Section 2.2: Derivative Rules from First Principles¶

Computing derivatives from the limit definition every time would be exhausting. Fortunately, a few key rules let us differentiate almost any function quickly.

But we won't just state these rules—we'll derive them from first principles. This ensures we understand not just what the rules are, but why they work.

Constants and Linear Functions¶

Constant Rule¶

If f(x) = c (a constant), what is f'(x)?

\[f'(x) = \lim_{h \to 0} \frac{f(x+h) - f(x)}{h} = \lim_{h \to 0} \frac{c - c}{h} = \lim_{h \to 0} \frac{0}{h} = 0\]

Result: The derivative of a constant is zero.

This makes sense: a horizontal line has slope 0 everywhere.

Constant Multiple Rule¶

If g(x) = c·f(x), what is g'(x)?

\[g'(x) = \lim_{h \to 0} \frac{c \cdot f(x+h) - c \cdot f(x)}{h} = \lim_{h \to 0} c \cdot \frac{f(x+h) - f(x)}{h} = c \cdot f'(x)\]

Result: (c·f)' = c·f'

Constants factor out of derivatives.

The Sum Rule¶

If h(x) = f(x) + g(x), what is h'(x)?

\[h'(x) = \lim_{k \to 0} \frac{[f(x+k) + g(x+k)] - [f(x) + g(x)]}{k}\]

Rearranging:

\[= \lim_{k \to 0} \frac{[f(x+k) - f(x)] + [g(x+k) - g(x)]}{k}\]

\[= \lim_{k \to 0} \frac{f(x+k) - f(x)}{k} + \lim_{k \to 0} \frac{g(x+k) - g(x)}{k}\]

\[= f'(x) + g'(x)\]

Result: (f + g)' = f' + g'

The derivative of a sum is the sum of derivatives. This is called linearity—differentiation is a linear operation.

The Power Rule¶

The power rule states: if f(x) = \(x^n\), then f'(x) = n·\(x^{n-1}\).

Let's prove this for positive integers using the binomial theorem.

Proof Using Binomial Expansion¶

For positive integer n:

\[(x+h)^n = \sum_{k=0}^{n} \binom{n}{k} x^{n-k} h^k = x^n + nx^{n-1}h + \frac{n(n-1)}{2}x^{n-2}h^2 + \cdots + h^n\]

The difference quotient:

\[\frac{(x+h)^n - x^n}{h} = \frac{nx^{n-1}h + \frac{n(n-1)}{2}x^{n-2}h^2 + \cdots + h^n}{h}\]

\[= nx^{n-1} + \frac{n(n-1)}{2}x^{n-2}h + \cdots + h^{n-1}\]

Taking the limit as h → 0:

\[\lim_{h \to 0} \left[ nx^{n-1} + \frac{n(n-1)}{2}x^{n-2}h + \cdots + h^{n-1} \right] = nx^{n-1}\]

All terms with h vanish, leaving only the first term.

Result: d/dx(\(x^n\)) = n·\(x^{n-1}\)

Examples¶

Function	Derivative
x¹	1·x⁰ = 1
x²	2x
x³	3x²
\(x^{10}\)	10x⁹

Extension to Negative and Fractional Powers¶

The power rule also works for negative and fractional exponents. Let's verify for n = -1:

We already proved: d/dx(1/x) = -1/x²

Using the power rule: d/dx(\(x^{-1}\)) = -1·\(x^{-2}\) = -1/x² ✓

For n = 1/2 (square root):

\[\frac{d}{dx}\sqrt{x} = \frac{d}{dx}x^{1/2} = \frac{1}{2}x^{-1/2} = \frac{1}{2\sqrt{x}}\]

This can be verified from the definition (more involved).

The Product Rule¶

Now things get interesting. If h(x) = f(x)·g(x), is h'(x) = f'(x)·g'(x)?

No! Let's see what actually happens.

Derivation¶

\[h'(x) = \lim_{k \to 0} \frac{f(x+k)g(x+k) - f(x)g(x)}{k}\]

The trick is to add and subtract a "bridge" term:

\[= \lim_{k \to 0} \frac{f(x+k)g(x+k) - f(x+k)g(x) + f(x+k)g(x) - f(x)g(x)}{k}\]

Grouping:

\[= \lim_{k \to 0} \frac{f(x+k)[g(x+k) - g(x)] + g(x)[f(x+k) - f(x)]}{k}\]

\[= \lim_{k \to 0} f(x+k) \cdot \frac{g(x+k) - g(x)}{k} + g(x) \cdot \lim_{k \to 0} \frac{f(x+k) - f(x)}{k}\]

Since f is continuous (differentiable functions are continuous):

\[\lim_{k \to 0} f(x+k) = f(x)\]

Therefore:

\[h'(x) = f(x) \cdot g'(x) + g(x) \cdot f'(x)\]

Result: (f·g)' = f·g' + f'·g

Intuition¶

Why isn't the derivative of a product the product of derivatives?

Think of a rectangle with sides f and g. Its area is f·g.

If both sides grow:

The area grows by f·Δg (original f, additional g)
The area grows by g·Δf (original g, additional f)
There's also a tiny Δf·Δg corner (negligible as Δ → 0)

Total growth ≈ f·Δg + g·Δf, which gives the product rule.

Example¶

Let h(x) = x²·sin(x). (We'll derive sin'(x) = cos(x) later.)

Using the product rule:

\[h'(x) = x^2 \cdot \cos(x) + \sin(x) \cdot 2x = x^2\cos(x) + 2x\sin(x)\]

The Quotient Rule¶

If h(x) = f(x)/g(x), what is h'(x)?

Derivation¶

We can derive this from the product rule by writing f/g = f · (1/g).

First, we need d/dx(1/g). Let's use the limit definition:

\[\frac{d}{dx}\left(\frac{1}{g(x)}\right) = \lim_{k \to 0} \frac{\frac{1}{g(x+k)} - \frac{1}{g(x)}}{k}\]

\[= \lim_{k \to 0} \frac{g(x) - g(x+k)}{k \cdot g(x) \cdot g(x+k)}\]

\[= \lim_{k \to 0} \frac{-[g(x+k) - g(x)]}{k} \cdot \frac{1}{g(x) \cdot g(x+k)}\]

\[= -g'(x) \cdot \frac{1}{g(x)^2} = -\frac{g'(x)}{g(x)^2}\]

Now apply the product rule to h = f · (1/g):

\[h' = f \cdot \left(-\frac{g'}{g^2}\right) + \frac{1}{g} \cdot f' = \frac{f'}{g} - \frac{f \cdot g'}{g^2}\]

\[= \frac{f' \cdot g - f \cdot g'}{g^2}\]

Result: (f/g)' = (f'g - fg')/g²

Memory Aid¶

Some remember this as "low d-high minus high d-low, over low squared":

\[\frac{d}{dx}\frac{f}{g} = \frac{g \cdot f' - f \cdot g'}{g^2}\]

The Exponential Function¶

The exponential function \(e^x\) is special: it's its own derivative.

What is e?¶

The number e ≈ 2.71828... is defined as:

\[e = \lim_{n \to \infty} \left(1 + \frac{1}{n}\right)^n\]

Or equivalently, e is the unique number such that:

\[\lim_{h \to 0} \frac{e^h - 1}{h} = 1\]

Derivative of \(e^x\)¶

\[\frac{d}{dx}e^x = \lim_{h \to 0} \frac{e^{x+h} - e^x}{h} = \lim_{h \to 0} \frac{e^x \cdot e^h - e^x}{h}\]

\[= e^x \cdot \lim_{h \to 0} \frac{e^h - 1}{h} = e^x \cdot 1 = e^x\]

Result: d/dx(\(e^x\)) = \(e^x\)

This is why \(e^x\) is so important in mathematics—it's the unique function (up to scaling) that equals its own derivative.

General Exponential¶

For \(a^x\) where a > 0:

Using \(a^x\) = \(e^{x·ln(a)}\) and the chain rule (next section):

\[\frac{d}{dx}a^x = a^x \cdot \ln(a)\]

The Natural Logarithm¶

If y = ln(x), what is dy/dx?

Derivation Using Inverse Functions¶

Since \(e^{ln(x)}\) = x, differentiate both sides:

\[e^{\ln(x)} \cdot \frac{d}{dx}\ln(x) = 1\]

\[x \cdot \frac{d}{dx}\ln(x) = 1\]

\[\frac{d}{dx}\ln(x) = \frac{1}{x}\]

Result: d/dx(ln x) = 1/x

Log of Other Bases¶

For log_a(x) = ln(x)/ln(a):

\[\frac{d}{dx}\log_a(x) = \frac{1}{x \cdot \ln(a)}\]

Summary of Rules¶

Rule	Formula	Derived From
Constant	(c)' = 0	Limit definition
Power	(\(x^n\))' = \(nx^{n-1}\)	Binomial theorem
Sum	(f+g)' = f'+g'	Linearity of limits
Product	(fg)' = fg' + f'g	Add-subtract trick
Quotient	(f/g)' = (f'g-fg')/g²	Product rule + reciprocal
Exponential	(\(e^x\))' = \(e^x\)	Definition of e
Logarithm	(ln x)' = 1/x	Inverse function

What's Missing: The Chain Rule¶

Notice we haven't handled compositions like sin(x²) or \(e^{-x²}\) or ln(1+x).

These require the chain rule, which is so important it gets its own section. The chain rule is the heart of automatic differentiation.

Exercises¶

Derive the power rule for n=3 by directly expanding (x+h)³ - x³.
Product rule practice: Find d/dx(x³·\(e^x\)).
Quotient rule practice: Find d/dx(x²/(1+x)).
Why the product rule?: Give a geometric argument for why (fg)' ≠ f'g'.
Verify log derivative: Using the limit definition, show that d/dx(ln x) = 1/x by computing the limit directly. (Hint: use the substitution k = h/x and the definition of e.)

Summary¶

We derived all fundamental derivative rules from the limit definition:

Constants disappear, sums split, constants factor out
The power rule handles polynomials
The product rule handles products (it's not just f'g')
The quotient rule is the product rule for reciprocals
\(e^x\) is its own derivative (remarkable!)
ln(x) differentiates to 1/x

With these rules, we can differentiate any polynomial, rational function, or expression involving exponentials and logarithms—as long as there's no function composition.

For compositions, we need the chain rule.