Essence of Derivatives

Derivative Notation

Lagrange Notation

Uses function symbols with primes to denote derivatives. If $y = f(x)$, then:

$$f'(x_0) \quad \text{or} \quad y'$$

Features:

• Directly associates with function names, making the correspondence between function and derivative clear
• Higher-order derivatives are indicated by the number of primes: second derivative $f''(x)$, third derivative $f'''(x)$
  Example:
  $f(x) = x^3$ has derivative:

$$f'(x) = 3x^2 \\ f''(x) = 6x$$

Applicable scenarios:

• Explicit function expressions (e.g., $f(x) = \sin x$)
• Theoretical proofs and formula derivations

Leibniz Notation

Uses differential symbols to represent derivatives. If $y = f(x)$, then:

$$\frac{dy}{dx} \bigg|_{x=x_0} \quad \text{or} \quad \frac{d}{dx}f(x)$$

Features:

• Intuitively shows the derivative as the limit of a ratio of differentials
• Higher-order derivatives are indicated by exponents: second derivative $\frac{d^2y}{dx^2}$
  Chain rule example:
  Let $y = \sin(u)$, $u = x^2$, then:

$$\frac{dy}{dx} = \frac{dy}{du} \cdot \frac{du}{dx} = \cos(u) \cdot 2x = 2x \cos(x^2)$$

Applicable scenarios:

• Implicit functions (e.g., $x^2 + y^2 = 1$)
• Multivariable calculus and physical equations

Newton Notation

Uses dots above variables to denote time derivatives. If displacement is $s(t)$, then:

$$\dot{s} = \frac{ds}{dt}, \quad \ddot{s} = \frac{d^2s}{dt^2}$$

Features:

• Concise notation, especially suitable for time derivatives
• For derivatives beyond third order, multiple dots are used (e.g., $\dddot{s}$ for third derivative)
  Kinematics example:
  Free fall motion $s(t) = \frac{1}{2}gt^2$, then:

$$\dot{s} = gt \quad (\text{velocity}) \\ \ddot{s} = g \quad (\text{acceleration})$$

Applicable scenarios:

• Classical mechanics and dynamics problems
• Systems of differential equations (e.g., vibration system $\ddot{x} + \omega^2x = 0$)

Comparison and Selection Principles

Notation Type	Advantages	Limitations
Lagrange	Clear function relationship	Cumbersome notation for higher-order derivatives
Leibniz	Intuitive reflection of differential nature	Must note it is not a fraction
Newton	Efficient for time derivatives	Only applicable to single-variable time functions

For example, in the heat equation, mixing different notations is more efficient:

$$\frac{\partial T}{\partial t} = \alpha \nabla^2 T \quad (\text{Leibniz for spatial derivatives + Newton for time derivative})$$

Definition

The derivative of a function is defined as the limit of the ratio of the increment of the function to the increment of the independent variable as the increment of the independent variable approaches zero. Mathematically:

\begin{eqnarray} f'(x_0) & = & \lim_{\Delta x \to 0}\frac{\Delta y}{\Delta x} = \lim_{\Delta x \to 0} \frac{f(x_0 + \Delta x) - f(x_0)}{\Delta x} \\\\ f'(x_0) & = & \lim_{x \to x_0} \frac{f(x) - f(x_0)}{x - x_0} \end{eqnarray}

When the limit exists, the function $y = f(x)$ is said to be differentiable at $x_0$, and this limit is called the derivative of $y = f(x)$ at point $x_0$.

Essentially, the definition of derivative is a limit problem

From the definition, it can be seen that the derivative studies the trend of change speed through the ratio of the change in function value to the change in independent variable.

「From a graphical perspective, the derivative is the slope of the tangent line to the function $y = f(x)$ at $x = x_0$」

Verify that the function $y = \ln{x} + 1$ at $x = 1$ has a tangent slope of $1$, so the tangent is $y - 1 = 1 \cdot (x - 1)$

And here $f'(x)$ at $x = 1$ is indeed equal to $1$

How to prove it more rigorously?

As can be seen in this illustration, when $\Delta x \to 0$, the points $(x_0, f(x_0))$ and $(x_0 + \Delta x, f(x_0 + \Delta x))$ approach a straight line. For this line, its slope is:

$$k = \frac{f(x_0 + \Delta x) - f(x_0)}{x_0 + \Delta x - x_0} = \frac{f(x_0 + \Delta x) - f(x_0)}{\Delta x}$$

Does that look familiar?

Differentiation Rules

Basic Elementary Function Derivatives

Function Type	Derivative Formula
Constant function	$(C)' = 0$
Power function	$(x^\mu)' = \mu x^{\mu-1}$
Exponential function	$(a^x)' = a^x \ln a$
Natural exponential function	$(e^x)' = e^x$
Logarithmic function	$(\log_a x)' = \frac{1}{x \ln a}$
Natural logarithmic function	$(\ln x)' = \frac{1}{x}$
Sine function	$(\sin x)' = \cos x$
Cosine function	$(\cos x)' = -\sin x$

The derivative formulas for basic elementary functions can of course be derived using the definition of the derivative.

Operations

Addition and Subtraction Rule

$$(u \pm v)' = u' \pm v'$$

Product Rule

$$(uv)' = u'v + uv'$$

Quotient Rule

$$\left(\frac{u}{v}\right)' = \frac{u'v - uv'}{v^2}$$

where $u, v$ are basic elementary functions

Chain Rule

For composite functions, the derivative is obtained using the chain rule:

$$\frac{d}{dx}[f(g(x))] = f'(g(x)) \cdot g'(x)$$

The chain rule states that the derivative of a composite function equals the derivative of the outer function with respect to the inner function multiplied by the derivative of the inner function. Intuitively, the rate of change of the rate of change equals the product of the individual rates of change.

Basic steps:

1. Identify the outer function and inner function in the composite function
2. Compute the derivative of the outer function, keeping the inner function unchanged
3. Compute the derivative of the inner function
4. Multiply the two derivatives to obtain the final result

Example 1: Find the derivative of $y = \sin(2x)$

Solution: Let $u = 2x$, then $y = \sin u$

• Derivative of outer function: $\frac{dy}{du} = \cos u$
• Derivative of inner function: $\frac{du}{dx} = 2$
• Applying chain rule: $\frac{dy}{dx} = \frac{dy}{du} \cdot \frac{du}{dx} = \cos u \cdot 2 = 2\cos(2x)$

Example 2: Find the derivative of $y = (3x + 2)^2$

Solution: Let $u = 3x + 2$, then $y = u^2$

• Derivative of outer function: $\frac{dy}{du} = 2u$
• Derivative of inner function: $\frac{du}{dx} = 3$
• Applying chain rule: $\frac{dy}{dx} = \frac{dy}{du} \cdot \frac{du}{dx} = 2u \cdot 3 = 6(3x + 2)$

Example 3: Find the derivative of $y = e^{3x^2+2}$

Solution: Let $u = 3x^2+2$, then $y = e^u$

• Derivative of outer function: $\frac{dy}{du} = e^u$
• Derivative of inner function: $\frac{du}{dx} = 6x$
• Applying chain rule: $\frac{dy}{dx} = \frac{dy}{du} \cdot \frac{du}{dx} = e^u \cdot 6x = 6x e^{3x^2+2}$

Chain Rule for Multiple Composite Functions

For multi-layered composite functions, the chain rule can be applied repeatedly:

$$\frac{d}{dx}[f(g(h(x)))] = f'(g(h(x))) \cdot g'(h(x)) \cdot h'(x)$$

Example: Find the derivative of $y = \ln\sqrt{x}$

Solution: Consider it as $y = \ln(x^{1/2})$

• Let $u = x^{1/2}$, then $y = \ln u$
• Derivative of outer function: $\frac{dy}{du} = \frac{1}{u}$
• Derivative of inner function: $\frac{du}{dx} = \frac{1}{2}x^{-1/2}$
• Applying chain rule: $\frac{dy}{dx} = \frac{dy}{du} \cdot \frac{du}{dx} = \frac{1}{u} \cdot \frac{1}{2}x^{-1/2} = \frac{1}{2x}$

Properties

Differentiability Condition

For a function $y = f(x)$, when $x = x_0$, the two-sided derivatives are:

$$\begin{align} \lim_{\Delta x \to 0^+} \frac{f(x_0 + \Delta x) - f(x_0)}{\Delta x} & = f'_{+}(x_0) \\\\ \lim_{\Delta x \to 0^-} \frac{f(x_0 + \Delta x) - f(x_0)}{\Delta x} & = f'_{-}(x_0) \end{align}$$

where $f'_{+}(x_0)$ and $f'_{-}(x_0)$ are the right-hand and left-hand derivatives of $f(x)$ at $x = x_0$, collectively called one-sided derivatives.

Furthermore, for $y = f(x)$, if at $x = x_0$ the one-sided derivatives exist and are equal:

$$f'_{+}(x_0) = f'_{-}(x_0)$$

then $f'(x)$ exists and equals the one-sided derivative value. (Sufficient and necessary condition)

Relationship between Differentiability and Continuity

From the above, it is easy to compare differentiability and continuity. The descriptions of these two properties are very similar, aren’t they?

Continuity
If a function $f(x)$ is continuous at $x = x_0$, then:
$f(x) = \lim_{x \to {x_0}^+}f(x) = \lim_{x \to {x_0}^-} f(x)$
Differentiability
If a function $f(x)$ is differentiable at $x = x_0$, then:
$f'(x) = f'_{+}(x_0) = f'_{-}(x_0)$

In the article Function Limits from the Perspective of Hyperreal Numbers, we have mentioned that the hyperreal perspective can better understand function limits. Similarly, we can better understand differentiability.

Here we first supplement continuity:

A function $f(x)$ is continuous at $x = x_0$ if and only if

$$\forall \epsilon > 0,\ \exists \delta > 0,\ \text{such that when } |x - x_0| < \delta \text{, we have } |f(x) - f(x_0)| < \epsilon.$$

That is, the result of applying the standard part function to all hyperreal numbers in a neighborhood centered at $x_0$ is the real number $f(x_0)$.

A continuous interval consists of real numbers, but these real numbers are accompanied by infinitely close hyperreal numbers in the hyperreal model. Continuity in the hyperreal perspective is manifested as the stability of function values under infinitesimal perturbations, rather than a ‘seamless connection’ in space.

Clearly, the definition of continuity: $f(x) = \lim_{x \to {x_0}^+}f(x) = \lim_{x \to {x_0}^-} f(x)$ essentially means that function $f(x)$ is continuous at $x_0$ if and only if for all hyperreal numbers $x$ infinitely close to $x_0$, $f(x)$ is infinitely close to $f(x_0)$. Continuity is a local property, and the continuity of a function on an interval must be verified at each point individually. Hence we often see the description ‘the function is everywhere continuous on an interval’.

Alright, let’s continue with differentiability.

Definition of derivative:

$$f'(x_0) = \lim_{x \to x_0} \frac{f(x) - f(x_0)}{x - x_0}$$

When the limit exists, the function is said to be differentiable. That is:

$$f'(x_0) \not= \infty$$

For the case $f'(x_0) = 0$, the numerator is a higher-order infinitesimal relative to the denominator. That is, $f(x) - f(x_0)$ approaches $0$ faster than $x - x_0$;

The expression for continuity at $x = x_0$ is essentially $f(x) - f(x_0)$ approaches 0.

But differentiability requires a stronger condition than continuity: $f(x) - f(x_0)$ must approach $0$ faster than $x - x_0$.

For the case $f'(x_0) = a, a \in \mathbb{R}$, the numerator and denominator are infinitesimals of the same order.

Then differentiability also requires a stronger condition: $f(x) - f(x_0)$ and $x - x_0$ must be infinitesimals of the same order.

Therefore:

$$\begin{align} f(x_0) \text{ differentiable } &\Rightarrow f(x_0) \text{ continuous} \\\\ f(x_0) \text{ continuous } &\not\Rightarrow f(x_0) \text{ differentiable} \end{align}$$

Continuity and Differentiability Analysis of a Function and Its Absolute Value

Continuity

Assume there is a function $f(x)$ that is continuous everywhere in its domain.

For $f(x)$, according to the range of its values, it can be summarized into three possible cases as shown in the figure below:

For $\mid f(x) \mid$, the corresponding cases are:

Clearly, if $f(x)$ is continuous at a point, then $\mid f(x) \mid$ is also continuous at that point. (After taking absolute value, hyperreal numbers in the neighborhood still approach that real point.)

So if $\mid f(x) \mid$ is continuous at a point, can we prove that $f(x)$ is also continuous at that point?

Conclusion:

$$\begin{align} f(x_0) \text{ continuous } &\Rightarrow \mid f(x_0) \mid \text{ continuous} \\\\ \mid f(x_0) \mid \text{ continuous } &\not\Rightarrow f(x_0) \text{ continuous} \end{align}$$

Differentiability

For $f(x)$, according to the range of its values, it can be summarized into three possible cases as shown in the figure below:

For $\mid f(x) \mid$, the corresponding cases are:

Regarding differentiability, as we just mentioned, differentiability requires a stronger condition than continuity, namely $f(x) - f(x_0)$ must approach $0$ faster than $x - x_0$.

This case satisfies continuity but does not satisfy $f'(x) = f'_{+}(x_0) = f'_{-}(x_0)$.

Conclusion:

$$\begin{cases} f(x_0) \text{ differentiable} \\\\ f(x_0) \not= 0 \end{cases} \Rightarrow \mid f(x_0) \mid \text{ differentiable}$$ $$\begin{cases} f(x_0) \text{ differentiable} \\\\ f(x_0) = 0 \\\\ f'(x_0) \not= 0 \end{cases} \Rightarrow \mid f(x_0) \mid \text{ not differentiable}$$

When $f(x_0) = 0$ and $f'(x_0) = 0$, the graph coincides with the x-axis, $\mid f'(x_0) \mid = 0$.

The necessity has been described in the proof of continuity above and will not be repeated.

It can be summarized as:

$$f(x_0) \text{ differentiable } \not\Leftrightarrow \mid f(x_0) \mid \text{ differentiable}$$

Parity Relationship between Derivative and Original Function

First, give the conclusion:

$$\begin{cases} \text{If } f(x) \text{ is a differentiable even function, then } f'(x) \text{ is an odd function} \\\\ \text{If } f(x) \text{ is a differentiable odd function, then } f'(x) \text{ is an even function} \end{cases}$$

Proof:
$\text{If } f(x) \text{ is a differentiable even function, then } f'(x) \text{ is an odd function}$

$$\begin{array}{c} f'(x) & = & \lim_{\Delta x \to 0} \frac{f(x + \Delta x) - f(x)}{\Delta x} \\\\ f'(-x) & = & \lim_{\Delta x \to 0} \frac{f(-x + \Delta x) - f(-x)}{\Delta x} \\\\ & = & \lim_{\Delta x \to 0} \frac{f(x - \Delta x) - f(x)}{\Delta x} \\\\ & = & \lim_{- \Delta x \to 0} \frac{f(x - \Delta x) - f(x)}{- \Delta x} \\\\ & = & -f'(x) \end{array}$$

The proof that if $f(x)$ is a differentiable odd function, then $f'(x)$ is an even function is similar to the above and will not be repeated.