Chapter 4 The Derivative

4.2 New Environment: The Bachmann–Landau Notation

Definition 4.2.1 ($o(1)$-mapping, $\mathcal{O}(h)$-mapping, $o(h)$-mapping).

Consider a mapping from some ball about the origin in one Euclidean space to a second Euclidean space,

$$\varphi : B(0_n, \epsilon) \rightarrow \mathbb{R}^m$$

where $n$ and $m$ are positive integers and $ε > 0$ is a positive real number. The mapping $\varphi$ is smaller than order $1$ if

for every $c > 0$, $|\varphi (h)| ≤ c$ for all small enough $h$.

The mapping $\varphi$ is of order $h$ if

for some $c > 0$, $|\varphi (h)| ≤ c|h|$ for all small enough $h$.

The mapping $\varphi$ is smaller than order $h$ if

for every $c > 0$, $|\varphi (h)| ≤ c|h|$ for all small enough $h$.

A mapping smaller than order $1$ is denoted $o(1)$, a mapping of order $h$ is denoted $\mathcal{O}(h)$, and a mapping smaller than order $h$ is denoted $o(h)$. Also $o(1)$ can denote the collection of $o(1)$-mappings, and similarly for $\mathcal{O}(h)$ and $o(h)$.

Proposition 4.2.3 (Dominance principle for the Landau spaces).

Let $\varphi$ be $o(1)$, and suppose that $|\psi (h)| ≤ |\varphi (h)|$ for all small enough $h$. Then also $ψ$ is $o(1)$. And similarly for $O(h)$ and for $o(h)$.

Proposition 4.2.4 (Vector space properties of the Landau spaces).

For every fixed domain-ball $B(0_n,ε)$ and codomain-space $\mathbb{R}^m$, the $o(1)$-mappings form a vector space, and $\mathcal{O}(h)$ forms a subspace, of which $o(h)$ forms a subspace in turn. Symbolically,

$$o(1) + o(1) = o(1), \mathbb{R}o(1) = o(1) \\ \mathcal{O}(h) + \mathcal{O}(h) = \mathcal{O}(h), \mathbb{R}\mathcal{O}(h) = \mathcal{O}(h) \\ o(h) + o(h) = o(h), \mathbb{R}o(h) = o(h) \\ $$

Proof:

$o(1)$:

Assume $\varphi , \psi$ are $o(1)$, for any given $c$, we can find $h_1$ such that when $h < h_1$, $|\varphi (h)| ≤ c/2$. Similarly, we can find $h_2$ such that when $h < h_2$, $|\psi (h)| ≤ c/2$.

Then let $h_0 = \min \{h_1, h_2\}$, when $h < h_0$,

$$|\varphi(h) + \psi(h)| \leq |\varphi(h)| + |\psi(h)| < c/2 + c/2 = c$$

For the scalor multiplication, given $\alpha$, we want to show $\alpha \varphi$ is also $o(1)$. For any given $c$, we can find $\delta > 0$, such that if $|h| \leq \delta$, $|\varphi (h)| \leq c/|\alpha|$.

So $|\alpha \varphi(h)| = |\alpha | |\varphi (h)| \leq |\alpha| |c/|\alpha| = c$

$\mathcal{O}(h)$:

Assume $\varphi , \psi$ are $\mathcal{O}(h)$. We can find $c_1, \delta_1$, if $|h| \leq \delta_1, \varphi (h) \leq c_1 |h|$. We can find $c_2, \delta_2$, if $|h| \leq \delta_2, \psi (h) \leq c_2 |h|$.

Let $c = 2 \cdot \max \{c_1, c_2\}, \delta = \min \{\delta_1, \delta_2\}$, when $|h| \leq \delta$,

$$|\varphi (h) + \psi (h)| \leq |\varphi (h)| + |\psi (h)| \leq c_1|h| + c_2 |h| \leq c |h|$$

For the scalor multiplication, given $\alpha$, we want to show $\alpha \varphi$ is also $\mathcal{O}(h)$.

Note $|\alpha \varphi(h)| = |\alpha | |\varphi (h)| \leq (|\alpha |c) |\varphi (h)|$, so $\alpha \varphi$ is also $\mathcal{O}(h)$.

$o(h)$:

Assume $\varphi , \psi$ are $o(h)$, for any given $c$, we can find $\delta_1$ such that when $h < \delta_1$, $|\varphi (h)| ≤ c|h|/2$. Similarly, we can find $\delta_2$ such that when $h < \delta_2$, $|\psi (h)| ≤ c|h|/2$.

Then let $\delta = \min \{\delta_1, \delta_2\}$, when $h < \delta$,

$$|\varphi(h) + \psi(h)| \leq |\varphi(h)| + |\psi(h)| < c|h|/2 + c|h|/2 = c|h|$$

For the scalor multiplication, given $\alpha$, we want to show $\alpha \varphi$ is also $o(h)$. For any given $c$, we can find $\delta > 0$, such that if $|h| \leq \delta$, $|\varphi (h)| \leq (c/|\alpha|)|h|$.

So $|\alpha \varphi(h)| = |\alpha | |\varphi (h)| \leq |\alpha| c/|\alpha||h| = c|h|$

$\square$

Proposition 4.2.5.

Every linear mapping is $\mathcal{O}(h)$. The only $o(h)$ linear mapping is the zero mapping.

Proof:

Let $T : \mathbb{R}^n \longrightarrow \mathbb{R}^m$, The unit sphere in $\mathbb{R}^n$ is compact.

$T$ is linear, then we can find matrix $A$, such that $b = T(a) = Aa$. Then $b_i = A_{i, .}a$, then $|b_i| = |⟨A_{i, .},a⟩| \leq |A_{i, .}||a|$. When $|a|$ is small, $|b_i|$ is also small. So each component is continuous. So $T$ is continuous.

Then the image of unit sphere is also compact. Since $||$ is also continuous, then we can find $c$, such that $|T(h_0)| \leq c$ for all $|h_0| = 1$.

Then for any $h$, $h = |h| h_0$, then $|T(|h| h_0)| = ||h|T(h_0)| \leq c|h|$.

Assume $T$ is non-zero, $|T(h_0)| = 2c > 0$. For any $h = |h| h_0$,

$$|T(h)| = |h| |T(h_0)| = 2c|h| > c|h|$$

$\square$

Proposition 4.2.6 (Product property for Landau functions).

Consider two scalar-valued functions and their product function,

$$\varphi, \psi , \varphi \psi : B(0_n, \epsilon) \rightarrow \mathbb{R}$$

If $\varphi$ is $o(1)$ and $ψ$ is $\mathcal{O}(h)$ then $\varphi ψ$ is $o(h)$. Especially, the product of two linear functions is $o(h)$.

Proof:

Given $c > 0$.

Since $ψ$ is $\mathcal{O}(h)$, we can find $d$ such that $\left| ψ (h) \right| \leq d |h|$ if $|h| \leq \delta_1$.

For $\varphi$, since it's $o(1)$, we can find $\delta_2$, if $|h| \leq \delta_2$, $\left| \varphi (h) \right| \leq c/d$.

Let $\delta = \min \{\delta_1, \delta_2\}$, if $|h| \leq \delta$, then $\left| ψ (h)\varphi (h) \right| \leq (c/d)(d|h|) = c|h|$.

$\square$

Proposition 4.2.7 (Composition properties of the Landau spaces).

The composition of $o(1)$-mappings is again an $o(1)$-mapping. Also, the composition of $\mathcal{O}(h)$-mappings is again an $\mathcal{O}(h)$-mapping. Furthermore, the composition of an $\mathcal{O}(h)$-mapping and an $o(h)$-mapping, in either order, is again an $o(h)$-mapping. Symbolically,

$$\begin{align*} o(o(1)) &= o(1)\\ \mathcal{O}(\mathcal{O}(h)) &= \mathcal{O}(h) \\ o(\mathcal{O}(h)) &= o(h) \\ \mathcal{O}(o(h)) &= o(h) \\ \end{align*}$$

That is, $o(1)$ and $\mathcal{O}(h)$ absorb themselves, and $o(h)$ absorbs $\mathcal{O}(h)$ from either side.

Proof:

$o(o(1)) = o(1)$

Assume $\psi, \varphi$ are both $o(1)$.

Given $c > 0$.

Since $ψ$ is $o(1)$, we can find $\delta_1$ , such that if $|h| \leq \delta_1$, $\left| ψ (h) \right| \leq c$.

Since $\varphi$ is $o(1)$, we can find $\delta_2$ , such that if $|h| \leq \delta_2$, $\left| \varphi (h) \right| \leq \delta_1$.

Let $\delta = \min \{\delta_1, \delta_2\}$, if $|h| \leq \delta$, $|\varphi (h)|\leq \delta_1$.

Then

$$|\psi (\varphi (h))| \leq c$$

---

$\mathcal{O}(\mathcal{O}(h)) = \mathcal{O}(h)$

Assume $\psi, \varphi$ are both $\mathcal{O}(h)$.

Since $ψ$ is $\mathcal{O}(h)$, we can find $c_1$ and $\delta_1$ , such that if $|h| \leq \delta_1$, $\left| ψ (h) \right| \leq c_1 |h|$

Since $\varphi$ is $\mathcal{O}(h)$, we can find $c_2$ and $\delta_2$, such that if $|h| \leq \delta_2$, $\left| \varphi (h) \right| \leq c_2 |h|$

Let $\delta = \min \{\delta_1 / c_2, \delta_2\}$, if $|h| \leq \delta$, $|\varphi (h)|\leq c_2 |h| \leq \delta_1$.

Then

$$|\psi (\varphi (h))| \leq c_1 |\varphi (h)| \leq c_1 c_2|h|$$

So we find $c_1 \cdot c_2$.

---

$o(\mathcal{O}(h)) = o(h)$

Assume $ψ$ is $\mathcal{O}(h)$ and $\varphi$ is $o(h)$.

Given $c > 0$.

Since $ψ$ is $\mathcal{O}(h)$, we can find $d$ and $\delta_1$ , such that if $|h| \leq \delta_1$, $\left| ψ (h) \right| \leq d |h|$

Since $\varphi$ is $o(h)$, we can find $\delta_2$, if $|h| \leq \delta_2$, $|\varphi (h)| \leq (c/d) |h|$.

So let $|h| \leq \min \{\delta_1, \delta_2 / d\}$, so $|\psi (h)| \leq d|h| \leq \delta_2$. Then

$$\left| \varphi (\psi (h)) \right| \leq (c/d)|\psi (h)| \leq (c/d) (d |h|) = c|h|$$

---

$\mathcal{O}(o(h)) = o(h)$

Assume $ψ$ is $\mathcal{O}(h)$ and $\varphi$ is $o(h)$.

Given $c > 0$.

Since $ψ$ is $\mathcal{O}(h)$, we can find $d$ and $\delta_1$ , such that if $|h| \leq \delta_1$, $\left| ψ (h) \right| \leq d |h|$

Since $\varphi$ is $o(h)$, we can find $\delta_2$, if $|h| \leq \delta_2$, $|\varphi (h)| \leq (c/d) |h|$.

So let $|h| \leq \min \{(d/c)\delta_1, \delta_2\}$, so $|\varphi (h)| \leq (c/d)|h| \leq \delta_1$.

$$|\psi (\varphi (h))| \leq d |\varphi (h)| \leq d ((c/d)|h|) = c|h|$$

$\square$

4.3 One-Variable Revisionism: The Derivative Redefined

The one-variable derivative as recalled at the beginning of the chapter,

$$f'(a) = \lim \limits_{h \to 0} \frac{ f(a+h) - f(a) }{h}$$

is a construction. To rethink the derivative, we should characterize it instead.

Consider

$$g(h) = f(a+h)−f(a)−th.$$

To reiterate, the idea that f has a tangent of slope t at $(a,f(a))$ has been normalized to the tidier idea that $g$ has slope $0$ at the origin:

To say that the graph of $g$ is horizontal at the origin is to say that for every positive real number $c$, however small, the region between the lines of slope $±c$ contains the graph of $g$ close enough to the origin.

that is

The intuitive condition for the graph of $g$ to be horizontal at the origin is precisely that $g$ is $o(h)$. The horizontal nature of the graph of $g$ at the origin connotes that the graph of $f$ has a tangent of slope $t$ at $(a,f(a))$.

Definition 4.3.1 (Interior point).

Let $A$ be a subset of $R^n$, and let a be a point of $A$. Then $a$ is an interior point of $A$ if some $ε$-ball about $a$ is a subset of $A$. That is, $a$ is an interior point of $A$ if $B(a,ε) ⊂ A$ for some $ε > 0$.

Definition 4.3.2 (Derivative).

Let $A$ be a subset of $R^n$, let $f : A → R^m$ be a mapping, and let $a$ be an interior point of $A$. Then $f$ is differentiable at $a$ if there exists a linear mapping $T_a : R^n → R^m$ satisfying the condition

$$\tag{4-1} f(a+h)−f(a)−Ta(h) \text{ is } o(h).$$

This $T_a$ is called the derivative of $f$ at $a$, written $Df_a$ or $(Df)_a$. When $f$ is differentiable at $a$, the matrix of the linear mapping $Df_a$ is written $f'(a)$ and is called the Jacobian matrix of $f$ at $a$.

Proposition 4.3.3 (Uniqueness of the derivative).

Let $f : A → R^m$ (where $A ⊂ R^n$) be differentiable at $a$. Then there is only one linear mapping satisfying the definition of $Df_a$.

Proof:

Assume we have $T, \tilde{T}$,

$$\begin{align*} f(a+h) - f(a) - T(h) &= o(h) \\ f(a+h) - f(a) - \tilde{T}(h) &= o(h) \\ & \Rightarrow \\ T(h) - \tilde{T}(h) &= o(h) \\ & \Rightarrow \\ (T - \tilde{T})(h) &= o(h) \end{align*}$$

Since the only linear $o(h)$ mapping is $0$ mapping, then $T = \tilde{T}$.

$\square$

Proposition 4.3.4.

If $f$ is differentiable at $a$ then $f$ is continuous at $a$.

Proof:

$$\begin{align*} f(a+h)−f(a) \\ &= f(a+h)−f(a)−T_a(h)+T_a(h) \\ &= o(h)+O(h) \\ &= O(h) \\ &= o(1). \end{align*}$$

$\square$

4.4 Basic Results and the Chain Rule

Proposition 4.4.1 (Derivatives of constant and linear mappings).

(1) Let $C : A → R^m$ (where $A ⊂ R^n$) be the constant mapping $C(x) = c$ for all $x ∈ A$, where $c$ is some fixed value in $R^m$. Then the derivative of $C$ at every interior point $a$ of $A$ is the zero mapping.

Proof:

$$\begin{align*} &C(a+h) - C(a) - 0(h) \\ &= c - c - 0 \\ &= 0 \\ &= o(h) \end{align*}$$

$\square$

(2) The derivative of a linear mapping $T : R^n → R^m$ at every point $a ∈ R^n$ is again $T$.

Proof:

$$\begin{align*} & T(a+h) - T(a) - T(h) \\ &= T(a) + T(h) - T(a) - T(h) \\ &= 0 \\ &= o(h) \\ \end{align*}$$

$\square$

Proposition 4.4.2 (Linearity of the derivative).

Let $f : A → R^m$ (where $A ⊂ R^n$) and $g : B → R^m$ (where $B ⊂ R^n$) be mappings, and let a be a point of $A∩B$. Suppose that $f$ and $g$ are differentiable at a with derivatives $Df_a$ and $Dg_a$. Then:

(1) The sum $f + g : A ∩ B → R^m$ is differentiable at $a$ with derivative $D(f +g)_a = Df_a +Dg_a$.

Proof:

$$\begin{align*} &(f+g)(a+h) - (f+g)(a) - (Df_a +Dg_a)(h) \\ &= (f(a+h) - f(a) - Df_a(h)) + (g(a+h) - g(a) - Dg_a(h)) \\ &= o(h) + o(h) \\ &= o(h) \end{align*}$$

$\square$

(2) For every $\alpha ∈ R$, the scalar multiple $\alpha f : A → Rm$ is differentiable at $a$ with derivative $D(\alpha f)_a = \alpha Df_a$.

Proof:

$$\begin{align*} &(\alpha f)(a+h) - (\alpha f)(a) - (\alpha Df_a)(h) \\ &= \alpha (f)(a+h) - \alpha (f)(a) - \alpha (Df_a)(h) \\ &= \alpha (f(a+h) - f(a) - (Df_a)(h)) \\ &= \alpha o(h) \\ &= o(h) \end{align*}$$

$\square$

Theorem 4.4.3 (Chain rule).

Let $f : A \rightarrow R^m$ (where $A ⊂ R^n$) be a mapping, let $B ⊂ R^m$ be a set containing $f(A)$, and let $g : B \rightarrow R^ℓ$ be a mapping. Thus the composition $g◦f : A \rightarrow R^ℓ$ is defined. If $f$ is differentiable at the point $a ∈ A$, and $g$ is differentiable at the point $f(a) ∈ B$, then the composition $g ◦f$ is differentiable at the point $a$, and its derivative there is

$$D(f \circ g)_a = Dg_{f(a)} \circ Df_a$$

In terms of Jacobian matrices, since the matrix of a composition is the product of the matrices, the chain rule is

$$(g ◦f)'(a) = g '(f(a))f'(a).$$

Proof:

For simplicity, we first take $a = 0_n$ and $f(a) = 0_m$.

$$\begin{align*} f(h) &= S(h) + o(h) \\ g(k) &= T(k) + o(k) \\ \end{align*}$$

Then

$$\begin{align*} g(f(h)) &= T(f(h)) + o(f(h)) \\ &= T(S(h) + o(h)) + o(S(h) + o(h)) \\ &= T(S(h)) + T(o(h)) + o(O(h) + o(h)) \\ &= T(S(h)) + O(o(h)) + o(O(h)) \\ &= T(S(h)) + o(h) + o(h) \\ &= T(S(h)) + o(h) \\ \end{align*}$$

Then for the general

$$\begin{align*} f(a + h) &= f(a) + S(h) + o(h) \\ g(f(a) + k) &= g(f(a)) + T(k) + o(k) \\ \end{align*}$$

Then

$$\begin{align*} g(f(a+h)) &= g(f(a) + S(h) + o(h)) \\ &= g(f(a)) + T(S(h) + o(h)) + o(S(h) + o(h)) \\ &= g(f(a)) + T(S(h)) + T(o(h)) + o(O(h) + o(h)) \\ &= g(f(a)) + T(S(h)) + O(o(h)) + o(O(h)) \\ &= g(f(a)) + T(S(h)) + o(h) + o(h) \\ &= g(f(a)) + T(S(h)) + o(h) \\ \end{align*}$$

$\square$

Lemma 4.4.4 (Derivatives of the product and reciprocal functions).

Define the product function,

$$p : \mathbb{R}^2 \longrightarrow \mathbb{R}, \qquad p(x,y) = xy,$$

and define the reciprocal function

$$r : \mathbb{R} - \{0\} \longrightarrow \mathbb{R}, \qquad r(x) = 1/x,$$

Then

(1) The derivative of $p$ at every point $(a,b) ∈ R^2$ exists and is

$$Dp_{(a,b)}(h, k) = ak + bh$$

Proof:

$$\begin{align*} (x+h)(y+k) - xy - (ak + bh) &= hk \\ \end{align*}$$

Since $|h| \leq |(h,k)|, |k| \leq |(h,k)|$, so $|h||k| \leq |(h,k)|^2 = o(h,k)$.

$\square$

(2) The derivative of $r$ at every nonzero real number a exists and is

$$Dr_a(h) = -h / a^2$$

Proof:

$$\begin{align*} r(a+h) - r(a) - Dr_a(h) &= 1/(a+h) - 1/a + h / a^2 \\ &= -h/(a+h)a + h / a^2 \\ &= h ( \frac{1}{a^2} - \frac{1}{a(a+h)}) \\ &= \frac{h}{a} \frac{h}{a(a+h)} \\ &= \frac{h^2}{a^2(a+h)} \\ \end{align*}$$

Since when $|h| < |a/2|$, $|a+h| \geq ||a| - |h|| > |a| - |a/2| = |a/2|$.

$$\left| \frac{h^2}{a^2(a+h)} \right| \leq \left| \frac{2h^2}{a^3} \right|$$

So it's $o(h)$.

$\square$

Proposition 4.4.5 (Multivariable product and quotient rules).

Let $f : A → R$ (where $A ⊂ \mathbb{R}^n$) and $g : B → \mathbb{R}$ (where $B ⊂ \mathbb{R}^n$) be functions, and let $f$ and $g$ differentiable at $a$. Then:

(1) $fg$ is differentiable at $a$ with derivative

$$D(fg)_a = f(a) Dg_a + g(a) Df_a$$

Proof:

Consider

$$\begin{align*} h : \mathbb{R}^n \longrightarrow \mathbb{R}^2 &\qquad h(a) = (f(a), g(a)) \\ p : \mathbb{R}^2 \longrightarrow \mathbb{R} &\qquad p(x, y) = xy \\ \end{align*}$$

Then $fg = p \circ h$.

$$\begin{align*} D(fg)_a &= D(p \circ h)_a \\ &= Dp_{h(a)} \circ Dh_a \\ \end{align*}$$

Note $Dh_a = (Df_a, Dg_a)$, so we have

$$\begin{align*} (Dp_{h(a)} \circ Dh_a)(k) &= Dp_{h(a)}(Dh_a(k)) \\ &= Dp_{h(a)}(Df_a(k), Dg_a(k)) \\ &= g(a)Df_a(k) + f(a) Dg_a(k) \\ &= (f(a) Dg_a + g(a) Df_a)(k) \\ \end{align*}$$

(2) If $g(a) \neq 0$ then $f/g$ is differentiable at a with derivative

$$D(\frac{f}{g})_a = \frac{ g(a)Df_a−f(a)Dg_a }{ g(a)^2 }$$

Proof:

Consider

$$\begin{align*} D(r \circ g)_a(k) &= (Dr_{g(a)} \circ Dg_a)(k) \\ &= Dr_{g(a)} (Dg_a(k)) \\ &= -\frac{Dg_a(k)}{g^2(a)} \\ &= (-\frac{Dg_a}{g^2(a)}) (k) \\ \end{align*}$$

So $D(r \circ g)_a = -\frac{Dg_a}{g^2(a)}$

$$\begin{align*} h : \mathbb{R}^n \longrightarrow \mathbb{R}^2 &\qquad h(a) = (f(a), 1/g(a)) \\ p : \mathbb{R}^2 \longrightarrow \mathbb{R} &\qquad p(x, y) = xy \\ \end{align*}$$

Then $f/g = p \circ h$.

$$\begin{align*} D(f/g)_a &= D(p \circ h)_a \\ &= Dp_{h(a)} \circ Dh_a \\ \end{align*}$$

Note $Dh_a = (Df_a, -\frac{Dg_a}{g^2(a)})$, so we have

$$\begin{align*} (Dp_{h(a)} \circ Dh_a)(k) &= Dp_{h(a)}(Dh_a(k)) \\ &= Dp_{h(a)}(Df_a(k), -\frac{Dg_a}{g^2(a)}(k)) \\ &= \frac{1}{g(a)} Df_a(k) - f(a) \frac{Dg_a}{g^2(a)}(k) \\ &= \left(\frac{g(a) Df_a - f(a) Dg_a}{g^2(a)}\right)(k) \\ \end{align*}$$

So

$$D\left(\frac{f}{g}\right)_a = \frac{ g(a)Df_a−f(a)Dg_a }{ g(a)^2 }$$

$\square$

Another example

Consider $f(x,y) = (x^2-y)(y+1)$ for all $(x,y) \in \mathbb{R}^2$

Let

$$\begin{align*} X(x,y) &= x \\ Y(x,y) &= y \\ \end{align*}$$

Then

$$f = \frac{ X^2 - Y }{ Y + 1 }$$

Let

$$\begin{align*} l &= X^2 - Y \\ g &= Y + 1 \\ a &= (x, y) \\ \end{align*}$$

Then note

$$\begin{align*} g(a) &= y + 1\\ g(a+(h,k)) - g(a) &= (y+k+1) - (y+1) = k\\ Dg_a(h,k) &= k \\ \end{align*}$$

Also note

$$\begin{align*} l(a) &= x^2 - y \\ l(a+(h, k)) - l(a) &= ((x+h)^2 - (y+k)) - (x^2 - y)\\ &= (2xh + h^2) - k \\ Dl_a(h,k) &= 2xh - k \end{align*}$$

Combine these together, we have

$$\begin{align*} Df_a(h,k) &= D\left( \frac{l}{g}\right)_a (h,k)\\ &= \frac{ g(a) Dl_a - l(a) Dg_a }{g^2(a)} (h,k) \\ &= \frac{ (y+1) (2xh-k) - (x^2 - y) k }{(y+1)^2} \\ &= \frac{2x}{y+1} h - \frac{x^2+1}{(y+1)^2} k \\ \end{align*}$$

This results are confirmed by the textbook.

Now we can follow the approach in the text book.

$$f = \frac{ X^2 - Y }{ Y + 1 }$$

$$\begin{align*} &Df_a(h, k) \\ \\ &\text{Use 4.4.5 (2) quotient rule} \\ &= \frac{ (Y+1)(a)D(X^2-Y)_a - (X^2-Y)(a) D(Y+1)_a }{(Y+1)^2(a)}(h, k)\\ \\ &\text{Use } (Y+1)(a) = y+1 \\ &= \frac{ (y+1)D(X^2-Y)_a - (X^2-Y)(a) D(Y+1)_a }{(y+1)^2}(h, k)\\ \\ &\text{Use } (X^2-Y)(a) = x^2-y \\ &= \frac{ (y+1)D(X^2-Y)_a - (x^2-y) D(Y+1)_a }{(y+1)^2}(h, k)\\ \\ &\text{Use 4.4.2 (Linearity of the derivative)} \\ &= \frac{ (y+1)(D(X^2)_a-DY_a) - (x^2-y) (DY_a+D1_a) }{(y+1)^2}(h, k)\\ \\ &\text{Use 4.4.1 Derivatives of constant and linear mappings} \\ &= \frac{ (y+1)(D(X^2)_a-Y) - (x^2-y) Y }{(y+1)^2}(h, k)\\ \\ &\text{Use 4.4.5 (a) product rule} \\ &= \frac{ (y+1)(2X(a)DX_a-Y) - (x^2-y) Y }{(y+1)^2}(h, k)\\ \\ &\text{Use 4.4.1 Derivatives of constant and linear mappings} \\ &= \frac{ (y+1)(2xX-Y) - (x^2-y) Y }{(y+1)^2}(h, k)\\ \\ &= \frac{ 2x(y+1)X - (x^2+1) Y }{(y+1)^2}(h, k)\\ \\ &= \frac{ 2x }{(y+1)}h - \frac{x^2+1}{(y+1)^2}k\\ \end{align*}$$

$\square$