The Derivative as a Linear Map
In one variable the derivative is a number, and the number happens to be the slope of the best linear approximation. In $n$ variables the number is gone and only the approximation survives. The derivative of $f$ at $a$ is the linear map that fits $f$ best near $a$, and every theorem in the next five units is a statement about that map.
- State differentiability as a limit that says "the error is $o(\|h\|)$".
- Show the derivative is unique, and that it forces continuity.
- Compute the Jacobian matrix and know when it represents a derivative.
- Give the standard example where all partials exist and $f$ is still not differentiable.
Partial derivatives are easy to compute and prove almost nothing. They look along $n$ directions and say nothing about the other infinitely many. The linear-map definition is the one that supports a chain rule, an inverse function theorem, and integration on manifolds.
Every gradient-based optimiser, every Newton step, and every backward pass in a neural network is this definition applied numerically. Automatic differentiation propagates linear maps, not numbers, which is exactly why it composes.
Fréchet gave this definition in 1911, and it works verbatim in infinite-dimensional normed spaces, which is where it is now mostly used. The weaker Gâteaux derivative — directional derivatives in every direction — came from the same period and is genuinely weaker: the two disagree on functions you meet in this unit.
The definition
Let $A\subseteq\mathbb{R}^n$ be open, $f:A\to\mathbb{R}^m$, and $a\in A$. Then $f$ is differentiable at $a$ if there is a linear map $Df(a):\mathbb{R}^n\to\mathbb{R}^m$ with
Equivalently $f(a+h)=f(a)+Df(a)h+r(h)$ where the remainder satisfies $r(h)/\|h\|\to0$. The whole content is that the error vanishes faster than linearly, not merely that it vanishes.
At most one linear map can satisfy the definition.
$f(a+h)-f(a)=Df(a)h+r(h)$. As $h\to0$ the first term goes to $0$ because linear maps on $\mathbb{R}^n$ are bounded, and the second goes to $0$ because it is $o(\|h\|)$.
Directional and partial derivatives
The directional derivative of $f$ at $a$ in direction $u$ is
when the limit exists. Taking $u=e_j$, the $j$-th standard basis vector, gives the partial derivative $\partial f/\partial x_j$.
If $f$ is differentiable at $a$ then $D_uf(a)=Df(a)u$ for every $u$.
Put $h=tu$ in the definition: $f(a+tu)-f(a)=tDf(a)u+r(tu)$, and $r(tu)/t\to0$ since $\|tu\|=|t|\,\|u\|$. Divide by $t$ and let $t\to0$.
So the derivative, a single linear map, packages every directional derivative at once. The converse is false, which is the point of the next box.
Let $f(x,y)=\dfrac{x^2y}{x^4+y^2}$ for $(x,y)\ne(0,0)$ and $f(0,0)=0$.
Directional derivatives probe along straight lines. Differentiability is a statement about every approach, including curved ones.
The Jacobian matrix
Write $f=(f_1,\dots,f_m)$. The Jacobian matrix of $f$ at $a$ is the $m\times n$ matrix $Jf(a)$ with entries $\partial f_i/\partial x_j$ evaluated at $a$.
If $f$ is differentiable at $a$, then $Jf(a)$ is the matrix of $Df(a)$ in the standard bases: column $j$ is $Df(a)e_j=\partial f/\partial x_j$.
If every $\partial f_i/\partial x_j$ exists on a neighbourhood of $a$ and is continuous at $a$, then $f$ is differentiable at $a$. Such an $f$ is called $C^1$.
This is the theorem you actually use. Nobody verifies the limit definition on a formula built from elementary functions — they check the partials are continuous and stop.
For $f(x,y)=(x^2-y^2,\,2xy)$, the Jacobian is $\begin{smallmatrix}2x&-2y\\2y&2x\end{smallmatrix}$ with determinant $4(x^2+y^2)$. The partials are polynomials, so $f$ is $C^1$ everywhere and the matrix genuinely is $Df$. Note the determinant vanishes only at the origin — a fact Unit 6 will turn into a statement about invertibility.
The gradient, and what it is not
For $f:A\to\mathbb{R}$, the gradient $\nabla f(a)$ is the vector representing the linear functional $Df(a)$ through the dot product: $Df(a)h=\nabla f(a)\cdot h$.
Among unit vectors $u$, the directional derivative $D_uf(a)=\nabla f(a)\cdot u$ is largest when $u$ points along $\nabla f(a)$, by Cauchy–Schwarz, and the largest value is $\|\nabla f(a)\|$.
• Calling the matrix of partials "the derivative" without checking
differentiability. The matrix can exist when the derivative does not.
• Testing differentiability along lines only. Curved approaches such as
$y=x^2$ break examples that all lines pass.
• Forgetting the domain must be open. At a boundary point the limit
$h\to0$ has no room to run in every direction.
• Writing $\nabla f$ for vector-valued $f$. The gradient is defined for
scalar-valued functions; the general object is the Jacobian.
• Treating $C^1$ as equivalent to differentiable. It is strictly
stronger: $x^2\sin(1/x)$ is differentiable everywhere with a discontinuous
derivative.
Practice
Original problems written for this guide in the style of Spivak and Munkres exercises. Write a full solution before opening the answer.
Show $f(x,y)=x^2+3xy$ is differentiable at $(1,2)$ and find $Df(1,2)$.
Solution
What is the derivative of a linear map $L$ at any point?
Answer
Let $f(x,y)=1$ when $xy=0$ and $f(x,y)=0$ otherwise. What are the partials at the origin? Is $f$ differentiable there?
Solution
Is $f(x)=\|x\|$ differentiable at $0$ in $\mathbb{R}^n$? Elsewhere?
Solution
Prove $f(x,y)=xy$ from $\mathbb{R}^2$ to $\mathbb{R}$ is differentiable with $Df(a,b)(h,k)=bh+ak$.
Solution
Give a continuous function on $\mathbb{R}^2$ that is not differentiable at the origin, with all partials existing there.
Solution
Show $f(x)=x^2\sin(1/x)$, $f(0)=0$, is differentiable everywhere but not $C^1$.
Solution
For $g(r,\theta)=(r\cos\theta,\,r\sin\theta)$, compute $Jg$ and its determinant.
Solution
Expanded topic checklist
Course coverage. This unit is part of the complete MAT257Y1 topic map. It should cover the following items before you move on:
- The definition. The $o(\|h\|)$ condition, uniqueness, and why differentiability forces continuity.
- Directional and partial derivatives. Recovered from $Df(a)$, but not sufficient to produce it.
- The Jacobian. Matrix of partials; the $C^1$ criterion that makes it a derivative.
- Gradient. Steepest ascent, and the inner product that turns a functional into a vector.
Spivak's Calculus on Manifolds opens its differentiation chapter with exactly this definition; Munkres' Analysis on Manifolds spends considerably longer on the $C^1$ theorem and is worth reading first. Check the assigned text against the current syllabus.
Differentiable gives continuous and gives all directional derivatives. Continuous partials give differentiable. Neither arrow reverses, and the counterexamples in this unit are the standard proofs of that.
Deep practice extensions
Use these after the main practice set. They are designed to force definition-level reasoning, theorem selection, and mixed application rather than routine calculation only.
State in one sentence what $Df(a)$ approximates.
Guide
Why must the domain be open in the definition?
Guide
What is the derivative of a bilinear map $B$ at $(a,b)$?
Guide
Why is $\nabla f(a)$ perpendicular to the level set through $a$?