2.4Basic Results and Chain Rule

Updated 23 Jun 2026Chapter (PDF)

[2.4.1]Theorem(Basic Differentiation Rules)#

Let $f: A \to \bR^m$ , where $A \subseteq \bR^m$ , be such that $f(a) = v_0$ for every $a \in A$ , where $v_0 \in \bR^m$ is fixed. Then $(Df)_a \equiv 0$ .
Let $T: \bR^n \to \bR^m$ be linear. Then $(DT)_a \equiv T$ .
The derivative of a function is additive and homogenous (i.e., linear).

Proof.

Note that $f(a + h) - f(a) - (Df)_a$ , where $(Df)_a$ , is equal to $(Df)_a$ . For $(Df)_a$ to be linear and $o(h)$ , we note that the derivative is forced to be the zero mapping.
Let $a \in \bR^n$ . Then note $T(a + h) - T(a) - (DT)_a = T(h) - (DT)_a$ must be $o(h)$ , but the subtraction of two linear maps is a linear map, and so $(DT)_a = T(h)$ and so $(DT)_a \equiv T$ .
Suppose $f: A \to \bR^m$ , where $A \subseteq \bR^n$ , is differentiable at $a \in \Int(A)$ . We seek to show that $\lambda f$ is differentiable at $a$ with derivative $(D \lambda f)_a = \lambda (Df)_a$ . Note that
$\lambda f(a + h) - \lambda f(a) - \lambda (Df)_a = \lambda(f(a + h) - f(a) - (Df)_a),$
a scalar multiple of an $o(h)$ map, which is also $o(h)$ (since $o(h)$ maps form a vector space and are thus closed under scalar multiplication).
Now suppose $g: B \to \bR^m$ , where $B \subseteq \bR^n$ , is also differentiable at our aforementioned $a \in \Int(A) \cap \Int(B)$ . Then we seek to show $[D(f + g)]_a = (Df)_a + (Dg)_a$ . Note that if $a \in \Int(A) \cap \Int(B)$ , then $a \in \Int(A \cap B)$ , so $f + g$ is differentiable at $a$ . Then
$(f + g)(a + h) - (f + g)(a) - (Df)_a - (Dg)_a = [f(a + h) - f(a) - (Df)_a] + [g(a + h) - g(a) - (Dg)_a]$
is the sum of two $o(h)$ maps, which is also $o(h)$ , completing the proof.

[2.4.2]Theorem(Chain Rule)#

Let $f: A \to \bR^m$ , where $A \subseteq \bR^n$ be a mapping. Let $g: B \to \bR^\ell$ , where $B \supseteq f(A)$ , be a mapping. Then the composition $g \circ f: A \to \bR^\ell$ is well defined. If $f$ is differentiable at some $a \in A$ , and $g$ is differentiable at $f(a) \in B$ , then $g \circ f$ is differentiable, such that

D(g \circ f)_a = D(g)_{f(a)} \circ D(f)_a.

With a choice of basis, the Jacobian is such that

(g \circ f)'(a) = g'(f(a)) f'(a).

Proof.

To avoid maintaining constant terms throughout multiple steps of algebra, we translate $f$ and $g$ such that $a$ is treated as the zero vector in $\bR^n$ and $\bR^m$ , as well as its image under $f$ and $g$ being the zero vector in $\bR^m$ and $\bR^\ell$ respectively.

Recalling that $f(a + h) = f(a) + D(f)_a[h] + o(h)$ , we may eliminate the constant term by subtracting $f(a)$ . With this in mind, define

\tilde f(h) = f(a + h) - f(a) \qquad \tilde g(k) = g(b + k) - g(b),

where $b = f(a) \in B$ . Then note that $\tilde f(\vb{0}_n) = \vb{0}_m$ and $\tilde g(\vb{0}_m) = \vb{0}_\ell$ . As a result,

\begin{align*} \tilde f(h) &= D(\tilde{f})_{\vb{0}_n}[h] + o(h) \\ &= f(a + h) - f(a) = D(f)_a[h] + o(h), \\ \\ \tilde g(k) &= D(\tilde{g})_{\vb{0}_m}[k] + o(k) \\ &= g(b + k) - g(b) = D(g)_b[k] + o(k), \\ \end{align*}

and so $D(\tilde{f})_{\vb{0}_n} = D(f)_a$ and $D(\tilde{g})_{\vb{0}_m} = D(g)_b$ . Now, note that

\begin{align*} \tilde g(\tilde f(h)) &= \tilde g(f(a + h) - f(a)) \\ &= g(b + f(a + h) - f(a)) - g(f(a)) \\ &= g(f(a + h)) - g(f(a)) \end{align*}

as desired, and an expansion analogous to our previous results reveal that $D(g \circ f)_a = D(\tilde g \circ \tilde f)_{\vb{0}_n}$ .

As a little aside, these results signify that differentiation is essentially translation-invariant. It is best to work in local coordinates where the problem simplifies nicely (our $\tilde f, \tilde g$ functions work on a translated coordinate plane), as opposed to global coordinates. Global coordinates are great for absolute data, but it may contain irrelevant data that can prove to be a hindrance.

In any case, we want to write $\tilde g(\tilde f(h))$ in the form $D(\tilde g \circ \tilde f)_{\vb{0}_n}[h] + o(h)$ . Expanding reveals

\begin{align*} \tilde g(\tilde f(h)) &= \tilde g(D(\tilde f)_{\vb{0}_n}[h] + o(h)) \\ &= D(\tilde{g})_{\vb{0}_m}\left[ D(\tilde f)_{\vb{0}_n}[h] + o(h) \right] + o\left(D(\tilde f)_{\vb{0}_n}[h] + o(h)\right) \\ &= [D(\tilde g)_{\vb{0}_m} \circ D(\tilde f)_{\vb{0}_n}](h) + D(\tilde g)_{\vb{0}_m}(o(h)) + o(o(h)) \\ &= [D(\tilde g)_{\vb{0}_m} \circ D(\tilde f)_{\vb{0}_n}](h) + o(h). \end{align*}

Thus,

D(g \circ f)_a = D(\tilde g \circ \tilde f)_{\vb{0}_n} = D(\tilde g)_{\vb{0}_m} \circ D(\tilde f)_{\vb{0}_n} = \boxed{D(g)_{f(a)} \circ D(f)_a}.

The product and quotient rules may be proven by differentiating $xy$ and $x/y$ , then using chain rule.