Tag: Probability

Profit Maximisation for Math People
When I had to learn economics in high school, I struggled tremendously due to its language-based emphasis (which meant tons of essay-writing with arguably subjective marking schemes).

Yet, I remember how one teaching intern scribbled some equations to prove them mathematically, and those equations were my only takeaways from the entire two years of economics. I’m convinced that economics without equations is gravely incomplete. Equivalently, what this means is that when we include the mathematics with the economic theory, the concepts therein start to illuminate meaning.

Here’s one of my favourite results in economics.

Theorem 1. A firm maximises profit when its marginal revenue (MC) equals its marginal cost (MC) and when the marginal cost is increasing.

Here’s a layperson’s justification for this theorem.

If the firm is producing at a quantity where MR > MC, then it can increase profit by increasing output. The reason is since the marginal revenue exceeds the marginal cost, additional output is adding more to profit than it is taking away. If the firm is producing at a quantity where MC > MR, then it can increase profit by reducing output.
Adapted from Lumen Learning

Technically, this explanation can make sense, but I was absolutely confused to pieces by it. Don’t even get me started on having to regurgitate this paragraph compounded with my atrocious handwriting when answering essay-based questions. Let’s define our terms then prove this mathematically.

Definition. Let $x$ denote the quantity of a product that a firm produces. Make the following notations:
- $R(x)$ : the revenue the firm earns by selling $x$ units of such a product,
- $C(x)$ : the cost the firm incurs by producing $x$ units of such a product,
- $\pi(x)$ : the profit the company earns by selling $x$ units of the product.
It is intuitive that $\pi = R - C$ . Note that we are adopting economists’ convention to denote profit by $\pi$ , not to be confused with the circle constant $3.14159\dots$ .

Equivalently, we define the marginal revenue function and marginal cost function by $\mathrm{MR} := R'$ and $\mathrm{MC} := C'$ respectively.

Now, onto the proof, using the derivative tests that we have previously developed. We will assume that $\mathrm{MR}$ is strictly decreasing, though that actually can be proven from more fundamental principles of economics.

Proof of Theorem 1. To maximise profit at the output level $c$ , we need $\pi'(c) = 0$ . Since $\pi = R - C$ , this implies that

$\displaystyle 0=\pi'(c) = (R-C)'(c) = R'(c) - C'(c) = \mathrm{MR}(c) - \mathrm {MC}(c).$

By algebra, $\mathrm{MR}(c) = \mathrm{MC}(c)$ . Assuming that $\mathrm{MC}$ is decreasing so that $\mathrm{MC}'(c) \leq 0$ ,

$\pi''(c) = \mathrm{MR}'(c) - \mathrm{MC}'(c) < 0 - 0 = 0.$

By the second derivative test, $c$ is a local maximum for $\pi$ .

We could have many more economics-contextualised applications of differential calculus, but perhaps another time.

For now, we shift gears and discuss the overrated l’Hôpital’s rule.

—Joel Kindiak, 31 Oct 24, 1402H
October 31, 2024

ai, machine-learning, math, mathematics, Probability
Technical Optimisation
One of the most practical uses of differential calculus is to optimise desired quantities, which amounts to either minimising a function $f(x)$ on some subset $I \subseteq \mathbb R$ , or maximising it. How do we do that?

Let’s first formulate the notion of a global minimum and local minimum respectively.

Definition 1. Let $I \subseteq \mathbb R$ and $f : I \to \mathbb R$ be a function.
- We say that $(c, f(c))$ is a global minimum point of $f$ on $I$ if for any $x \in I$ , $f(x) \geq f(c)$ .
- We say that $(c, f(c))$ is a local minimum point of $f$ if there exists $\delta > 0$ such that $(c, f(c))$ is a global minimum point on $(c - \delta , c + \delta) \cap I$ .
To abbreviate, we will say that $c$ is a global (resp. local) minimum point of $f$ .

In a symmetric manner, we can define the global maximum and local maximum respectively.

We say that $c$ is a global (reap. local) extremum if it is either a global (resp. local) minimum or a global (resp. local) maximum. Since differentiable functions on an interval are continuous on that interval, by the extreme value theorem, we will always obtain a global extremum. In fact, we have the following symmetrical result.

Lemma 1. Suppose that $c$ is a global (resp. local) maximum for a function $f$ . Then $c$ is a global (reap. local) minimum for $-f$ .

Proof. Employ the equivalence of $f(x) \geq f(c) \iff -f(x) \leq -f(c)$ .

What about a local extremum?

Theorem 1. Let $f$ be a function that is differentiable at $c$ . Suppose $c$ is a local extremum at $f$ . Then $f'(c) = 0$ .

Proof. The proof of this result is essentially the same as that of Rolle’s theorem.

Since global and local extrema constitute all possible extremum points, we have the critical value test.

Theorem 2. Let $f: [a, b] \to \mathbb R$ be a continuous function that is differentiable on $(a, b)$ . Then $f$ obtains global extrema either at $a, b$ or interior points $c$ such that $f'(c)$ is undefined or $f'(c) = 0$ . These possible points are known as critical points.

Does the converse hold? That is, does $f'(c) = 0$ imply that $c$ is a local extremum? Not if we consider $c = 0$ and the function $f(x) = x^3$ . How then do we determine the nature of this stationary point?

Suppose $f'(c) = 0$ . Denote $f'(c^-) \leq 0$ (resp. $f'(c^+) \geq 0$ ). to mean that there exists $\delta > 0$ such that for any $x \in (c-\delta, c)$ (resp. $x \in (c, c + \delta)$ ), $f'(x) \leq 0$ (resp. $f'(x) \geq 0$ ). We can use the first derivative test to answer our aforementioned question.

Theorem 3 (First Derivative Test). Let $f : [a, b] \to \mathbb R$ be differentiable at $c \in (a, b)$ . Suppose $f'(c) = 0$ .
- If $f'(c^-) \leq 0$ and $f'(c^+) \geq 0$ , then $c$ is a local minimum of $f$ .
- If $f'(c^-) \geq 0$ and $f'(c^+) \leq 0$ , then $c$ is a local maximum of $f$ .
- If $f'(c^-), f'(c^+) \geq 0$ or $f'(c^-),f'(c^+) \geq 0$ , then $c$ is not a local extremum. In this case, we say that $c$ is a stationary point of inflection.
Proof. We will prove the first result.
- Since $f'(c^-) \leq 0$ , find $\delta_1 > 0$ such that for $x \in (c - \delta_1, c)$ , $f'(x) \leq 0$ .
- Since $f'(c^+) \geq 0$ , find $\delta_2 > 0$ such that for $x \in (c, c + \delta_2)$ , $f'(x) \geq 0$ .
Define $\delta := \min\{\delta_1,\delta_2\}$ . We claim that for any $x \in (c-\delta, c+\delta)$ , $f(x) \geq f(c)$ .

Suppose $x < c$ . By the mean value theorem, there exists

$c_1 \in (x,c) \subseteq (c-\delta, c) \subseteq (c-\delta_1,c),$

implying $f'(c_1) \leq 0$ , such that

$\displaystyle f(c) = f(x) + f'(c_1) ( c - x) \leq f(x) + 0 \cdot 0 = f(x).$

A similar argument for $x > c$ yields

$f(x) = f(c) + \underbrace{(\text{stuff})}_{\geq 0} \underbrace{(x-c)}_{> 0} \geq 0.$

Can we have a simpler computation for this process? Yes, but this does not help us with inflection points. Furthermore, we can make no conclusion if $f''(c) = 0$ .

Theorem 4 (Second Derivative Test). Let $f : [a, b] \to \mathbb R$ be twice-differentiable at $c \in (a, b)$ . Suppose $f'(c) = 0$ .
- If $f''(c) > 0$ , then $c$ is a local minimum of $f$ .
- If $f''(c) < 0$ , then $c$ is a local maximum of $f$ .
- If $f''(c) = 0$ , the test is inconclusive.
Proof. We will prove the first result since the second result follows by symmetry.

By the definition of differentiation, for sufficiently small $h > 0$ ,

$\displaystyle \frac{f'(c+h)}{h}=\frac{f'(c+h) - f'(c)}{h} = f''(c) + o(h) > 0.$

Hence, $f'(c+h) > (f''(c) + o(h)) \cdot h > 0$ , which after taking $h \to 0^+$ , implies that $f'(c^+) \geq 0$ . Likewise, from the left side,

$\displaystyle \frac{-f'(c-h)}{h}=\frac{f'(c) - f'(c-h)}{h} = f''(c) + o(h) > 0,$

which yields $-f'(c^-) \leq 0 \iff f'(c^-) \geq 0$ .

For the last result, consider the functions $x^4$ for $0$ being a local minimum, $-x^4$ for $0$ being a local maximum, and $x^3$ for $0$ being a stationary point of inflection since in all cases, $f''(0) = 0$ .

In the next post, we will use this test to maximise profit.

—Joel Kindiak, 31 Oct 24, 1316H
October 31, 2024

math, mathematics, physics, Probability, sine
Proving Volumes

In this post, we use integral calculus to prove, or at least derive, several famous formulae in geometry.

Definition 1. The volume of a prism with constant base area $A$ and height $h$ is defined by $Ah$ .

Corollary 1. The volume of a cylinder with base radius $r$ and height $h$ is $\pi r^2 h$ .

We can use the prism to derive the volume of a cone (more generally, a pyramid).

Definition 2. Let $A(x)$ be a continuous function defined on $[a , b]$ that denotes the cross-sectional area of a solid at $x$ . Define $x_k = a + (b-a)k/n$ and $\Delta x = (b-a)/n$ . Its volume can be computed using the limit

$\displaystyle V := \lim_{n \to \infty} \sum_{k=1}^n A(x_k) \Delta x = \int_a^b A(x)\,\mathrm dx.$

Theorem 1. The volume of a pyramid with base area $A$ and height $h$ is $\frac 13 Ah$ .

Proof. Using similar triangles and similar areas, we can consider the continuous function $A(x) = Ax^2/h^2$ on $[0, h]$ . Using the integral definition of volume,

$V = \displaystyle \int_0^h A(x)\, \mathrm dx = \int_0^h \frac{Ax^2}{h^2}\, \mathrm dx = \frac{A}{h^2} \int_0^h x^2\, \mathrm dx = \frac{A}{h^2} \cdot \frac{h^3}{3} = \frac 13 Ah.$

Corollary 2. The volume of a cone with base radius $r$ and height $h$ is $\frac 13 \pi r^2h$ .

Proof. Contextualise the previous result to $A = \pi r^2$ .

In fact, whenever $A(x) = \pi (f(x))^2$ corresponds to a circle with radius $f(x)$ , we obtain the volume of revolution formula.

Theorem 2 (Volume of Revolution). Let $f(x) \geq 0$ be a continuous function on $[a, b]$ . The volume of revolution $V$ of $f(x)$ about the $x$ -axis is

$\displaystyle V = \pi \int_a^b (f(x))^2\, \mathrm dx.$

We can also prove the formula for the volume of a sphere using a similar integration technique.

Theorem 3. The volume of a sphere with radius $r$ is $\frac 43 \pi r^3$ .

Proof. Recall that the upper-half semicircle with radius $r$ has equation $y = \sqrt{r^2 -x^2}$ . Consider the function $A(x) = \pi y^2 = \pi (r^2-x^2)$ defined on $[-r, r]$ . Using the integral definition of volume,

$\displaystyle \begin{aligned} \int_{-r}^r A(x)\,\mathrm dx &= \pi \int_{-r}^r (r^2-x^2)\, \mathrm dx \\ &= \pi \left[ r^2x - \frac{x^3}{3} \right]_{-r}^r \\ &= \pi \left(2r^3 - \frac{2r^3}{3}\right) = \frac{4\pi r^3}{3}. \end{aligned}$

What about the surface area of a sphere? We first need the surface area of a cone.

Lemma 1. The surface area of a cone with slant height $l$ and radius $r$ is given by $\pi r l$ .

Proof. The cone can be thought of as a sector with radius $l$ and arc length $2 \pi r$ , yielding an area of

$\displaystyle \frac{ 2 \pi r }{ 2 \pi l } \cdot \pi l^2 = \pi r l.$

Remark 1. The angle that this sector corresponds to is $2\pi r/l$ .

Next, we require the surface area of a frustum.

Lemma 2. The surface area of a frustum with radii $r < R$ and slant height $l$ is given by $\pi (r + R) l$ .

Proof. Let the total slant height of the completed cone be $s + l$ . By similar triangles,

$\displaystyle \frac{r}{s} = \frac{R}{s+l} \quad \iff \quad s = \frac{rl}{R-r}.$

By subtracting surface areas of the smaller cone from the larger cone, we have a surface area of

$\displaystyle \pi R(s+l) - \pi rs = \pi (R-r)s + \pi Rl = \pi rl + \pi Rl = \pi(r+R)l.$

This helps us define the surface area of revolution.

Definition 3. Let $f(x) \geq 0$ be a continuous function on $[a, b]$ . The surface area of revolution $S$ is defined by

$\displaystyle S := \lim_{n \to \infty} \sum_{k=1}^n \pi (r_k+R_k) l_k,$

where $r_k = f(x_{k-1}), R_k = f(x_k), l_k = \sqrt{(R_k - r_k)^2 + \Delta x^2}$ .

Since $r_k+R_k \to 2f(x_k)$ and $l_k \to \sqrt{1+(f'(x_k))^2}\Delta x$ , taking $n \to \infty$ yields

$\displaystyle S = \int_a^b 2 \pi f(x) \sqrt{1+(f'(x))^2}\, \mathrm dx.$

Setting $y = \sqrt{r^2-x^2}$ and computing integrals therefore yields the surface area of a sphere.

Theorem 4. The surface area of a sphere with radius $r$ is given by $4 \pi r^2$ .

—Joel Kindiak, 31 Oct 24, 0229H

October 31, 2024

geometry, machine-learning, math, mathematics, Probability
A Fascinating Multiples Problem
Problem 1. Let $I \subseteq \mathbb Z$ be a subset with the following properties:
- $I \backslash \{0\} \neq \emptyset$ .
- For any $x,y \in I$ , $x+y \in I$ .
- For any $x \in I$ and $\alpha \in \mathbb Z$ , $\alpha x \in I$ .
Prove that there exists $n_0 \in \mathbb Z$ such that $I = \{kn_0 : k \in \mathbb Z\} \equiv n_0 \mathbb Z$ .

(Click for Solution)

Solution. By the first property, there exists some nonzero $n \in I$ . By the third property, $-n = (-1)n \in I$ . Thus, without loss of generality, suppose $n > 0$ .

By the well-ordering principle, there exists some unique $n_0 > 0$ such that

$\displaystyle n_0 = \min \{m \in I : m > 0\}.$

Now we prove that $I = n_0 \mathbb Z$ .

For $(\supseteq)$ , fix $u \in n_0 \mathbb Z$ . Then there exists $k \in \mathbb Z$ such that $u = k n_0 \in I$ .

For $(\subseteq)$ , fix $v \in I$ . By the division algorithm, find integers $q, r$ with $0 \leq r < n_0$ such that $v = qn_0 + r$ .

We claim that $r = 0$ by way of contradiction. Suppose otherwise that $1 \leq r < n_0$ . Then by the third and second properties respectively, $r = v + (-q)n_0 \in I$ .

However, by the definition of $n_0$ , we have $n_0 \leq r < n_0$ , a contradiction.

Therefore, $r = 0$ , which yields $v = qn_0 \in n_0 \mathbb Z$ .

—Joel Kindiak, 31 Oct 24, 0105H
October 31, 2024

machine-learning, math, mathematics, Probability, research
What is an Angle?

As many people know intuitively, the number $\pi \approx 3.14$ is defined to be the ratio of a circle’s circumference to its diameter. Equivalently, we can consider it to be $1/2$ the circumference of a circle with radius $1$ (i.e. a unit circle). But what is the circumference of a circle?

To answer that question, we first recall that a circle has cartesian equation $x^2 + y^2 = 1$ . For the top half of the circle, we have the equation $y = \sqrt{1-x^2} =: f(x)$ . What is the length of this curve?

Let’s partition the base $[-1, 1]$ into $n$ equal subintervals with $x$ -coordinates given by $x_k = -1 + 2k/n$ for $k = 0,1,\dots,n$ .

We can approximate the curve by considering the lengths $L_k$ of each secant line connecting $(x_k, f(x_k))$ to $(x_{k+1}, f(x_{k+1}))$ .

Denoting $\Delta x_k := x_{k+1} - x_k = 2/n$ and $\Delta y_k := f(x_{k+1}) - f(x_{k})$ , each $L_k$ can be computed using Pythagoras’ theorem via

$\displaystyle L_k := \sqrt{{\Delta x_k}^2 + {\Delta y_k}^2} = \sqrt{1 + \left( \frac{\Delta y_k}{\Delta x_k} \right)^2 } \Delta x_k$

By definition,

$\displaystyle \lim_{n \to \infty} \frac{\Delta y_k}{\Delta x_k} = \lim_{\Delta x_k\to 0} \frac{f(x_{k}+\Delta x_k) - f(x_{k})}{\Delta x_k} = f'(x_k).$

Therefore, the arc length is defined, whenever $f$ is continuously differentiable on $[a, b]$ , as

$\displaystyle \lim_{n \to \infty} \sum_{k=1}^n L_k = \lim_{n \to \infty} \sum_{k=1}^n \sqrt{1 + \left( \frac{\Delta y_k}{\Delta x_k} \right)^2 } \Delta x_k = \int_a^b \sqrt{1+ (f'(x))^2}\, \mathrm dx.$

Here is the formal definition for the arc length.

Definition 1. Let $f : [a, b] \to \mathbb R$ be continuously differentiable, perhaps except at the endpoints. Assuming the integral exists, the arc length of $f$ is defined by

$\displaystyle \int_a^b \sqrt{1+ (f'(x))^2}\, \mathrm dx.$

Setting $f(x) = \sqrt{1-x^2}$ and simplifying, one obtains the length of the upper-half of the unit circle, defined by $\pi$ , as follows.

Definition 2. $\displaystyle \pi := \int_{-1}^1 \frac{1}{ \sqrt{1-x^2} }\, \mathrm dx$ .

This is where we obtain $\pi$ as the angle corresponding to half of a turn, and $2\pi$ corresponding to a full turn. Strictly speaking, this is an improper integral, but we will explore this later on.

This actually opens up an interesting discussion on what an angle is. Intuitively, it should be measuring the “separation” between two lines.

How do we formalise the angle rigorously?

Our definition of $\pi$ , really, is the technical formulation of an arc length. The Greek letter $\theta$ is usually used to denote an angle, and in that spirit, we will define the function

$\displaystyle \theta : [-1, 1] \to [0, \pi],\quad \theta(r) = \int_{r}^1 \frac{1}{ \sqrt{1-x^2} }\, \mathrm dx.$

We can verify that this function is continuous and strictly decreasing in $r$ , and therefore, has an inverse $\theta^{-1}$ . A principal angle in $[0,\pi]$ , therefore, is (rather anticlimactically), any real number of the form $\theta(r)$ . It is acute if $0 < \theta(r) < \pi/2$ and obtuse if $\pi/2 < \theta(r) < \pi$ .

Any other angle is therefore a real number of the form $k\pi + \theta(r)$ , where $k$ is an integer, and $\theta(r)$ is a principal angle. Observe that $\theta(1) = 0$ and $\theta(-1) = \pi$ . Using arguments involving symmetry, we could even establish that $\theta(0) = \pi/2$ .

In fact, the way we defined $\theta$ is intimately connected with the usual trigonometric functions. We will now formally define them here.

Definition 3. Let $\alpha$ be an acute angle. The cosine and sine of $\alpha$ respectively are defined by

$\displaystyle \cos(\alpha) := \theta^{-1}(\alpha),\quad \sin(\alpha) := \sqrt{1 - \cos^2(\alpha)}.$

Almost by construction, we obtain $\sin^2(\alpha) + \cos^2(\alpha) = 1$ . Furthermore, we recover the usual characterisations of $\sin(\alpha) = \text{(opposite)}/\text{(hypotenuse)}$ , $\cos(\alpha) = \text{(adjacent)}/\text{(hypotenuse)}$ , and $\cos(\alpha) = \sin(\pi/2-\alpha)$ (this yields $\cos(\alpha) = \sin(\pi/2 + \alpha)$ .

From this definition and its immediate corollaries, we can derive all relevant formulae for $\sin$ and $\cos$ , including that they have inverses on $[-\pi/2,\pi/2]$ and $[0,\pi]$ respectively. In fact, this definition confirms our suspicion that $\theta = \cos^{-1} \equiv \arccos$ , since $\cos^{-1}(1) = 0, \cos^{-1}(0) = \pi/2, \cos^{-1}(-1) = \pi$ .

Most importantly: at last, we have the foundational logical grounds for trigonometry, from which we define $\tan$ and the co-circular functions $\sec$ , $\csc$ , $\cot$ .

In the next few posts, we will finally prove that $\sin' = \cos$ . The plan initially was to do this in a discussion in real analysis, but it turns out that we have the necessary tools for the job.

—Joel Kindiak, 27 Oct 24, 1109H

October 27, 2024

Calculus, math, mathematics, physics, Probability
How to Reverse the Product Rule

Let’s bring our attention to one function: $\ln$ .

Theorem 1. $\displaystyle \ln'(x) = 1/x$ .

Proof. Coupling $\ln = \exp^{-1}$ with $\exp' = \exp$ , the inverse function theorem yields

$\displaystyle \ln'(x) = (\exp^{-1})'(x) = \frac{1}{\exp'(\exp^{-1}(x))} = \frac{1}{\exp(\exp^{-1}(x))} = \frac 1x.$

This, in fact, needed to be proven before we derive the useful implications of the reverse chain rule.

How can we calculate $\displaystyle \int \ln(x)\, \mathrm dx$ ?

Theorem 2. $\displaystyle \int \ln(x)\,\mathrm dx = x(\ln(x) - 1) + C$ .

It is trivial to differentiate the right-hand side to get $\ln(x)$ . The real question is how we discovered the right-hand side in the first place.

We need to use integration by parts, which I like to call the reverse product rule.

Theorem 3 (Reverse Product Rule). Let $u\equiv u(x),v\equiv v(x)$ be known differentiable functions. Then

$\displaystyle \int u\, \mathrm dv = uv - \int v\, \mathrm du,$

where we formally abbreviated $\mathrm du = u'(x)\,\mathrm dx$ and $\mathrm dv = v'(x)\,\mathrm dx$ .

Proof. Reverse the product rule $(uv)' = v u' + u v'$ .

For pedagogical utility, I introduce my students to IS-ID notation:

$\displaystyle \int v'u\, \mathrm dx = \underbrace{v}_{\text{I}}\underbrace{u}_{\text{S}} - \int \underbrace{v}_{\text{I}} \underbrace{u'}_{\text{D}}\, \mathrm dx.$

Just like the reverse chain rule, the utility of this rule does not arise in its theory (which is immediate), but in its pervasive applications.

The crucial trick here is that $u$ , $v$ , and the integral of $vu'$ must be known. Let’s investigate the integral of question:

$\displaystyle \int \ln(x)\,\mathrm dx.$

Here, if we allow $u = 1$ and $v' = \ln(x)$ , then $u' = 0$ , but what is $v$ ? It’s unknown! This choice of $u,v'$ therefore would be useless for this question.

One useful heuristic would be LIATE, which, roughly speaking, summarises the varied difficulty of the knowledge of $v$ . For this blogpost, we will not reference LIATE, but may detail it in the future as an exercise.

Let’s instead work with the choice $v' = 1$ and $u = \ln(x)$ . Integrating $v'$ yields $v = x$ , and differentiating $u$ yields $u' = 1/x$ . Now we know all the required information, and can make (at least partial) progress in computing the integral.

Proof of Theorem 2. Using the reverse product rule (i.e. integrating by parts),

$\displaystyle \begin{aligned} \int 1 \cdot \ln(x)\, \mathrm dx &= \underbrace{x}_{\text{I}}\underbrace{\ln(x)}_{\text{S}} - \int \underbrace{x}_{\text{I}} \underbrace{\frac 1x}_{\text{D}}\, \mathrm dx \\ &= x \ln(x) - \int 1\, \mathrm dx \\ &= x \ln(x) - x + C,\end{aligned}$

where the integration after the first step is trivial.

Alternate Proof of Theorem 2. Alternatively, make the substitution

$u = \ln(x) \quad \Rightarrow \quad \mathrm dx = e^u\, \mathrm du.$

Then by Theorem 3 below,

$\displaystyle \int \ln(x)\, \mathrm dx = \int ue^u\, \mathrm du = (u-1)e^u + C = x(\ln(x) - 1) + C.$

The power of these integration techniques, therefore, is to help us compute integrals that were, up till now, out of reach by our naïve knowledge of reverse-differentiation. The case of integrating $\ln(x)$ can be generalised.

Corollary 1. Suppose $u(x)$ is a known differentiable function such that the anti-derivative of $xu'(x)$ is known. Then

$\displaystyle \int u(x)\, \mathrm dx = xu(x) - \int xu'(x)\, \mathrm dx.$

This next integral explores the situation when both $u$ and $v$ are known, but not necessarily the integral of $vu'$ .

Theorem 3. $\displaystyle \int xe^x\, \mathrm dx = (x-1)e^x + C$ .

Proof. There are two ways to integrate this expression:

$\begin{aligned} \int x \cdot e^x\, \mathrm dx &= \underbrace{\frac{x^2}{2}}_{\text{I}}\underbrace{e^x}_{\text{S}} - \int \underbrace{\frac{x^2}{2}}_{\text{I}} \underbrace{e^x}_{\text{D}}\, \mathrm dx = \frac 12 x^2 e^x - \frac 12 \int x^2 e^x\, \mathrm dx, \\ \int x \cdot e^x\, \mathrm dx &= \underbrace{e^x}_{\text{I}}\underbrace{x}_{\text{S}} - \int \underbrace{e^x}_{\text{I}} \underbrace{1}_{\text{D}}\, \mathrm dx = xe^x - \int e^x\, \mathrm dx. \end{aligned}$

Observe that the first integral is hard to compute, while the second integral gives us the desired answer. The choice of $u,v'$ therefore directly affects the outcome of our answer.

Integrating $\tan^{-1}$ is an interesting problem involving both the reverse product rule and the reverse chain rule.

Theorem 4. $\displaystyle \int \tan^{-1}(x)\, \mathrm dx = x\tan^{-1}(x) - \frac 12 \ln(1+x^2) + C$ .

Proof. Integrating by parts then integrating by substitution,

$\begin{aligned} \int 1 \cdot \tan^{-1}(x)\, \mathrm dx &= \underbrace{x}_{\text{I}}\underbrace{\tan^{-1}(x)}_{\text{S}} - \int \underbrace{x}_{\text{I}} \underbrace{\frac{1}{1 + x^2}}_{\text{D}}\, \mathrm dx \\ &= x \tan^{-1}(x) - \frac 12 \int \frac{2x}{1+x^2}\, \mathrm dx \\ &= x \tan^{-1}(x) - \frac 12 \int \frac{(1+x^2)'}{1+x^2}\, \mathrm dx \\ &= x \tan^{-1}(x) - \frac 12 \ln(1+x^2) + C. \end{aligned}$

Finally, it is entirely possible to apply the reverse product rule twice to obtain a sensible answer.

Theorem 5. For $a, b > 0$ ,

$\displaystyle \int e^{ax} \cos(bx)\, \mathrm dx = \frac{ e^{ax}}{a^2+b^2}\cdot (a \cos(bx) + b \sin(bx)) + C.$

Proof. Integrating by parts once (in this special case, the choice of $u,v'$ doesn’t really affect the answer),

$\displaystyle \begin{aligned} \int e^{ax} \cos(bx)\, \mathrm dx &= \underbrace{ \frac{e^{ax}}{a} }_{\text I} \underbrace{ \cos(bx) }_{\text S} - \int \underbrace{ \frac{e^{ax}}{a} }_{\text I} \underbrace{ (-b\sin(bx)) }_{\text D}\, \mathrm dx \\ &= \frac{e^{ax}}{a}\cos(bx) + \frac{b}{a} \int e^{ax} \sin(bx)\, \mathrm dx. \end{aligned}$

Integrating by parts a second time,

$\displaystyle \begin{aligned} \int e^{ax} \sin(bx)\, \mathrm dx &= \underbrace{ \frac{e^{ax}}{a} }_{\text I} \underbrace{ \sin(bx) }_{\text S} - \int \underbrace{ \frac{e^{ax}}{a} }_{\text I} \underbrace{ (b\cos(bx)) }_{\text D}\, \mathrm dx \\ &= \frac{e^{ax}}{a} \sin(bx) - \frac{b}{a} \int e^{ax} \cos(bx)\, \mathrm dx. \end{aligned}$

Back-substituting,

$\displaystyle \begin{aligned} \int e^{ax} \cos(bx)\, \mathrm dx &= \frac{e^{ax}}{a} \cos(bx) + \frac{b}{a} \left( \frac{e^{ax}}{a} \sin(bx) - \frac{b}{a} \int e^{ax} \cos(bx)\, \mathrm dx \right). \end{aligned}$

By algebra,

$\displaystyle \left(1 + \frac{b^2}{a^2}\right) \int e^{ax} \cos(bx)\, \mathrm dx = \frac{ e^{ax} }{a^2} \left( a\cos(bx) + b\sin(bx) \right).$

The result follows by dividing by $(1 + b^2/a^2)$ , then adding the arbitrary constant $+C$ .

And that is how we reverse the product rule to solve even more integration problems—integration by parts, which I call the IS-ID technique.

What good is integration for? How about even defining angles in the first place? What do I mean? Next post.

—Joel Kindiak, 26 Oct 24, 1753H

October 26, 2024

math, mathematics, physics, Probability, representation-theory
Why u-Substitution is So Awesome
Students are introduced to “ $u$ -substitution” as yet another exam-question solving technique. That is not wrong, and does make for many challenging questions.

But the reason it is so is massive—the chain rule helps us discover new integrals that we would not have known otherwise.

Let’s first state $u$ -substitution, or what I’d like to call, the reverse chain rule.

Theorem 1 (Reverse Chain Rule). Let $f,g$ be (known) differentiable functions. Then letting $u := g(x),$

$\displaystyle \int f'(g(x))g'(x)\, \mathrm dx = f(g(x))+C = f(u) + C = \int f'(u)\, \mathrm du.$

Proof. By the chain rule,

$\displaystyle \frac{\mathrm d}{\mathrm d x} (f(g(x))) = f'(g(x)) g'(x) \quad \Rightarrow \quad \int f'(g(x))g'(x)\, \mathrm dx = f(g(x)) + C.$

The power of the reverse chain rule arises not from its theory—it’s right there—but from its applications. For instance, whenever we have an integrand (i.e. the function being integrated) of the form $f'/f$ , we obtain a very useful integral.

Theorem 2. Let $f$ be a known differentiable function. Then

$\displaystyle \int \frac{f'(x)}{f(x)}\, \mathrm dx = \ln|{f(x)}| + C.$

Proof. While the result is straightforward, a computation helps us evaluate the integral more conveniently. Let $u = f(x)$ . Then

$\displaystyle \frac{\mathrm du}{\mathrm dx} = f'(x) \quad \iff \quad \mathrm du = f'(x)\, \mathrm dx .$

Here, the notation on the right-hand side is purely formal and not a well-defined mathematical notion. However, this helps us compute the $u$ -substitution:

$\begin{aligned} \displaystyle \int \frac{f'(x)}{f(x)}\, \mathrm dx &= \int \frac{1}{f(x)}\, f'(x)\, \mathrm dx \\ &= \int \frac 1u\, \mathrm du \\ &= \ln|u| + C \\ &= \ln|{f(x)}| + C. \end{aligned}$

This helps us integrate $\tan$ . Recall that $\tan' = \sec^2$ . It is definitely not the case that

$\displaystyle \int \tan(x)\, \mathrm dx = \sec^2(x) + C.$

But, we do have the following result:

Corollary 1. $\displaystyle \int \tan(x)\, \mathrm dx = -{\ln|{\cos(x)}|} + C$ .

Proof. The immediate proof is by letting $f(x) = \cos(x)$ in Theorem 1.

Fleshing out the details, define $u = \cos(x) \Rightarrow \mathrm du = -\sin(x)\,\mathrm dx$ so that

$\displaystyle \begin{aligned} \int \tan(x)\, \mathrm dx &= \int \frac{\sin(x)}{\cos(x)}\, \mathrm dx \\ &= -\int \frac{1}{\cos(x)}\cdot -\sin(x)\, \mathrm dx \\ &= -\int \frac{1}{u}\, \mathrm du \\ &= -{\ln|u|} + C \\ &= -{\ln|{\cos(x)}|} + C. \end{aligned}$

We also obtain integrals for rational functions of the form $(\text{constant})/(\text{linear})$ and $(\text{constant})/(\text{quadratic})$ .

Technically, much more is true, but we will illustrate these simpler cases here, and expand on the “much more” in a future post.

Corollary 2. $\displaystyle \int \frac{1}{x^2 - 1}\, \mathrm dx = \frac 1{2} \ln \left| \frac{x-1}{x+1} \right| + C$ .

Proof. By writing $x^2 - 1 = (x-1)(x+1)$ and decomposing using partial fractions,

$\displaystyle \frac 1{x^2-1} = \frac 1{(x-1)(x+1)} = \frac 1{2}\left(\frac{1}{x-1} - \frac{1}{x+1}\right).$

Integrating on both sides,

$\displaystyle \begin{aligned} \int \frac 1{x^2-1}\,\mathrm dx &= \frac 1{2}\int\left(\frac{1}{x-1} - \frac{1}{x+1}\right)\, \mathrm dx \\ &= \frac 1{2}\left(\int\frac{1}{x-1}\, \mathrm dx - \int\frac{1}{x+1}\, \mathrm dx\right). \end{aligned}$

To integrate each component, we observe that $(x \pm 1)' = 1$ . Therefore, Theorem 2 yields

$\displaystyle \begin{aligned} \int \frac{1}{x \pm 1}\, \mathrm dx = \int \frac{(x \pm 1)'}{x \pm 1}\, \mathrm dx = \ln|x \pm 1| + C. \end{aligned}$

Thus, completing our integration,

$\displaystyle \begin{aligned} \int \frac 1{x^2-1}\,\mathrm dx &= \frac 1{2}\left(\int\frac{1}{x-1}\, \mathrm dx - \int\frac{1}{x+1}\, \mathrm dx\right) \\ &= \frac 1{2} \left( (\ln|x-1| + C_1) - (\ln|x-1| + C_2) \right) \\ &= \frac 1{2} \ln \left|\frac{x-1}{x+1}\right| + \frac{C_1 - C_2}{2} = \frac 1{2} \ln \left|\frac{x-1}{x+1}\right| + C.\end{aligned}$

where we abbreviated the constant $C := (C_1 - C_2)/2$ .

As a side-note, this abbreviation also explains why in each iteration of integration, we only need one $C$ , since all of the respective arbitrary constants $C_i$ coalesce into a single arbitrary constant $C$ .

Letting $u$ be a linear function also proves to be incredibly useful in integrating such functions.

Theorem 3. Let $f$ be a known differentiable function and $a,b$ be constants with $a \neq 0$ . Then

$\displaystyle \int f'(ax+b)\, \mathrm dx = \frac 1a f(ax+b) + C.$

Proof. Make the substitution $u = ax+b$ . Then the notation $\mathrm du = a\, \mathrm dx$ yields

$\displaystyle \begin{aligned} \int f'(ax+b)\, \mathrm dx &= \frac 1a \cdot \int f'(ax+b) \cdot a \, \mathrm dx \\ &= \frac 1a \int f'(u) \, \mathrm du \\ &= \frac 1a f(u) + C \\ &= \frac 1a f(ax+b) + C. \end{aligned}$

This helps us take advantage of simpler integrals to compute more complicated integrals.

Corollary 3. For any $a > 0$ , $\displaystyle \int \frac{1}{a^2 - x^2}\, \mathrm dx = \frac 1{2a} \ln \left|\frac {a-x}{a+x}\right| + C$ .

Proof. Make the substitution $x = au$ . Then the notation $\mathrm dx = a\, \mathrm du$ yields

$\displaystyle \begin{aligned} \int \frac{1}{a^2 - x^2}\, \mathrm dx &= \int \frac{1}{a^2 - (au)^2}\cdot a\, \mathrm du \\ &= a \int \frac{1}{a^2 - a^2 u^2}\, \mathrm du \\ &= -\frac {1}{a} \int \frac{1}{u^2 - 1}\, \mathrm du \\ &= -\frac 1{2a} \ln \left| \frac{u-1}{u+1}\right| + C \\ &= \frac 1{2a} \ln \left| \frac{a+x}{a-x}\right| + C,\end{aligned}$

where the final equality is left as an exercise in algebruh.

Finally, we can use known trigonometric substitutions by taking advantage of the Pythagorean identity.

Corollary 4. $\displaystyle \int \frac 1{1+x^2}\, \mathrm dx = \tan^{-1}(x) + C$ .

Proof. Make the substitution $x = \tan(u)$ . Then the notation $\mathrm dx = \sec^2(u)\, \mathrm du$ yields

$\displaystyle \begin{aligned} \int \frac{1}{1 + x^2}\, \mathrm dx &= \int \frac{1}{1 + \tan^2(u)} \cdot \sec^2(u)\, \mathrm du \\ &= \int \frac{1}{\sec^2(u)} \cdot \sec^2(u)\, \mathrm du \\ &= \int 1\, \mathrm du = u + C \\ &= \tan^{-1}(x) + C. \end{aligned}$

Sidenote: One might wonder if this could be used to derive the formula $(\tan^{-1}(x))' = 1/(1+x^2)$ . This does not work, since we needed to assume that $u = \tan^{-1}(x)$ is differentiable. If that were the case, then computation just reiterates the differentiation property.

This result, in particular, helps us integrate the reciprocal of non-factorisable quadratic functions.

Corollary 5. For any $a > 0$ , $\displaystyle \int \frac{1}{a^2 + x^2}\, \mathrm dx = \frac 1{a} \tan^{-1}\left(\frac xa\right) + C$ .

Proof. Make the substitution $x = au$ . Then the notation $\mathrm dx = a\, \mathrm du$ yields

$\displaystyle \begin{aligned} \int \frac{1}{a^2 + x^2}\, \mathrm dx &= \int \frac{1}{a^2 + (au)^2}\cdot a\, \mathrm du \\ &= \int \frac{a}{a^2(1 + u^2)}\, \mathrm du \\ &= \frac 1{a} \int \frac{1}{1 + u^2}\, \mathrm du \\ &= \frac 1{a} \tan^{-1}(u) + C \\ &= \frac 1{a} \tan^{-1}\left( \frac{x}{a} \right) + C. \end{aligned}$

Corollary 6. For any $a,b > 0$ , $\displaystyle \int \frac{1}{ a + b u^2 }\, \mathrm dx = \frac 1{\sqrt{ab}} \tan^{-1}\left(\frac {x \sqrt b}{\sqrt a}\right) + C$ .

Proof. Make the substitution $u = x\sqrt b$ . Then the notation $\displaystyle \mathrm dx = \frac 1{\sqrt b} \, \mathrm du$ yields

$\displaystyle \begin{aligned} \int \frac{1}{a + bx^2}\, \mathrm dx &= \int \frac{1}{(\sqrt a)^2 + u^2}\cdot \frac 1{\sqrt b}\, \mathrm du \\ &= \frac 1{\sqrt b} \int \frac{1}{(\sqrt a)^2 + u^2}\, \mathrm du \\ &= \frac 1{\sqrt b} \cdot \frac 1{\sqrt a} \tan^{-1}\left( \frac{u}{\sqrt a} \right) + C \\ &= \frac 1{\sqrt{ab}} \tan^{-1}\left( \frac{x\sqrt b}{\sqrt a} \right) + C. \end{aligned}$

Applications of similar techniques yields the following integral.

Theorem 4. For any $a > 0$ , $\displaystyle \int \frac{1}{\sqrt{a^2-x^2}}\, \mathrm dx = \sin^{-1}\left(\frac xa\right) + C$ .

Finally, no discussion on the reverse chain rule is complete without the notorious integral of $\sec$ .

Theorem 5. $\displaystyle \int \sec(x)\, \mathrm dx = \ln|{\sec(x) + \tan(x)}| + C$ .

Proof. By *ingenuity*, make the substitution $u = \sec(x) + \tan(x)$ . Then

$\displaystyle \frac{\mathrm du}{\mathrm dx} = (\sec(x)\tan(x) + \sec^2(x)) \quad \Rightarrow \quad \frac 1u\, \mathrm du = \sec(x)\, \mathrm dx$

yields the integration

$\begin{aligned} \displaystyle \int \sec(x)\, \mathrm dx = \int \frac 1u\, \mathrm du = \ln|u| + C = \ln|{\sec(x) + \tan(x)}| + C. \end{aligned}$

Applying similar techniques yields the following integral:

Theorem 6. For any $a > 0$ ,

$\displaystyle \int \frac{1}{\sqrt{x^2 \pm a^2}}\, \mathrm dx = \ln \left| x + \sqrt{x^2 \pm a^2} \right| + C$ .

Proof. Left as an exercise in *ingenuity*. Just kidding. But yes, an exercise in the reverse chain rule. Here are the hints:
- For $\displaystyle \int \frac{1}{\sqrt{x^2 + a^2}}\, \mathrm dx$ , make the substitution $x = a \tan u$ .
- For $\displaystyle \int \frac{1}{\sqrt{x^2 - a^2}}\, \mathrm dx$ , make the substitution $x = a \sec u$ .
Then invoke Theorem 5 and perform some bookkeeping via algebruh.

Next time, we reverse the product rule.

—Joel Kindiak, 25 Oct 24, 1405H
October 25, 2024

geometry, math, mathematics, physics, Probability
The Trifecta of Differentiation

In differentiation, three results are used repeatedly to establish a slew of commonly-used derivatives: the chain rule, the product rule, and the quotient rule.

The chain rule is the key result that establishes the other two, which we have proven before, and will re-state for clarification.

Theorem 1 (Chain Rule). Let $f,g: \mathbb R \to \mathbb R$ be functions. Suppose $f$ is differentiable at $c$ and $g$ is differentiable at $f(c)$ . Then $g \circ f$ is differentiable at $c$ with derivative $(g \circ f)'(c) = g'(f(c))f'(c)$ .

With the chain rule, we obtain the product rule as a corollary.

Theorem 2 (Product Rule). Let $f,g$ be functions differentiable at some $c$ . Then the product $fg$ defined by

$(fg)(x) := f(x)g(x)$

is differentiable at $c$ with derivative

$(fg)'(c) = f'(c)g(c) + g'(c)f(c).$

Proof. Apply linearity and the chain rule on the function

$fg = \frac 12 ((f + g)^2 - f^2 - g^2).$

The quotient rule then follows as an immediate corollary of both the chain rule and the product rule via $f/g = f \cdot 1/g$ .

Theorem 3 (Quotient Rule). Let $f,g$ be functions differentiable at some $c$ . If $g(c) neq 0$ , then the quotient $f/g$ defined by

$\displaystyle \left( \frac fg \right)(x) := \frac{f(x)}{g(x)}$

is differentiable at $c$ with derivative

$\displaystyle \left( \frac fg \right)'(x) := \frac{f'(c) g(c) - f(c) g'(c)}{{g(c)}^2}.$

These properties, coupled with the starting points $\sin' = \cos$ and $\exp' = \exp$ , suffice to prove most commonly used derivatives in high school and even in freshmen calculus.

To properly define the functions $\sin$ and $\exp$ , however, requires more thought. We will explore them in real analysis.

For now, let’s use the quotient rule to prove that $\tan' = \sec^2$ .

Theorem 4. We have $\tan' = \sec^2$ .

Proof. Writing $\tan(x) = \sin(x)/{\cos(x)}$ ,

$\displaystyle \begin{aligned} \frac{\mathrm d}{\mathrm d x}(\tan(x)) &= \frac{\mathrm d}{\mathrm d x} \left( \frac{\sin(x)}{\cos(x)} \right) \\ &= \frac{\cos(x) \cdot \cos(x) - \sin(x) \cdot (-{\sin(x)})}{\cos^2(x)} \\ &= \frac{\cos^2(x) + \sin^2(x)}{\cos^2(x)}\\ &= \frac{1}{\cos^2(x)} \\ &= \sec^2(x). \end{aligned}$

—Joel Kindiak, 20 Oct 24, 1504H

October 22, 2024

Calculus, math, mathematics, physics, Probability
Differentiating Inverses
In this calculus series, we will assume that the functions $\sin$ and $\exp(x) \equiv e^x$ have been well-defined. These allow for definitions of common trigonometric functions like $\cos$ .

We want to differentiate their inverses $\sin^{-1}$ and $\ln$ . Here is the definition of the inverse of a function.

Definition 1. A function $f$ defined on $I \subseteq \mathbb R$ has an inverse if for any $r, s \in I$ , $f(r) = f(s) \Rightarrow r = s$ .

However, a technical discussion on the properties of $\sin$ and $\exp$ will take us beyond the focus of differentiation. For this discussion, we will make the following assumptions:
- $\sin$ restricted to $[-\pi/2,\pi/2]$ has an inverse $\sin^{-1}$ .
- $\exp$ has an inverse $\ln$ .
- $\sin$ is differentiable on $\mathbb R$ and $\sin' = \cos$ .
- $\exp$ is differentiable on $\mathbb R$ and $\exp' = \exp$ .
We will rigorously define these functions more carefully, and prove these properties, when studying the big brother of calculus that grounds it, known as real analysis.

Before we can prove the inverse function theorem, which allows us to differentiate inverses, we first need some technical tools. The first is the famous intermediate value theorem, which can only be fully proven using real-analytic tools, and thus is omitted from this discussion.

Theorem 1 (Intermediate Value Theorem). Let $f : [a, b] \to \mathbb R$ be continuous. Suppose $f(a) \neq f(b)$ without loss of generality. Then for any $f(a) < m < f(b)$ , there exists $c \in (a, b)$ such that $f(c) = m$ .

With some effort, we can derive the general case by first proving the special case $[a, b] = [0, 1]$ , $f(a) < 0 < f(b)$ , and $m = 0$ . We will relegate that in a discussion on real analysis.

Using the intermediate value theorem, we will first show that any interval remains an interval.

Lemma 1. Let $f$ be continuous on $\mathbb R$ . For any $a < b$ , define $I := (a, b)$ . Then for any $c, d \in f(I)$ with $c < d$ , if $c < y < d$ , then $y \in f(I)$ .

Proof. Fix $c, d \in f(I)$ with $c < d$ . Find inputs $r, s \in (a, b)$ such that $f(r) = c < d = f(s)$ . Suppose $r < s$ without loss of generality.

Fix $f(r) < y < f(s)$ . Since $f$ is continuous on $[r, s]$ , by the intermediate value theorem, there exists $t \in (r, s)$ such that $y = f(t) \in f(I)$ .

We will next show that the inverse of any continuous function must be continuous.

Lemma 2. Let $f$ be a real-valued function that is continuous on $(a, b)$ . Suppose $f$ has an inverse $f^{-1}$ . Then $f^{-1}$ is continuous on the interval $f((a, b))$ .

Proof. Fix $c \in f((a, b))$ . By the intermediate value theorem, find $d \in (a, b)$ such that $c = f(d) \iff f^{-1}(c) = d$ . Since $f$ is continuous at $d$ ,

$f^{-1}(c + o(1)) = f^{-1}(f(d) + o(1)) = f^{-1}(f(d + o(1))) = d + o(1) .$

Finally, here is the inverse function theorem in one dimension.

Theorem 2 (Inverse Function Theorem). Let $f$ be differentiable on $(a, b)$ . Suppose $f$ has an inverse $f^{-1}$ on $(a, b)$ . Then for any $c \in f((a, b))$ such that $f'(f(c)) \neq 0$ , $f^{-1}$ is differentiable at $c$ and

$\displaystyle (f^{-1})'(c) = \frac{1}{f'(f^{-1}(c))}.$

The theorem for higher-dimensions takes a lot more effort to define and prove. For this post, we will omit that.

Proof. The key component of this proof is Carathéodory’s theorem, which we have proven previously.

Let $y = f(x) \iff x = f^{-1}(y)$ and $c = f(d) \iff d = f^{-1}(c)$ . Then

$\displaystyle \begin{aligned} f^{-1}(y) - f^{-1}(c) &= \frac{f^{-1}(y) - f^{-1}(c)}{y-c} \cdot (y-c) \\ &= \frac{1}{\frac{y-c}{f^{-1}(y) - f^{-1}(c)}} \cdot (y-c) \\ &= \frac 1{\frac{f(x) - f(d)}{x-d}} \cdot (y-c). \end{aligned}$

By Carathéodory’s theorem, find a continuous function $\varphi$ such that

$f(x) - f(d) = \varphi(x) (x-d)$

with $(\varphi \circ f^{-1})(c) = \varphi(d) = f'(d) = f'(f^{-1}(c))$ .

Substituting,

$\displaystyle \begin{aligned} f^{-1}(y) - f^{-1}(c) &= \frac 1{\frac{f(x) - f(d)}{x-d}} \cdot (y-c) = \frac 1{(\varphi \circ f^{-1}) (x)} (y-c). \end{aligned}$

By Caratheodory’s theorem, define the continuous function $\psi := 1/{(\varphi \circ f^{-1})}$ , which satisfies $(f^{-1})'(c) = \psi(c) = 1/{f'(f^{-1}(c))}$ , to complete the result.

As a consequence, we obtain the differentiability of $\sin^{-1}$ on $(-\pi/2,\pi/2)$ and $\ln = \exp'$ on $(0,\infty)$ . To obtain their derivatives $(\sin^{-1})'(x) = 1/{\sqrt{1-x^2}}$ and $\ln'(x) = 1/x$ , either apply the inverse function theorem, or simply use the chain rule.

To illustrate this technique, let’s prove that $\ln'(x) = 1/x$ .

Theorem 3. $\ln'(x) = 1/x$ .

Proof. By the inverse function theorem,

$\displaystyle \ln'(x) = (\exp^{-1})'(x) = \frac{1}{\exp'(\exp^{-1}(x))} = \frac{1}{\exp(\exp^{-1}(x))} = \frac 1x.$

The derivative of $\ln$ helps us obtain one of the most lovely results in differentiation.

Theorem 3. Let $r$ be a nonzero real number. Then for positive $x$ , $(x^r)' = rx^{r-1}$ .

Proof. Writing $x^r = e^{r \ln(x)} \equiv \exp(r\ln(x))$ and applying the chain rule,

$\displaystyle \begin{aligned} (x^r)' &= (\exp(r\ln(x)))' \\ &= \exp'(r\ln(x)) (r\ln(x))' \\ &= \exp(r \ln(x)) \cdot r \cdot \frac 1x \\ &= x^r \cdot \frac rx = rx^{r-1}. \end{aligned}$

—Joel Kindiak, 21 Oct 24, 1126H
October 21, 2024

machine-learning, math, mathematics, Probability, python