Chapter 4 The Derivative

4.2 New Environment: The Bachmann–Landau Notation

4.3 One-Variable Revisionism: The Derivative Redefined

4.3.1.

Let $T : R^n → R^m$ be a linear mapping. Show that for every $ε$ > 0, the behavior of $T$ on $B(0_n,ε)$ determines the behavior of $T$ everywhere.

Proof:

Given any $x$, let $\lambda = 2|x|/\epsilon$ and $x = \lambda x_0$. Then $|x_0| = \epsilon / 2$, so $x_0 \in B(0_n,ε)$.

$$T(x) = T( \lambda x_0) = \lambda T(x_0)$$

So $x$ is determined by $x_0$.

$\square$

4.3.2.

Give a geometric interpretation of the derivative when $n = m = 2$. Give a geometric interpretation of the derivative when $n = 1$ and $m = 2$.

Solution:

When $n = m = 2$, we can think that there are 2 graphs of Figure 4.4. So the derivative is 2 planes.

When $n = 1$ and $m = 2$, the derivative is a line.

$\square$

4.3.3.

Let $f : A → R^m$ (where $A ⊂ R^n$) have component functions $f_1,...,f_m$, and let a be an interior point of $A$. Let $T : R^n → R^m$ be a linear mapping with component functions $T_1,...,T_m$. Using the componentwise nature of the $o(h)$ condition, established in Section 4.2, prove the component-wise nature of differentiability: $f$ is differentiable at a with derivative $T$ if and only if each component $f_i$ is differentiable at a with derivative $T_i$.

Proof:

$$\begin{align*} f(a+h) - f(a) - T(h) \text{ is } o(h) \\ & \Longleftrightarrow \\ (f_1(a+h) - f_1(a) - T_1(h), \cdots, f_n(a+h) - f_n(a) - T_n(h)) \text{ is } o(h) \\ & \Longleftrightarrow \\ f_i(a+h) - f_i(a) - T_i(h) \text{ is } o(h) \\ & \text{componentwise nature of the } o(h) \end{align*}$$

4.3.4.

Let $f(x,y) = (x^2 −y^2, 2xy)$. Show that $Df_{(a,b)}(h,k) = (2ah−2bk, 2bh+2ak)$ for all $(a,b) ∈ R^2$. (By the previous problem, you may work component-wise.)

Proof:

First component:

$$\begin{align*} &f(a+h, b+k) - f(a, b) - (2ah−2bk) \\ &=((a+h)^2-(b+k)^2) - (a^2 - b^2) - (2ah−2bk) \\ &= (2ah + h^2) - (2bk + k^2) - (2ah−2bk) \\ &= h^2 - k^2 \\ &= o(h) \end{align*}$$

Second component:

$$\begin{align*} & f(a+h, b+k) - f(a, b) - (2ah+2bk) \\ &= 2(a+h)(b+k) - 2ab - (2ah+2bk) \\ &= hk \\ &= o(h) \end{align*}$$

$\square$

4.3.5

Let $g(x,y) = xe^y$. Show that $Dg_{(a,b)}(h,k) = he^b +kae^b$ for all $(a,b) ∈ R^2$. (Note that because $e^0 = 1$ and because the derivative of the exponential function at 0 is 1, the one-variable characterizing property says that $e^k −1 = k +o(k)$.)

Proof:

$$\begin{align*} &g(a+h,b+k) - g(a,b) - (he^b +kae^b) \\ &= (a+h)e^{b+k} - ae^b - (he^b + kae^b) \\ &= (a+h)(e^{b+k}-e^b) - kae^b \\ &= (a+h)e^b(k + o(k)) - kae^b \\ &= h e^b k + ae^b o(k) + h e^b o(k) \\ &= o(h) + o(k) + o(h) \\ &= o(h) \end{align*}$$

$\square$

4.3.6.

Show that if $f : R^n → R^m$ satisfies $|f(x)| ≤ |x|^2$ for all $x ∈ R^n$ then $f$ is differentiable at $0^n$.

Proof:

Note $|f(0)| \leq |0|^2 = 0$. We will show $Df_a(0) = 0$

$$\begin{align*} & |f(h) - f(0) - 0 \cdot h| \\ &= |f(h)| \\ & \leq |h|^2 \\ &= o(h) \end{align*}$$

So $Df_a(0) = 0$.

$\square$

4.3.7

Show that the function $f(x,y) = |xy|$ for all $(x,y) ∈ R^2$ is not differentiable at $(0,0)$. (First see what $Df_{(0,0)}(h,0)$ and $Df_{(0,0)}(0,k)$ need to be.)

Proof:

First assume $Df_{(0,0)}$ exists and it's $(\alpha, \beta)$.

We first check $Df_{(0,0)}(h,0)$

$$\begin{align*} & f(h,0) - f(0,0) - (\alpha, \beta)(h,0)^T \\ &= 0 - 0 - \alpha h \\ &= -\alpha h \end{align*}$$

For $Df_{(0,0)}(h,0)h$ to be $o(h)$, we must have $\alpha = 0$. Similarly, we must have $\beta = 0$.

So if $Df_{(0,0)}$ exists, it can only be $0$.

Then we check

$$\begin{align*} & f(h,h) - f(0,0) - Df_{(0,0)} h \\ &= |h| - 0 - 0 \\ &= |h| \\ & = \mathcal{O}(h) \\ & \neq o(h) \end{align*}$$

This shows $Df_{(0,0)}$ cannot be $0$, then we have a contradiction.

So $f$ is not differentiable at $(0,0)$

$\square$

4.4 Basic Results and the Chain Rule

4.4.1.

Prove part (2) of Proposition 4.4.1.

Proof: See ch04 notes.

$\square$

4.4.2.

Prove part (2) of Proposition 4.4.2.

Proof: See ch04 notes.

$\square$

4.4.3.

Prove part (2) of Lemma 4.4.4.

Define the reciprocal function

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

Then 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 $o(h)$.

$\square$

4.4.4

Prove the quotient rule.

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

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

$\square$

4.4.5.

Let $f(x,y,z) = xyz$. Find $Df_{(a,b,c)}$ for arbitrary $(a,b,c) ∈ R^3$. (Hint: $f$ is the product $XYZ$, where $X$ is the linear function $X(x,y,z) = x$ and similarly for $Y$ and $Z$.)

Solution:

$$\begin{align*} f(x,y,z) &= XYZ \\ Df_{(a,b,c)} &= D(XYZ)_{(a,b,c)} \\ &= X(a,b,c)D(YZ)_{(a,b,c)} + (YZ)(a,b,c) DX_{(a,b,c)} \\ &= a (Y(a,b,c) DZ_{(a,b,c)} + Z(a,b,c) DY_{(a,b,c)}) + bcX \\ &= abZ + acY + bcX \\ \end{align*}$$

$\square$

4.4.6.

Define $f(x,y) = xy^2/(y − 1)$ on $\{(x,y) \in R^2 : y \neq 1\}$. Find $Df_{(a,b)}$ where $(a,b)$ is a point in the domain of $f$.

Solution:

Define $X(x, y) = x, Y(x,y) = y$, then we have:

$$\begin{align*} &Df_{a,b}(h,k) \\ &= \frac{ (Y-1)(a,b) D(XY^2)_{(a,b)} - (XY^2)(a,b)D(Y-1)_{(a,b)} }{ (Y-1)^2(a,b) }(h,k) \\ &= \frac{ (b-1)(X(a,b)D(Y^2)_{(a,b)}+Y^2(a,b)DX_{(a,b)}) - ab^2 Y }{ (b-1)^2 }(h,k) \\ &= \frac{ (b-1)(2abY + b^2X) - ab^2 Y }{ (b-1)^2 }(h,k) \\ &= \left( \frac{b^2}{b-1}X + \frac{ab^2-2ab}{(b-1)^2}Y \right) (h,k) \end{align*}$$

$\square$

4.4.7

(A generalization of the product rule.) Recall that a function

$$f : \mathbb{R}^n \times \mathbb{R}^n \longrightarrow \mathbb{R}$$

is called bilinear if for all $x,x',y,y' ∈ \mathbb{R}^n$ and all $\alpha \in \mathbb{R}$,

$$\begin{align*} f(x+x',y) &= f(x,y)+f(x',y), \\ f(x,y +y') &= f(x,y)+f(x,y'), \\ f(αx,y) &= αf(x,y) = f(x,αy). \\ \end{align*}$$

(a) Show that if $f$ is bilinear then $f(h,k)$ is $o(h,k)$.

Proof:

Let

$$\begin{align*} h &= h_1 e_1 + \cdots + h_n e_n \\ k &= k_1 e_1 + \cdots + k_n e_n \\ \end{align*}$$

Then

$$\begin{align*} f(h,k) &= \sum_{i = 1}^{n} \sum_{j = 1}^{n} h_i k_j f(e_i, e_j) \end{align*}$$

We can find a $M$ such that $|f(e_i, e_j)| \leq M$ for all $i, j$. We also know $|h_i| \leq |h| \leq |(h,k)|, |k_j| \leq |k| \leq |(h,k)|$ for all $i, j$.

So we have

$$\begin{align*} |f(h,k)| &= \left| \sum_{i = 1}^{n} \sum_{j = 1}^{n} h_i k_j f(e_i, e_j) \right| \\ & \leq \sum_{i = 1}^{n} \sum_{j = 1}^{n} |h_i k_j f(e_i, e_j)| \\ & \leq |h||k| n^2 M \\ & \leq n^2 M |(h,k)|^2 \\ &= o(h, k) \end{align*}$$

$\square$

(b) Show that if $f$ is bilinear then $f$ is differentiable with $Df_{(a,b)}(h,k) = f(a,k)+f(h,b)$.

Proof:

First of all, we need to show $Df_{(a,b)}(h,k)$ is a linear function of $(h,k)$.

Given $(h_1,k_1)$ and $(h_2,k_2)$

$$\begin{align*} & Df_{(a,b)}((h_1,k_1) + (h_2,k_2)) \\ &= Df_{(a,b)}(h_1+h_2,k_1+k_2) \\ &= f(a, k_1+k_2) + f(h_1+h_2, b) \\ &= (f(a,k_1)+f(h_1,b)) + (f(a,k_2)+f(h_2,b)) \\ &= Df_{(a,b)}(h_1,k_1) + Df_{(a,b)}(h_2,k_2) \\ \end{align*}$$

Similarly given $\alpha \in \mathbb{R}$,

$$\begin{align*} & Df_{(a,b)}(\alpha(h,k)) \\ &= Df_{(a,b)}(\alpha h, \alpha k) \\ &= f(a, \alpha k) + f(\alpha h, b) \\ &= \alpha (f(a,k)+f(h,b)) \\ &= \alpha Df_{(a,b)}(h,k) \end{align*}$$

Then finally, we have

$$\begin{align*} &f(a+h, b+k) - f(a, b) - (f(a,k)+f(h,b)) \\ &=(f(a,b) + f(h, b) + f(a, k) + f(h, k)) - f(a, b) - (f(a,k)+f(h,b)) \\ &= f(h,k) \\ &= o(h, k) \end{align*}$$

So $f$ is differentiable with $Df_{(a,b)}(h,k) = f(a,k)+f(h,b)$.

$\square$

(c) What does this exercise say about the inner product?

Solution:

Since inner product is a bilinear function, then it's differentiable at any point of $(a,b)$.

$\square$

A little comments for part (a):

It took me a long way to reach to the solution above. Initially, I was thinking to use the following way.

Assume $h = |h| r_0, k = |k| r_1$, where $r_0, r_1 \in \mathbb{R}^n$ and $|r_0| = |r_1| = 1$. Then I want to show $f(h,k) = |h||k|f(r_0, r_1)$, as long as I can bound $|f(r_0, r_1)|$, then it's done.

So my next thought is to prove $f$ is continuous.

Given $(x_0, y_0)$ and $(x_1, y_1)$, I want to use triangle inequality. Ideally, I would like to have the following, but it's not correct.

$$\begin{align*} \left| f(x_1, y_1) - f(x_0, y_0) \right| &= \left| f(x_0, y_1 - y_0) + f(x_1 - x_0, y_0) \right| \\ \end{align*}$$

So I stuck here. I guess I was kinda inspired by how to prove a linear map $T$ is continuous which I learned from Gemini.

$\square$

4.4.8. (A bigger generalization of the product rule.)

A function

$$f : \mathbb{R}^n \times \cdots \times \mathbb{R}^n \longrightarrow \mathbb{R}$$

(there are $k$ copies of $\mathbb{R}^n$) is called multilinear if for each $j ∈ {1,...,k}$, for all $x_1,...,x_j,x_j',...,x_k ∈ \mathbb{R}^n$ and all $\alpha ∈ \mathbb{R}$,

$$\begin{align*} f(x_1,...,x_j +x_j',...,x_k) &= f(x_1,...,x_j,...,x_k)+f(x_1,...,x_j',...,x_k) \\ f(x_1,...,αx_j,...,x_k) &= αf(x_1,...,x_j,...,x_k) \end{align*}$$

(a) Show that if $f$ is multilinear and $a_1,...,a_k,h_1,...,h_k ∈ \mathbb{R}^n$ then for $i,j ∈ \{1,...,k\}$ (distinct), $f(a_1,...,h_i,...,h_j,...,a_k)$ is $o(h_1,...,h_k)$. (Use the previous problem.)

Proof:

Let $g(h_i, h_j) = f(a_1,...,h_i,...,h_j,...,a_k)$ then $g(h_i, h_j)$ is a bilinear function.

Since $g(h_i, h_j)$ is $o(h_i, h_j)$ from the exercise 4.4.7 and since $|(h_i, h_j)| \leq |(h_1,...,h_k)|$, then we can conclude $g(h_i, h_j)$ is $o(h_1,...,h_k)$.

$\square$

(b) Show that if $f$ is multilinear then $f$ is differentiable with

$$Df_{(a_1 ,...,a_k)}(h_1,...,h_k) = \sum_{j = 1}^{k} f(a_1,...,a_{j−1},h_j,a_{j+1},...,a_k).$$

Proof:

The linearity of $Df_{(a_1 ,...,a_k)}$ is similarly to exercise 4.4.7 (b). So we skip here.

We would like to first prove: given a multilinear function $f(h_1, h_2, a_3, \cdots, a_k)$ and any $c > 0$, for any $|a_i| \leq M$, we can find $h > 0$ such that if $|(h_1, h_2)| < h$,

$$|f(h_1, h_2, a_3, \cdots, a_k)| < c|(h_1, h_2)|$$

We use induction on $k$. When $k = 3$, assume

$$a_3 = x_1 e_1 + \cdots + x_n e_n$$

Then

$$\begin{align*} & |f(h_1, h_2, a_3)| \\ &= \left| \sum_{i = 1}^{n} x_i f(h_1, h_2, e_i) \right| \\ &\leq \sum_{i = 1}^{n} \left| x_i f(h_1, h_2, e_i) \right| \\ &= \sum_{i = 1}^{n} |x_i| |f(h_1, h_2, e_i)| \\ & \leq M \sum_{i = 1}^{n} |f(h_1, h_2, e_i)| \end{align*}$$

For each $i$ we can find $k_i$ such that if $|(h_1, h_2)| < k_i$, then $|f(h_1, h_2, e_i)| < c/(nM) \cdot |(h_1, h_2)|$. Let $h = \min k_i$, then if $|(h_1, h_2)| < h$,

$$M \sum_{i = 1}^{n} |f(h_1, h_2, e_i)| < M \cdot n \cdot c/(nM) \cdot |(h_1, h_2)| = c|(h_1, h_2)|$$

Then assume the statement is true for $3 \leq k \leq m$.

We will prove when $k = m+1$, assume

$$a_{m+1} = x_1 e_1 + \cdots + x_n e_n$$

then

$$\begin{align*} &|f(h_1, h_2, a_3, \cdots, a_m, a_{m+1})| \\ & \leq \sum_{i = 1}^{n} \left| x_i f(h_1, h_2, a_3, \cdots, a_m, e_i) \right| \\ & = \sum_{i = 1}^{n} \left| x_i | | f(h_1, h_2, a_3, \cdots, a_m, e_i) \right| \\ & \leq M \sum_{i = 1}^{n} | f(h_1, h_2, a_3, \cdots, a_m, e_i) | \\ \end{align*}$$

Let $g_i(h_1, h_2, a_3, \cdots, a_m) = f(h_1, h_2, a_3, \cdots, a_m, e_i)$ and $g_i$ is a multilinear function with only $k = m$ parameters.

Now we can apply the induction, for each $g_i$, we can find $k_i$ such that if $|(h_1, h_2)| < k_i$, then $|g_i(h_1, h_2, a_3, \cdots, a_m)| < c/(nM) \cdot |(h_1, h_2)|$. Let $h = \min k_i$, then if $|(h_1, h_2)| < h$,

$$M \sum_{i = 1}^{n} |g_i(h_1, h_2, a_3, \cdots, a_m)| < M \cdot n \cdot c/(nM) \cdot |(h_1, h_2)| = c|(h_1, h_2)|$$

So in summary, we proved given a multilinear function $f(h_1, h_2, a_3, \cdots, a_k)$ and any $c > 0$, we can find $h > 0$, as long as $|a_i| \leq M$, and $|(h_1, h_2)| < h$,

$$|f(h_1, h_2, a_3, \cdots, a_k)| < c|(h_1, h_2)|$$

The above statement can be generalized to given a multilinear function $f(a_1, a_2, a_3, \cdots, a_k)$ and any $c > 0$, we can find $h > 0$, as long as $|a_i| \leq M$, and $|(h_i, h_j)| < h$, we have

$$|f(a_1,...,h_i,...,h_j,...,a_k)| < c |(h_i, h_j)| \leq c |(h_1, \cdots, h_k)|$$

Then note that

$$\begin{align*} & f(a_1+h_1, \cdots, a_k+h_k) \\ &= f(a_1, \cdots, a_k) \\ &+ \sum_{j = 1}^{k} f(a_1,...,a_{j−1},h_j,a_{j+1},...,a_k) \\ &+ \text{ items with at least 2 } h_i, h_j \end{align*}$$

From the previous induction and the generalization, as long as $|h_i| \leq M, |a_j| \leq M$, then items with at least 2 $h_i, h_j$ are $o(h_1,...,h_k)$, so

$$\begin{align*} &f(a_1+h_1, \cdots, a_k+h_k) - f(a_1, \cdots, a_k) - Df_{(a_1 ,...,a_k)}(h_1,...,h_k) \\ &= \text{ items with at least 2 } h_i, h_j \\ &= o(h_1,...,h_k) \end{align*}$$

$\square$

Another much cleaner approach suggested by Gemini is to prove this first

$$|f(a_1, \cdots, a_k)| \leq C |a_1| \cdots| a_k|$$

Assume

$$a_i = x_{i1} e_1 + \cdots + x_{in} e_n$$

Then

$$\begin{align*} &|f(a_1, \cdots, a_k)| \\ & \leq \sum \limits_{j_1 = 1}^{n} \cdots \sum \limits_{j_k = 1}^{n} |x_{j_1,1}|\cdots|x_{j_k,k}||f(e_1, \cdots, e_k)| \\ & \leq |a_1| \cdots| a_k| \sum \limits_{j_1 = 1}^{n} \cdots \sum \limits_{j_k = 1}^{n} |f(e_1, \cdots, e_k)| \end{align*}$$

$\square$

(c) When $k = n$, what does this exercise say about the determinant?

Proof:

It means the determinant is differentiable at any point $(a_1, \cdots, a_n)$.

$\square$

4.5 Calculating the Derivative

4.5.1.

Explain why in the discussion beginning this section the tangent plane $\mathcal{P}$ consists of all points $(a,b,f(a,b)) + (h,k,T(h,k))$ where $T(h,k) = ϕ'(a)h+ψ'(b)k$.

4.5.2

This exercise shows that all partial derivatives of a function can exist at and about a point without being continuous at the point. Define $f : \mathbb{R}^2 \longrightarrow \mathbb{R}$ by

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

(a) Show that $D_1f(0,0) = D_2f(0,0) = 0$.

Proof:

$$\begin{align*} D_1f(0,0) &= \lim \limits_{t \to 0} \frac{ f(t,0) - f(0,0) }{t} \\ &= \lim \limits_{t \to 0} \frac{\frac{2t \cdot 0}{t^2 + 0^2}-0}{t} \\ &= \lim \limits_{t \to 0} \frac{0-0}{t} \\ &= 0 \end{align*}$$

Similarly $D_{2}f_{}(0,0) = 0$.

$\square$

(b) Show that $D_1f(a,b)$ and $D_2f(a,b)$ exist and are continuous at all other $(a,b) ∈ \mathbb{R}^2$.

Proof:

$$\begin{align*} D_1f(a,b) &= \frac{2b}{a^2+b^2} -\frac{(2ab)(2a)}{(a^2+b^2)^2} \\ &= \frac{2a^2b + 2b^3 - 4a^2b}{(a^2+b^2)^2} \\ &= \frac{2b^3 - 2a^2b}{(a^2+b^2)^2} \end{align*}$$

Similarly,

$$\begin{align*} D_2f(a,b) &= \frac{2a^3 - 2b^2a}{(b^2+a^2)^2} \end{align*}$$

It is a continuous function.

$\square$

(c) Show that $D_1f$ and $D_2f$ are discontinuous at $(0,0)$.

Proof:

If we approach $(0,0)$ with $a=b$, then

$$\begin{align*} D_1f(a,b) &= \frac{2b^3 - 2a^2b}{(a^2+b^2)^2} \\ &= \frac{2a^3 - 2a^3}{(a^2+a^2)^2} \\ &= 0 \end{align*}$$

If we approach $(0,0)$ with $b=2a$, then

$$\begin{align*} D_1f(a,b) &= \frac{2b^3 - 2a^2b}{(a^2+b^2)^2} \\ &= \frac{16a^3 - 4a^3}{(a^2+4a^2)^2} \\ &= \frac{12}{25a} \\ & \rightarrow \infty \end{align*}$$

So $D_1f$ is not continuous at $(0,0)$. Similarly, $D_2f$ is not continuous at $(0,0)$.

$\square$

4.5.3

Define $f : \mathbb{R} \longrightarrow \mathbb{R}$ by

$$f(x) = \begin{cases} x^2 \sin \frac{1}{x} &\text{if } x \neq 0\\ 0 &\text{if } x = 0\\ \end{cases}$$

Show that $f'(x)$ exists for all $x$ but that $f'$ is discontinuous at $0$. Explain how this disproves the converse of Theorem 4.5.3.

Proof:

When $x \neq 0$,

$$\begin{align*} f'(x) &= 2x \sin \frac{1}{x} - x^2 \cos \frac{1}{x} (\frac{1}{x^2}) \\ &= 2x \sin \frac{1}{x} - \cos \frac{1}{x} \end{align*}$$

When $x = 0$,

$$\begin{align*} f'(0) &= \lim \limits_{x \to 0} \frac{x^2 \sin \frac{1}{x} - 0}{x} \\ &= \lim \limits_{x \to 0} x \sin \frac{1}{x} \\ &= 0 \end{align*}$$

Consider $a_n = \frac{1}{2 \pi n}, b_n = \frac{1}{(2n+1) \pi }$, then $f'(a_n) = -1, f'(b_n) = 1$.

So $f'(x)$ is not continuous at $0$.

$\square$

4.5.4.

Discuss the derivatives of the following mappings at the following points.

(a) $f(x,y) = \frac{x^2 - y}{y+1}$ on $\{(x,y) \in R^2 : y \neq −1\}$ at generic $(a,b)$ with $b \neq −1$.

Proof:

$$\begin{align*} D_{1}f_{}(a,b) &= \frac{2a}{b+1} \end{align*}$$

$$\begin{align*} D_{2}f_{}(a, b) &= \frac{-1}{b+1} - \frac{a^2-b}{(b+1)^2} \\ &= -\frac{a^2+1}{(b+1)^2} \end{align*}$$

So

$$Df(a,b) = \left[ \frac{2a}{b+1}, -\frac{1+a^2}{(b+1)^2} \right]$$

$\square$

(b) $f(x,y) = \frac{xy^2}{y-1}$ on $\{(x,y) ∈ \mathbb{R}^2 : y \neq 1\}$ at generic $(a,b)$ with $b \neq 1$.

Proof:

$$\begin{align*} D_{1}f_{}(a,b) &= \frac{b^2}{b-1} \end{align*}$$

$$\begin{align*} D_{2}f_{}(a,b) &= \frac{2ab}{b-1} - \frac{ab^2}{(b-1)^2} \\ &= \frac{ab^2-2ab}{(b-1)^2} \end{align*}$$

So

$$Df(a,b) = \left[ \frac{b^2}{b-1}, \frac{ab^2-2ab}{(b-1)^2} \right]$$

$\square$

(c)

$$f(x,y)= \begin{cases} \frac{xy}{\sqrt[]{x^2+y^2}} &\text{if } (x,y) \neq (0,0)\\ 0 &\text{if } (x,y) = (0,0)\\ \end{cases}$$

at generic $(a,b) \neq (0,0)$ and at $(0,0)$.

Proof:

First consider $(a,b) \neq (0,0)$.

$$\begin{align*} D_{1}f_{}(a,b) &= \frac{b}{\sqrt[]{a^2+b^2}} - \frac{2a^2b}{2(a^2+b^2)^{3/2}} \\ &= \frac{2b(a^2+b^2)-2a^2b}{2(a^2+b^2)^{3/2}} \\ &= \frac{b^3}{(a^2+b^2)^{3/2}} \end{align*}$$

Similarly,

$$\begin{align*} D_{2}f_{}(a,b) &= \frac{a^3}{(a^2+b^2)^{3/2}} \end{align*}$$

So

$$Df(a,b)= \left[ \frac{b^3}{(a^2+b^2)^{3/2}}, \frac{a^3}{(a^2+b^2)^{3/2}} \right]$$

Then note $D_1f(a,b)$ is not continuous at $(0,0)$. If we approach $(0,0)$ with $a = b$, then we get

$$\lim \limits_{(a,b) \to (0,0)} D_1f(a,b) = \frac{1}{2\sqrt[]{2}}$$

But if we approach $(0,0)$ with $a = 2b$, then we get

$$\lim \limits_{(a,b) \to (0,0)} D_1f(a,b) = \frac{1}{5\sqrt[]{5}}$$

Similarly, $D_2f(a,b)$ is not continuous at $(0,0)$.

Now, assume $f$ is differentiable at $(0,0)$ and $Df(0,0) = [c,d]$, and we approach approach $(0,0)$ with $a = b = h$

$$\begin{align*} f(h,h) - f(0,0) - Df(0,0)(h,h) = \frac{1}{\sqrt[]{2}}|h| - (c+d)h = o(|h|) \end{align*}$$

So if $h > 0$, then $c+d = \frac{1}{\sqrt[]{2}}$ and if $h < 0$, then $c+d = -\frac{1}{\sqrt[]{2}}$. This is not possible.

So $f$ does not have derivative at $(0,0)$.

$\square$

4.5.5.

For what differentiable mappings $f : A \longrightarrow \mathbb{R}^m$ is $f'(a)$ a diagonal matrix for all $a ∈ A$? (A diagonal matrix is a matrix whose $(i,j)$th entries for all $i \neq j$ are $0$.)

Solution:

It means

$$D_{j}f_{i}(a) = \begin{cases} 0 &\text{if } i \neq j \\ f_i'(a_i) &\text{if } i = j \\ \end{cases}$$

So the format is like

$$f(a) = (f_1(a_1), \cdots, f_n(a_n))$$

Basically, $f_i$ is only one-variable function and is only determined by $a_i$.

$\square$

4.5.6.

Show that if $z = f(xy)$ then $x$, $y$, and z satisfy the differential equation $x \cdot z_x−y \cdot z_y = 0$.

Solution:

Let

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

So $z = f (g(x,y)) = (f \circ g)(x,y)$

$$\begin{align*} z_x(a,b) &= D_1(f \circ g)(a,b) \\ &= D_1 f(g(a,b)) \cdot D_1g(a,b) \\ &= D_1 f(g(a,b)) \cdot b \end{align*}$$

Similarly

$$\begin{align*} z_y(a,b) &= D_2(f \circ g)(a,b) \\ &= D_1 f(g(a,b)) \cdot D_2g(a,b) \\ &= D_1 f(g(a,b)) \cdot a \end{align*}$$

So we have

$$a \cdot z_x(a,b) = ab D_1 f(g(a,b)) = b \cdot z_y(a,b)$$

i.e.

$$x \cdot z_x−y \cdot z_y = 0$$

$\square$

4.5.7.

Let $w = F(xz,yz)$. Show that $x \cdot w_x +y \cdot w_y = z \cdot w_z$.

Proof:

Let

$$g : \mathbb{R}^3 \longrightarrow \mathbb{R}^2, \qquad (x,y,z) \longrightarrow (xz, yz)$$

Then $w = F(g(x,y,z)) = (F \circ g)(x,y,z)$

$$\begin{align*} w_x(a,b,c) &= D_{1}(F \circ g)_{1}(a,b,c) \\ &= D_{1}F_{1}(g(a,b,c))D_{1}g_{1}(a,b,c)+ D_{2}F_{1}(g(a,b,c))D_{1}g_{2}(a,b,c) \\ &= D_{1}F_{1}(g(a,b,c)) \cdot c + D_{2}F_{1}(g(a,b,c)) \cdot 0 \\ &= D_{1}F_{1}(g(a,b,c)) \cdot c \end{align*}$$

$$\begin{align*} w_y(a,b,c) &= D_{2}(F \circ g)_{1}(a,b,c) \\ &= D_{1}F_{1}(g(a,b,c))D_{2}g_{1}(a,b,c)+ D_{2}F_{1}(g(a,b,c))D_{2}g_{2}(a,b,c) \\ &= D_{1}F_{1}(g(a,b,c)) \cdot 0 + D_{2}F_{1}(g(a,b,c)) \cdot c \\ &= D_{2}F_{1}(g(a,b,c)) \cdot c \end{align*}$$

$$\begin{align*} w_z(a,b,c) &= D_{3}(F \circ g)_{1}(a,b,c) \\ &= D_{1}F_{1}(g(a,b,c))D_{3}g_{1}(a,b,c)+ D_{2}F_{1}(g(a,b,c))D_{3}g_{2}(a,b,c) \\ &= D_{1}F_{1}(g(a,b,c)) \cdot a + D_{2}F_{1}(g(a,b,c)) \cdot b \\ \end{align*}$$

So

$$(x \cdot w_x)(a,b,c) = ac D_{1}F_{1}(g(a,b,c)) \\ (y \cdot w_y)(a,b,c) = bc D_{2}F_{1}(g(a,b,c)) \\ (z \cdot w_z)(a,b,c) = ac D_{1}F_{1}(g(a,b,c)) + bc D_{2}F_{1}(g(a,b,c))$$

Therefore

$$x \cdot w_x +y \cdot w_y = z \cdot w_z$$

$\square$

4.5.8.

If $z = f(ax+by)$, show that $bz_x = az_y$.

Proof:

Let

$$g : \mathbb{R}^2 \longrightarrow \mathbb{R}^1, \qquad (x,y) \longrightarrow ax+by$$

So $z = f(g(x,y)) = (f \circ g)(x,y)$.

$$\begin{align*} z_x(c,d) &= D_{1}(f \circ g)_{1}(c,d) \\ &= D_{1}f_{1}(g(c,d))D_{1}g_{1}(c,d) \\ &= D_{1}f_{1}(g(c,d)) \cdot a \end{align*}$$

$$\begin{align*} z_y(c,d) &= D_{2}(f \circ g)_{1}(c,d) \\ &= D_{1}f_{1}(g(c,d))D_{2}g_{1}(c,d) \\ &= D_{1}f_{1}(g(c,d)) \cdot b \end{align*}$$

So we have

$$b z_x(c,d) = abD_{1}f_{1}(g(c,d)) = a z_y(c,d)$$

So we have $bz_x = az_y$.

$\square$

4.5.9.

The function $f : \mathbb{R}^2 \longrightarrow \mathbb{R}$ is called homogeneous of degree $k$ if $f(tx,ty) = t^kf(x,y)$ for all scalars $t$ and vectors $(x,y)$. Letting $f_1$ and $f_2$ denote the first and second partial derivatives of $f$, show that such $f$ satisfies the differential equation

$$xf_1(x,y)+yf_2(x,y) = kf(x,y).$$

(Hint: First differentiate the homogeneity condition with respect to $t$, viewing $x$ and $y$ as fixed but generic; the derivative of one side will require the chain rule. Second, since the resulting condition holds for all scalars $t$, it holds for any particular $t$ of your choosing.)

Proof:

Let

$$g : \mathbb{R}^3 \longrightarrow \mathbb{R}^2, \qquad g(x,y,t) \longrightarrow (tx, ty)$$

Then $f(tx,ty) = f(g(x,y,t)) = (f \circ g)(x,y,t)$,

$$\begin{align*} D_{3}(f \circ g)(a,b,c) &= D_{1}f(g(a,b,c)) D_{3}g_{1}(a,b,c) + D_{2}f(g(a,b,c)) D_{3}g_{2}(a,b,c) \\ &= D_{1}f(g(a,b,c)) \cdot a + D_{2}f(g(a,b,c)) \cdot b \end{align*}$$

On the other hand since $f(tx,ty) = t^kf(x,y)$, then

$$D_{3}(f \circ g)(a,b,c) = kc^{k-1}f(a,b)$$

If we plug $c = 1$ in both equation, then for any $a,b$ we get

$$a D_{1}f(a, b) + b D_{2}f(a,b) = k f(a,b)$$

Then that's exactly

$$xf_1(x,y)+yf_2(x,y) = kf(x,y)$$

$\square$

4.5.10

Let

$$f : \mathbb{R}^2 \longrightarrow \mathbb{R}$$

be a function such that for all $(x,y) ∈ \mathbb{R}^2$, the integral

$$F : \mathbb{R}^2 \longrightarrow \mathbb{R}, \qquad F(x,y) = \int_{v=0}^{y} f(x,v) dv$$

exists and is differentiable with respect to $x$, its partial derivative with respect to $x$ being obtained by passing the $x$-derivative through the $v$-integral,

$$\begin{align*} \frac{\partial F(x,y)}{\partial x} &= \frac{\partial }{\partial x} \int_{v=0}^{y} f(x,v) dv \\ &= \lim \limits_{h \to 0} \frac{ \int_{v=0}^{y}f(x+h,v)dv - \int_{v=0}^{y}f(x,v)dv }{h} \\ &= \lim \limits_{h \to 0} \int_{v=0}^{y} \frac{ f(x+h,v) - f(x,v) }{h} dv \\ & \overset{!}{=} \int_{v=0}^{y} \lim \limits_{h \to 0} \frac{ f(x+h,v) - f(x,v) }{h} dv \\ &= \int_{v=0}^{y} \frac{ \partial f }{\partial x} (x,v) dv \end{align*}$$

(The "!" step requires justification, but under reasonable circumstances it can be carried out.) Define a function

$$G : \mathbb{R} \longrightarrow \mathbb{R}, \qquad G(x) = \int_{v=0}^{x}f(x,v)dv$$

Thus $x$ affects $G$ in two ways: as a parameter for the integrand, and as the upper limit of integration. What is $dG(x)/dx$?

Solution:

Note $G(x) = F(x, x)$, now we can consider

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

Now we have $G(x) = (F \circ g)(x)$, so

$$\begin{align*} \frac{dG}{dx}(a) &= D_{1}(F \circ g)_{1}(a) \\ &= D_{1}F_{1}(g(a)) D_{1}g_{1}(a) + D_{2}F_{1}(g(a)) D_{1}g_{2}(a) \\ &= D_{1}F_{1}(a, a) \cdot 1 + D_{2}F_{1}(a, a) \cdot 1 \\ &= \left( \int_{v=0}^{a} \frac{ \partial f }{\partial x} (a,v) dv \right) + f(a,a) \end{align*}$$

$\square$

4.6 Higher-Order Derivatives

4.6.2

Suppose $u$, as a function of $x$ and $y$, satisfies the differential equation $u_{xx} − u_{yy} = 0$. Make the change of variables $x = s + t, y= s− t$. What corresponding differential equation does $u$ satisfy when viewed as a function of $s$ and $t$? (That is, find a nontrivial relation involving at least one of $u, u_s, u_t, u_{ss}, u_{tt},$ and $u_{st}$.)

Solution:

$$\begin{align*} u_s &= u_x x_s + u_y y_s \\ &=u_x + u_y \\ u_t &= u_x x_t + u_y y_t \\ &= u_x - u_y \end{align*}$$

$$\begin{align*} u_{ss} &= u_{xx}x_s + u_{xy} y_s + u_{yx} x_s + u_{yy} y_s \\ &= u_{xx} + u_{xy} + u_{yx} + u_{yy} \\ &= u_{xx} + 2 u_{xy} + u_{yy} \end{align*}$$

$$\begin{align*} u_{st} &= u_{xx}x_t + u_{xy} y_t + u_{yx} x_t + u_{yy} y_t \\ &= u_{xx} - u_{xy} + u_{yx} - u_{yy} \\ &= u_{xx} - u_{yy} \\ &= 0 \end{align*}$$

$$\begin{align*} u_{tt} &= u_{xx}x_t + u_{xy} y_t - u_{yx} x_t - u_{yy} y_t \\ &= u_{xx} - u_{xy} - u_{yx} + u_{yy} \\ &= u_{xx} - 2 u_{xy} + u_{yy} \end{align*}$$

So the function is $u_{st}$

$\square$

4.6.3. (The wave equation)

(a) Let $c$ be a constant, tacitly understood to denote the speed of light. Let $x$ and $t$ denote a space variable and a time variable, and introduce variables

$$p =x+ct, \qquad q = x-ct.$$

Show that a quantity $w$, viewed as a function of $x$ and $t$, satisfies the wave equation,

$$c^2 w_{xx} = w_{tt}$$

if and only if it satisfies the equation

$$w_{pq} = 0$$

Proof:

$\Rightarrow$

$$\begin{align*} x &= \frac{p+q}{2} \\ t &= \frac{p-q}{2c} \\ x_p = 1/2 &, x_q = 1/2 \\ t_p = 1/(2c) &, t_q = -1/(2c) \end{align*}$$

$$\begin{align*} w_p &= w_x x_p + w_t t_p \\ &= \frac{1}{2} w_x + \frac{1}{2c} w_t \\ \end{align*}$$

$$\begin{align*} w_{pq} &= \frac{1}{2} w_{xx}x_q + \frac{1}{2} w_{xt} t_q + \frac{1}{2c} w_{tx}x_q + \frac{1}{2c} w_{tt}t_q \\ &= \frac{1}{4} w_{xx} - \frac{1}{4c} w_{xt} + \frac{1}{4c} w_{tx} - \frac{1}{4c^2} w_{tt} \\ &= 0 \end{align*}$$

$\Leftarrow$

$$\begin{align*} p_x = 1, &\qquad p_t = c \\ q_x = 1, &\qquad q_t = -c \\ \end{align*}$$

$$\begin{align*} w_x &= w_p p_x + w_q q_x \\ &= w_p + w_q \\ w_t &= w_p p_t + w_q q_t \\ &= w_p c - w_q c \\ w_{xx} &= w_{pp} p_x + w_{pq} q_x + w_{qp} p_x + w_{qq} q_x \\ &= w_{pp} + w_{pq} + w_{qp} + w_{qq} \\ &= w_{pp} + w_{qq} \\ w_{tt} &= c (w_{pp} p_t + w_{pq} q_t - w_{qp} p_t - w_{qq} q_t) \\ &= c (w_{pp} c - w_{pq} c - w_{qp} c + w_{qq} c) \\ &= c^2 (w_{pp} - w_{pq} - w_{qp} + w_{qq}) \\ &= c^2 (w_{pp} + w_{qq}) \\ \end{align*}$$

So $w_{tt} = c^2 w_{xx}$ as required.

$\square$