting partial derivatives to zero, and the Hessian matrix of second partials determines their nature.
• Curves and Surfaces: Tangent vectors and normal vectors to surfaces are computed using partial derivatives of parametrisations.
• Quantum Mechanics
• Graph Theory
• Classical Mechanics
• Electromagnetism