Once the derivative is a linear map, the chain rule stops being a formula to memorise and becomes the only thing it could be: the derivative of a composition is the composition of the derivatives. Higher derivatives then arrive as multilinear maps, and the surprise of the unit is that mixed partials commute — but only under a hypothesis, and there is a function that shows why.
Almost nothing you differentiate is given directly; it is built by composing simpler maps. A chain rule that is a clean algebraic statement about linear maps is what lets you differentiate anything assembled from pieces, in any number of variables, without a new argument each time.
Backpropagation is the chain rule applied right-to-left through a composition of hundreds of maps; the cost saving over left-to-right is entirely a statement about matrix shapes. Change of variables in physics is the same identity read the other way.
Clairaut asserted the equality of mixed partials in 1740; a hypothesis-free version is false, and Schwarz gave the continuity condition that repairs it in 1873. Peano's counterexample from the 1890s is the one still used, and it is in this unit.
Let $f:A\to\mathbb{R}^m$ be differentiable at $a$ with $A$ open in $\mathbb{R}^n$, and let $g:B\to\mathbb{R}^p$ be differentiable at $b=f(a)$ with $f(A)\subseteq B$ open. Then $g\circ f$ is differentiable at $a$ and
In matrices, $J(g\circ f)(a)=Jg(b)\,Jf(a)$: a $p\times m$ times an $m\times n$, giving $p\times n$. The shapes only fit one way, which is a useful memory aid.
Note where the proof would break if we only knew directional derivatives: $k(h)$ does not travel along a line.
Let $u(r,\theta)=f(r\cos\theta,r\sin\theta)$. Then $Ju=Jf\cdot Jg$ with $g$ the polar map of Unit 4, giving
Every "chain rule in several variables" formula from a first calculus course is one row of one matrix product.
$f$ is $C^k$ on an open set if all partial derivatives up to order $k$ exist and are continuous there; $C^\infty$ (or smooth) if this holds for every $k$. Since $Df$ is itself a map into the space of matrices, $D^2f(a)$ is a linear map into linear maps — equivalently a bilinear map $\mathbb{R}^n\times\mathbb{R}^n\to\mathbb{R}^m$.
For scalar $f$, the Hessian at $a$ is the $n\times n$ matrix $Hf(a)$ with entries $\partial^2f/\partial x_i\partial x_j$. It represents the bilinear form $(h,k)\mapsto D^2f(a)(h,k)$.
If $\partial^2f/\partial x\,\partial y$ and $\partial^2f/\partial y\,\partial x$ exist near $a$ and are continuous at $a$, they agree at $a$. Hence the Hessian of a $C^2$ function is symmetric.
$f(x,y)=xy\dfrac{x^2-y^2}{x^2+y^2}$ with $f(0,0)=0$.
Both mixed partials exist at the origin and differ. Neither is continuous there, so Clairaut is not contradicted — it is illustrated.
If $f$ is $C^2$ on a ball around $a$, then for $h$ in that ball
and consequently $f(a+h)=f(a)+\nabla f(a)\cdot h+\tfrac12h^{T}Hf(a)h+o(\|h\|^2)$.
The proof is the one-variable Taylor theorem applied to $\varphi(t)=f(a+th)$ on $[0,1]$; the chain rule supplies $\varphi'(t)=\nabla f(a+th)\cdot h$ and $\varphi''(t)=h^{T}Hf(a+th)h$.
$f(x,y)=x^2+4xy+y^2$ has $\nabla f(0,0)=(0,0)$ and constant Hessian $\begin{smallmatrix}2&4\\4&2\end{smallmatrix}$. The expansion at the origin reproduces $f$ exactly with zero remainder, as it must: every derivative of order three and higher vanishes.
• Writing the chain rule in the wrong order. $Df(a)\,Dg(b)$ usually
does not even have compatible shapes.
• Applying Clairaut without continuity of the second partials. Peano's
function is the standing counterexample.
• Assuming $C^1$ implies $C^2$. Each level of smoothness is a genuinely
stronger hypothesis.
• Treating the Hessian as a linear map. It is a bilinear form; the matrix
looks the same but the change-of-basis rule does not.
• Expanding to second order and calling the remainder $o(\|h\|)$. It is
$o(\|h\|^2)$, and the sharper statement is the one you need.
Original problems written for this guide in the style of Spivak and Munkres exercises. Write a full solution before opening the answer.
With $f(t)=(t,t^2)$ and $g(x,y)=x^2y$, compute $(g\circ f)'(1)$ two ways.
If $f$ is invertible near $a$ with differentiable inverse, what is $D(f^{-1})(f(a))$?
For $u(x)=\varphi(\|x\|)$ on $\mathbb{R}^n$ minus the origin, compute $\nabla u$.
Verify Clairaut for $f(x,y)=x^3y^2+e^{xy}$.
Suppose $f$ is differentiable and homogeneous of degree $d$, meaning $f(tx)=t^df(x)$ for $t>0$. Show $x\cdot\nabla f(x)=df(x)$.
Find the Hessian of $f(x,y)=x^3-3xy^2$ at the origin and describe the critical point.
If $\gamma$ is a differentiable curve inside the level set $\{f=c\}$, what is $\nabla f(\gamma(t))\cdot\gamma'(t)$?
Show $\Delta u=u_{rr}+\tfrac1ru_r+\tfrac1{r^2}u_{\theta\theta}$.
Course coverage. This unit is part of the complete MAT257Y1 topic map. It should cover the following items before you move on:
Spivak proves the chain rule in about a page; Munkres separates the partial-derivative bookkeeping from the linear-map argument and is easier to follow the first time. Confirm the assigned text and problem numbers against the current syllabus.
Chain rule builds every later computation. Clairaut makes the Hessian symmetric, which is what lets Unit 8 diagonalise it. Taylor is the bridge between the derivative and the classification of critical points.
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.
Explain by shapes why $Dg(b)\,Df(a)$ and not the reverse.
Which hypothesis fails in Peano's example?
To what order is the second-order Taylor remainder small?
Is a composition of $C^k$ maps $C^k$?