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Mathematics · Ch 12 — Limits and Derivatives

Algebra of Derivative of Functions

12.5.1

Algebra of Derivative of Functions

The Algebra of Derivatives: Why Limits Make It Work

Derivatives are defined through limits — the derivative of a function ff at a point xx is lim⁡h→0f(x+h)−f(x)h\lim_{h \to 0} \frac{f(x+h)-f(x)}{h}. Because limits themselves obey algebraic rules (the limit of a sum is the sum of the limits, and so on), it is natural that derivatives follow similar rules. This section collects those rules into a single theorem and then uses them to build derivatives of standard functions step by step.

Theorem 5: The Four Basic Rules of Differentiation

Let ff and gg be two functions whose derivatives exist over a common domain. Then:

  1. Sum Rule The derivative of the sum of two functions equals the sum of their derivatives.

    ddx[f(x)+g(x)]=ddxf(x)+ddxg(x)\frac{d}{dx}\big[f(x) + g(x)\big] = \frac{d}{dx}f(x) + \frac{d}{dx}g(x)

  2. Difference Rule The derivative of the difference of two functions equals the difference of their derivatives.

    ddx[f(x)−g(x)]=ddxf(x)−ddxg(x)\frac{d}{dx}\big[f(x) - g(x)\big] = \frac{d}{dx}f(x) - \frac{d}{dx}g(x)

  3. Product Rule The derivative of the product of two functions is given by:

    ddx[f(x)⋅g(x)]=f(x)⋅ddxg(x)+g(x)⋅ddxf(x)\frac{d}{dx}\big[f(x) \cdot g(x)\big] = f(x) \cdot \frac{d}{dx}g(x) + g(x) \cdot \frac{d}{dx}f(x)

  4. Quotient Rule If g(x)≠0g(x) \neq 0, the derivative of the quotient of two functions is:

    ddx[f(x)g(x)]=g(x)⋅ddxf(x)−f(x)⋅ddxg(x)[g(x)]2\frac{d}{dx}\left[\frac{f(x)}{g(x)}\right] = \frac{g(x) \cdot \frac{d}{dx}f(x) - f(x) \cdot \frac{d}{dx}g(x)}{[g(x)]^2}

    Note

    The proofs of these four rules follow directly from the corresponding limit laws (limit of a sum, limit of a product, etc.). The textbook states these proofs are not given here, but the logic is: each derivative is a limit of a difference quotient, and the algebraic manipulation of that quotient mirrors the limit algebra you already know.

A Handy Notation: The Leibnitz Form

To make the product and quotient rules easier to remember, let u=f(x)u = f(x) and v=g(x)v = g(x). Then:

Product rule (Leibnitz rule):

(uv)′=u′v+uv′(uv)' = u'v + uv'

Quotient rule:

(uv)′=u′v−uv′v2\left(\frac{u}{v}\right)' = \frac{u'v - uv'}{v^2}

Tip

The order in the product rule matters: "first times derivative of second plus second times derivative of first." A common mnemonic is "first d-second plus second d-first."

The Derivative of f(x)=xf(x) = x

Before applying the rules, we need a starting point. The simplest non-constant function is f(x)=xf(x) = x. Using the definition:

f′(x)=lim⁡h→0(x+h)−xh=lim⁡h→0hh=lim⁡h→01=1f'(x) = \lim_{h \to 0} \frac{(x+h) - x}{h} = \lim_{h \to 0} \frac{h}{h} = \lim_{h \to 0} 1 = 1

So the derivative of xx is the constant function 11.

Example: Derivative of f(x)=10xf(x) = 10x (Ten Terms of xx)

We can compute this in two ways, both illustrating the sum rule and the product rule.

Method 1 — Using the sum rule:

Write f(x)=x+x+⋯+xf(x) = x + x + \dots + x (ten terms). Then:

f′(x)=ddx(x)+ddx(x)+⋯+ddx(x)(ten times)f'(x) = \frac{d}{dx}(x) + \frac{d}{dx}(x) + \dots + \frac{d}{dx}(x) \quad (\text{ten times})

Each ddx(x)=1\frac{d}{dx}(x) = 1, so:

f′(x)=1+1+⋯+1=10f'(x) = 1 + 1 + \dots + 1 = 10

Method 2 — Using the product rule:

Write f(x)=10xf(x) = 10x as u⋅vu \cdot v where u(x)=10u(x) = 10 (a constant function) and v(x)=xv(x) = x.

We know u′(x)=0u'(x) = 0 (derivative of a constant is zero) and v′(x)=1v'(x) = 1.

By the product rule:

f′(x)=u′v+uv′=0⋅x+10⋅1=10f'(x) = u'v + uv' = 0 \cdot x + 10 \cdot 1 = 10

Both methods give the same result, as they must.

Watch out

A constant function has derivative zero. Do not confuse "the derivative of 10x10x" with "the derivative of 1010" — the xx factor makes all the difference.

Derivative of f(x)=x2f(x) = x^2

Write x2x^2 as x⋅xx \cdot x. Using the product rule with u=xu = x, v=xv = x:

ddx(x2)=ddx(x⋅x)=x⋅ddx(x)+x⋅ddx(x)=x⋅1+x⋅1=2x\frac{d}{dx}(x^2) = \frac{d}{dx}(x \cdot x) = x \cdot \frac{d}{dx}(x) + x \cdot \frac{d}{dx}(x) = x \cdot 1 + x \cdot 1 = 2x

This matches the pattern we expect: the derivative of x2x^2 is 2x2x.

Theorem 6: Derivative of xnx^n for Positive Integer nn

ddx(xn)=nxn−1\frac{d}{dx}(x^n) = n x^{n-1}

This is the power rule, and the textbook proves it in two ways.

›Proof

Proof 1 — Using the definition and the binomial theorem:

By definition:

f′(x)=lim⁡h→0(x+h)n−xnhf'(x) = \lim_{h \to 0} \frac{(x+h)^n - x^n}{h}

Expand (x+h)n(x+h)^n using the binomial theorem:

(x+h)n=xn+(n1)xn−1h+(n2)xn−2h2+⋯+(nn−1)xhn−1+hn(x+h)^n = x^n + \binom{n}{1} x^{n-1} h + \binom{n}{2} x^{n-2} h^2 + \dots + \binom{n}{n-1} x h^{n-1} + h^n

Subtract xnx^n:

(x+h)n−xn=(n1)xn−1h+(n2)xn−2h2+⋯+(nn−1)xhn−1+hn(x+h)^n - x^n = \binom{n}{1} x^{n-1} h + \binom{n}{2} x^{n-2} h^2 + \dots + \binom{n}{n-1} x h^{n-1} + h^n

Factor out hh:

(x+h)n−xn=h[(n1)xn−1+(n2)xn−2h+⋯+(nn−1)xhn−2+hn−1](x+h)^n - x^n = h \left[ \binom{n}{1} x^{n-1} + \binom{n}{2} x^{n-2} h + \dots + \binom{n}{n-1} x h^{n-2} + h^{n-1} \right]

Now the difference quotient becomes:

(x+h)n−xnh=(n1)xn−1+(n2)xn−2h+⋯+(nn−1)xhn−2+hn−1\frac{(x+h)^n - x^n}{h} = \binom{n}{1} x^{n-1} + \binom{n}{2} x^{n-2} h + \dots + \binom{n}{n-1} x h^{n-2} + h^{n-1}

Take the limit as h→0h \to 0. Every term containing hh vanishes, leaving only the first term:

f′(x)=(n1)xn−1=nxn−1f'(x) = \binom{n}{1} x^{n-1} = n x^{n-1}

…

Theorem 5

Theorem 6: The Power Rule for Positive Integer Exponents

ddx(xn)=nxn−1for any positive integer n\frac{d}{dx}(x^n) = n x^{n-1} \quad \text{for any positive integer } n

The theorem states that if f(x)=xnf(x) = x^n where nn is a positive integer, then the derivative exists for all real xx and equals nxn−1n x^{n-1}. The only hypothesis is that n∈Nn \in \mathbb{N} (the set of positive integers). The domain of the derivative is all real numbers — the same as the domain of ff itself.

Complete Proof

The textbook gives two independent proofs. Both are rigorous; the first uses the binomial theorem directly from the limit definition, while the second uses mathematical induction together with the product rule. We present both in full.

›Proof

Proof 1 (Using the limit definition and binomial theorem)

By the definition of the derivative,

f′(x)=lim⁡h→0f(x+h)−f(x)h=lim⁡h→0(x+h)n−xnh.f'(x) = \lim_{h \to 0} \frac{f(x+h) - f(x)}{h} = \lim_{h \to 0} \frac{(x+h)^n - x^n}{h}.

Expand (x+h)n(x+h)^n using the binomial theorem:

(x+h)n=(n0)xn+(n1)xn−1h+(n2)xn−2h2+⋯+(nn−1)xhn−1+(nn)hn.(x+h)^n = \binom{n}{0} x^n + \binom{n}{1} x^{n-1} h + \binom{n}{2} x^{n-2} h^2 + \cdots + \binom{n}{n-1} x h^{n-1} + \binom{n}{n} h^n.

Since (n0)=1\binom{n}{0} = 1 and (n1)=n\binom{n}{1} = n, this becomes

(x+h)n=xn+nxn−1h+(n2)xn−2h2+⋯+nxhn−1+hn.(x+h)^n = x^n + n x^{n-1} h + \binom{n}{2} x^{n-2} h^2 + \cdots + n x h^{n-1} + h^n.

Subtract xnx^n from both sides:

(x+h)n−xn=nxn−1h+(n2)xn−2h2+⋯+nxhn−1+hn.(x+h)^n - x^n = n x^{n-1} h + \binom{n}{2} x^{n-2} h^2 + \cdots + n x h^{n-1} + h^n.

Factor out hh from every term on the right-hand side:

(x+h)n−xn=h(nxn−1+(n2)xn−2h+⋯+nxhn−2+hn−1).(x+h)^n - x^n = h \left( n x^{n-1} + \binom{n}{2} x^{n-2} h + \cdots + n x h^{n-2} + h^{n-1} \right).

Now substitute this into the limit:

f′(x)=lim⁡h→0h(nxn−1+(n2)xn−2h+⋯+nxhn−2+hn−1)h.f'(x) = \lim_{h \to 0} \frac{h \left( n x^{n-1} + \binom{n}{2} x^{n-2} h + \cdots + n x h^{n-2} + h^{n-1} \right)}{h}.

Cancel hh (valid for h≠0h \neq 0, which is all that matters in the limit):

f′(x)=lim⁡h→0(nxn−1+(n2)xn−2h+⋯+nxhn−2+hn−1).f'(x) = \lim_{h \to 0} \left( n x^{n-1} + \binom{n}{2} x^{n-2} h + \cdots + n x h^{n-2} + h^{n-1} \right).

As h→0h \to 0, every term that contains a factor of hh vanishes. The only term that survives is the first one, nxn−1n x^{n-1}, which has no hh in it. Therefore,

f′(x)=nxn−1.f'(x) = n x^{n-1}.

This completes the proof.

Proof 2 (By mathematical induction and the product rule)

Base case: n=1n = 1. We have f(x)=xf(x) = x. From the limit definition,

f′(x)=lim⁡h→0(x+h)−xh=lim⁡h→0hh=lim⁡h→01=1.f'(x) = \lim_{h \to 0} \frac{(x+h) - x}{h} = \lim_{h \to 0} \frac{h}{h} = \lim_{h \to 0} 1 = 1.

Since 1=1⋅x01 = 1 \cdot x^{0}, the formula ddx(x)=1⋅x0=1\frac{d}{dx}(x) = 1 \cdot x^{0} = 1 holds. So the statement is true for n=1n = 1.

Induction hypothesis: Assume that for some positive integer kk, we have

ddx(xk)=kxk−1.\frac{d}{dx}(x^k) = k x^{k-1}.

Induction step: Prove the statement for n=k+1n = k+1. Write xk+1=x⋅xkx^{k+1} = x \cdot x^k. Apply the product rule with u(x)=xu(x) = x and v(x)=xkv(x) = x^k:

ddx(xk+1)=ddx(x⋅xk)=(ddx(x))⋅xk+x⋅(ddx(xk)).\frac{d}{dx}(x^{k+1}) = \frac{d}{dx}(x \cdot x^k) = \left( \frac{d}{dx}(x) \right) \cdot x^k + x \cdot \left( \frac{d}{dx}(x^k) \right).

We know ddx(x)=1\frac{d}{dx}(x) = 1 from the base case, and by the induction hypothesis ddx(xk)=kxk−1\frac{d}{dx}(x^k) = k x^{k-1}. Substituting:

ddx(xk+1)=1⋅xk+x⋅(kxk−1)=xk+kxk=(k+1)xk.\frac{d}{dx}(x^{k+1}) = 1 \cdot x^k + x \cdot (k x^{k-1}) = x^k + k x^k = (k+1) x^k.

This is exactly the formula for n=k+1n = k+1. By the principle of mathematical induction, the statement holds for all positive integers nn. …

Theorem 6

Theorem 6: The Power Rule for Positive Integer Exponents

ddx(xn)=nxn−1for any positive integer n\frac{d}{dx}(x^n) = n x^{n-1} \quad \text{for any positive integer } n

The theorem states that if f(x)=xnf(x) = x^n where nn is a positive integer, then the derivative exists for all real xx and equals nxn−1n x^{n-1}. The only hypothesis is that n∈Nn \in \mathbb{N} (the set of positive integers). The domain of the derivative is all real numbers — the same as the domain of ff itself.

Complete Proof

The textbook gives two independent proofs. Both are rigorous; the first uses the binomial theorem directly from the limit definition, while the second uses mathematical induction together with the product rule. We present both in full.

›Proof

Proof 1 (Using the limit definition and binomial theorem)

By the definition of the derivative,

f′(x)=lim⁡h→0f(x+h)−f(x)h=lim⁡h→0(x+h)n−xnh.f'(x) = \lim_{h \to 0} \frac{f(x+h) - f(x)}{h} = \lim_{h \to 0} \frac{(x+h)^n - x^n}{h}.

Expand (x+h)n(x+h)^n using the binomial theorem:

(x+h)n=(n0)xn+(n1)xn−1h+(n2)xn−2h2+⋯+(nn−1)xhn−1+(nn)hn.(x+h)^n = \binom{n}{0} x^n + \binom{n}{1} x^{n-1} h + \binom{n}{2} x^{n-2} h^2 + \cdots + \binom{n}{n-1} x h^{n-1} + \binom{n}{n} h^n.

Since (n0)=1\binom{n}{0} = 1 and (n1)=n\binom{n}{1} = n, this becomes

(x+h)n=xn+nxn−1h+(n2)xn−2h2+⋯+nxhn−1+hn.(x+h)^n = x^n + n x^{n-1} h + \binom{n}{2} x^{n-2} h^2 + \cdots + n x h^{n-1} + h^n.

Subtract xnx^n from both sides:

(x+h)n−xn=nxn−1h+(n2)xn−2h2+⋯+nxhn−1+hn.(x+h)^n - x^n = n x^{n-1} h + \binom{n}{2} x^{n-2} h^2 + \cdots + n x h^{n-1} + h^n.

Factor out hh from every term on the right-hand side:

(x+h)n−xn=h(nxn−1+(n2)xn−2h+⋯+nxhn−2+hn−1).(x+h)^n - x^n = h \left( n x^{n-1} + \binom{n}{2} x^{n-2} h + \cdots + n x h^{n-2} + h^{n-1} \right).

Now substitute this into the limit:

f′(x)=lim⁡h→0h(nxn−1+(n2)xn−2h+⋯+nxhn−2+hn−1)h.f'(x) = \lim_{h \to 0} \frac{h \left( n x^{n-1} + \binom{n}{2} x^{n-2} h + \cdots + n x h^{n-2} + h^{n-1} \right)}{h}.

Cancel hh (valid for h≠0h \neq 0, which is all that matters in the limit):

f′(x)=lim⁡h→0(nxn−1+(n2)xn−2h+⋯+nxhn−2+hn−1).f'(x) = \lim_{h \to 0} \left( n x^{n-1} + \binom{n}{2} x^{n-2} h + \cdots + n x h^{n-2} + h^{n-1} \right).

As h→0h \to 0, every term that contains a factor of hh vanishes. The only term that survives is the first one, nxn−1n x^{n-1}, which has no hh in it. Therefore,

f′(x)=nxn−1.f'(x) = n x^{n-1}.

This completes the proof.

Proof 2 (By mathematical induction and the product rule)

Base case: n=1n = 1. We have f(x)=xf(x) = x. From the limit definition,

f′(x)=lim⁡h→0(x+h)−xh=lim⁡h→0hh=lim⁡h→01=1.f'(x) = \lim_{h \to 0} \frac{(x+h) - x}{h} = \lim_{h \to 0} \frac{h}{h} = \lim_{h \to 0} 1 = 1.

Since 1=1⋅x01 = 1 \cdot x^{0}, the formula ddx(x)=1⋅x0=1\frac{d}{dx}(x) = 1 \cdot x^{0} = 1 holds. So the statement is true for n=1n = 1.

Induction hypothesis: Assume that for some positive integer kk, we have

ddx(xk)=kxk−1.\frac{d}{dx}(x^k) = k x^{k-1}.

Induction step: Prove the statement for n=k+1n = k+1. Write xk+1=x⋅xkx^{k+1} = x \cdot x^k. Apply the product rule with u(x)=xu(x) = x and v(x)=xkv(x) = x^k:

ddx(xk+1)=ddx(x⋅xk)=(ddx(x))⋅xk+x⋅(ddx(xk)).\frac{d}{dx}(x^{k+1}) = \frac{d}{dx}(x \cdot x^k) = \left( \frac{d}{dx}(x) \right) \cdot x^k + x \cdot \left( \frac{d}{dx}(x^k) \right).

We know ddx(x)=1\frac{d}{dx}(x) = 1 from the base case, and by the induction hypothesis ddx(xk)=kxk−1\frac{d}{dx}(x^k) = k x^{k-1}. Substituting:

ddx(xk+1)=1⋅xk+x⋅(kxk−1)=xk+kxk=(k+1)xk.\frac{d}{dx}(x^{k+1}) = 1 \cdot x^k + x \cdot (k x^{k-1}) = x^k + k x^k = (k+1) x^k.

This is exactly the formula for n=k+1n = k+1. By the principle of mathematical induction, the statement holds for all positive integers nn. …

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