Derivatives

The Derivative as a Limit

Key Point (Definition 1): the derivative of $f$ at a point $a$ is

$$f'(a) = \lim_{h \to 0} \frac{f(a+h) - f(a)}{h}$$

provided the limit exists. It measures how fast $f$ changes at $a$.

Geometric reading: the quotient $\frac{f(a+h) - f(a)}{h}$ is the slope of the chord $PQ$; as $h \to 0$, $Q$ slides into $P$ and the chord becomes the tangent: $f'(a) = \tan\psi$, the tangent's slope. Physical reading: with $s(t)$ a position function, $s'(t)$ is the instantaneous velocity — exactly the falling-body computation of Section 1.

Key Point (Definition 2 — first principles): letting the point vary gives the derivative function

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

also written $\frac{dy}{dx}$, $\frac{d}{dx}f(x)$ or $D(f(x))$.

Four first-principles classics: $f(x) = 10x$ gives $f' = 10$; $f(x) = x^2$ gives $\frac{(x+h)^2 - x^2}{h} = 2x + h \to 2x$; a constant gives 0; and $f(x) = \frac{1}{x}$ gives $\frac{-1}{x(x+h)} \to -\frac{1}{x^2}$.

The Algebra of Derivatives and the Standard Table

Key Point (Theorem 5 — rules): with $u = f(x)$, $v = g(x)$ differentiable:

$$(u \pm v)' = u' \pm v' \qquad (uv)' = u'v + uv' \qquad \left(\frac{u}{v}\right)' = \frac{u'v - uv'}{v^2} \;(v \ne 0)$$

The product rule is Leibnitz's rule — note it is not $u'v'$.

The power rule (Theorem 6): $\frac{d}{dx}x^n = nx^{n-1}$ — provable from the binomial theorem, since $(x+h)^n - x^n = h(nx^{n-1} + \cdots)$; true in fact for all real $n$. Combined with the sum rule it differentiates every polynomial: $\frac{d}{dx}(a_nx^n + \cdots + a_1x + a_0) = na_nx^{n-1} + \cdots + a_1$.

Trigonometric derivatives: using $\sin A - \sin B$ factorisation and $\frac{\sin h}{h} \to 1$:

$$\frac{d}{dx}\sin x = \cos x \qquad \frac{d}{dx}\cos x = -\sin x \qquad \frac{d}{dx}\tan x = \sec^2 x$$

and by the quotient rule, $\frac{d}{dx}\cot x = -\mathrm{cosec}^2 x$, $\frac{d}{dx}\sec x = \sec x \tan x$, $\frac{d}{dx}\mathrm{cosec}\,x = -\mathrm{cosec}\,x\cot x$.

[Board Tip] "From first principles" in a question is an instruction, not a suggestion — write the $\lim_{h \to 0}$ quotient and evaluate it. Quoting the answer from the standard table scores zero on such questions, however correct it is.

Solved Examples

Example 1: Derivative at a point

(i) Find $f'(2)$ for $f(x) = 3x$. (ii) For $f(x) = 2x^2 + 3x - 5$, find $f'(-1)$ and verify $f'(0) + 3f'(-1) = 0$.

Solution:

Step 1 — (i) Write the quotient. $\frac{f(2+h) - f(2)}{h} = \frac{3(2+h) - 6}{h} = \frac{3h}{h} = 3$.

Step 2 — (i) Limit. Constantly 3: $f'(2) = 3$.

Step 3 — (ii) Two point-derivatives. At $-1$: $\frac{f(-1+h) - f(-1)}{h} = \frac{2h^2 - h}{h} = 2h - 1 \to -1$. At $0$: $\frac{2h^2 + 3h}{h} = 2h + 3 \to 3$.

Step 4 — (ii) Verify. $f'(0) + 3f'(-1) = 3 + 3(-1) = 0$ ✓. ∎

Takeaway: At a specific point, expand $f(a + h)$, cancel $f(a)$, divide by $h$ — the surviving expression's limit is the derivative.

Example 2: sin x at 0 and constants

(i) Find the derivative of $\sin x$ at $x = 0$. (ii) Find the derivative of $f(x) = 3$ at $x = 0$ and $x = 3$.

Solution:

Step 1 — (i) The quotient is an old friend. $\frac{\sin(0+h) - \sin 0}{h} = \frac{\sin h}{h}$.

Step 2 — (i) Limit. The fundamental limit: 1.

Step 3 — (ii) Constants never move. $\frac{3 - 3}{h} = 0$ for every $h$, at every point: derivative 0.

Takeaway: $(\sin x)' = \cos x$ evaluated at 0 gives $\cos 0 = 1$ — consistent with the fundamental limit, as it must be.

Example 3: Quick point evaluations

Find: (i) the derivative of $x^2 - 2$ at $x = 10$; (ii) of $x$ at $x = 1$; (iii) of $99x$ at $x = 100$.

Solution:

Step 1 — (i). $f'(x) = 2x$: $f'(10) = 20$.

Step 2 — (ii). $f'(x) = 1$ everywhere; at 1, still 1.

Step 3 — (iii). $f'(x) = 99$ everywhere — the flashy point $x = 100$ changes nothing.

Takeaway: Linear functions have constant derivatives; a dramatic evaluation point is usually a distraction test.

Example 4: First principles, four flavours

Differentiate from first principles: (i) $x^3 - 27$; (ii) $(x-1)(x-2)$; (iii) $\frac{1}{x^2}$; (iv) $\frac{x+1}{x-1}$.

Solution:

Step 1 — (i) Cube expansion. $\frac{(x+h)^3 - x^3}{h} = 3x^2 + 3xh + h^2 \to 3x^2$ (the $-27$ dies immediately).

Step 2 — (ii) Expand first. $(x-1)(x-2) = x^2 - 3x + 2$: the quotient collapses to $2x + h - 3 \to 2x - 3$.

Step 3 — (iii) Combine fractions. $\frac{1}{h}\left(\frac{1}{(x+h)^2} - \frac{1}{x^2}\right) = \frac{-(2x + h)}{x^2(x+h)^2} \to -\frac{2x}{x^4} = -\frac{2}{x^3}$.

Step 4 — (iv) Common denominator. $\frac{1}{h}\left(\frac{x+h+1}{x+h-1} - \frac{x+1}{x-1}\right) = \frac{-2}{(x+h-1)(x-1)} \to \frac{-2}{(x-1)^2}$.

Takeaway: First principles always ends the same way — algebra until $h$ factors out, then let $h \to 0$ in what remains.

Example 5: A telescoping check

For $f(x) = \frac{x^{100}}{100} + \frac{x^{99}}{99} + \cdots + \frac{x^2}{2} + x + 1$, prove $f'(1) = 100 f'(0)$.

Solution:

Step 1 — Differentiate term-by-term. $\frac{d}{dx}\frac{x^k}{k} = x^{k-1}$: so $f'(x) = x^{99} + x^{98} + \cdots + x + 1$ — one hundred terms.

Step 2 — Evaluate both points. $f'(1) = \underbrace{1 + 1 + \cdots + 1}_{100} = 100$; $f'(0) = 1$ (only the constant term survives).

Step 3 — Conclude. $f'(1) = 100 = 100 f'(0)$. ∎

Takeaway: The coefficients $\frac{1}{k}$ are DESIGNED to cancel against the power rule — recognise engineered sums before grinding.

Example 6: A polynomial evaluated at 1

Find the derivative of $1 + x + x^2 + \cdots + x^{50}$ at $x = 1$.

Solution:

Step 1 — Differentiate. $f'(x) = 1 + 2x + 3x^2 + \cdots + 50x^{49}$.

Step 2 — Evaluate at 1. $f'(1) = 1 + 2 + 3 + \cdots + 50$.

Step 3 — Sum. $\frac{50 \times 51}{2} = 1275$.

Takeaway: Differentiating a geometric-style sum at $x = 1$ produces an arithmetic series — two chapters shaking hands.

Example 7: Power rule and quotient rule

Differentiate: (i) $6x^{100} - x^{55} + x$; (ii) $\frac{x+1}{x}$.

Solution:

Step 1 — (i) Term-by-term. $6 \cdot 100x^{99} - 55x^{54} + 1 = 600x^{99} - 55x^{54} + 1$.

Step 2 — (ii) Quotient rule. $u = x+1$, $v = x$: $\frac{u'v - uv'}{v^2} = \frac{x - (x+1)}{x^2}$.

Step 3 — (ii) Simplify. $-\frac{1}{x^2}$. (Faster: $\frac{x+1}{x} = 1 + \frac{1}{x}$, whose derivative is $-\frac{1}{x^2}$ at sight.)

Takeaway: Rewrite before you differentiate — splitting $\frac{x+1}{x}$ avoids the quotient rule entirely.

Example 8: sin x and tan x from first principles

Compute the derivatives of (i) $\sin x$ and (ii) $\tan x$.

Solution:

Step 1 — (i) Factor the difference of sines. $\frac{\sin(x+h) - \sin x}{h} = \frac{2\cos\left(x + \frac{h}{2}\right)\sin\frac{h}{2}}{h}$.

Step 2 — (i) Split and limit. $\cos\left(x + \frac{h}{2}\right) \cdot \frac{\sin(h/2)}{h/2} \to \cos x \cdot 1 = \cos x$.

Step 3 — (ii) Combine tangents. $\frac{\tan(x+h) - \tan x}{h} = \frac{\sin h}{h\cos(x+h)\cos x} \to \frac{1}{\cos^2 x} = \sec^2 x$.

Takeaway: Both proofs run through the SAME two tools: a sum-to-product identity and $\frac{\sin\theta}{\theta} \to 1$.

Example 9: Products and double angles

Differentiate: (i) $\sin^2 x$; (ii) $\sin 2x$; (iii) $\cot x$.

Solution:

Step 1 — (i) Leibnitz on $\sin x \cdot \sin x$. $\cos x\sin x + \sin x\cos x = 2\sin x\cos x = \sin 2x$.

Step 2 — (ii) Write $\sin 2x = 2\sin x\cos x$ and Leibnitz again. $2(\cos^2 x - \sin^2 x) = 2\cos 2x$.

Step 3 — (iii) Quotient rule on $\frac{\cos x}{\sin x}$. $\frac{-\sin^2 x - \cos^2 x}{\sin^2 x} = -\frac{1}{\sin^2 x} = -\mathrm{cosec}^2 x$.

Takeaway: Without the chain rule (a Class 12 tool), double angles are handled by product-form rewrites — and the answers preview the chain rule's pattern.

Example 10: Algebraic workouts

Differentiate: (i) $(5x^3 + 3x - 1)(x - 1)$; (ii) $x^{-3}(5 + 3x)$; (iii) $x^5(3 - 6x^{-9})$.

Solution:

Step 1 — (i) Leibnitz. $(15x^2 + 3)(x - 1) + (5x^3 + 3x - 1)(1) = 20x^3 - 15x^2 + 6x - 4$.

Step 2 — (ii) Expand instead. $5x^{-3} + 3x^{-2}$: derivative $-15x^{-4} - 6x^{-3}$.

Step 3 — (iii) Expand again. $3x^5 - 6x^{-4}$: derivative $15x^4 + 24x^{-5}$.

Takeaway: The power rule covers negative exponents too — expanding often beats the product rule when one factor is a power of $x$.

Example 11: The trig family

Differentiate: (i) $\cos x$ from first principles; (ii) $\sec x$; (iii) $5\sec x + 4\cos x$; (iv) $3\cot x + 5\,\mathrm{cosec}\,x$.

Solution:

Step 1 — (i) Difference of cosines. $\frac{\cos(x+h) - \cos x}{h} = \frac{-2\sin\left(x + \frac{h}{2}\right)\sin\frac{h}{2}}{h} \to -\sin x$.

Step 2 — (ii) Quotient rule on $\frac{1}{\cos x}$. $\frac{0 \cdot \cos x - 1 \cdot(-\sin x)}{\cos^2 x} = \frac{\sin x}{\cos^2 x} = \sec x\tan x$.

Step 3 — (iii) and (iv) Linearity. $5\sec x\tan x - 4\sin x$; and $-3\,\mathrm{cosec}^2 x - 5\,\mathrm{cosec}\,x\cot x$.

Takeaway: All six trig derivatives trace back to $(\sin x)'$ and $(\cos x)'$ plus the quotient rule — derive once, then trust the table.

Example 12: Linear combinations

Differentiate: (i) $5\sin x - 6\cos x + 7$; (ii) $2\tan x - 7\sec x$.

Solution:

Step 1 — (i) Differentiate each term. $5\cos x - 6(-\sin x) + 0 = 5\cos x + 6\sin x$.

Step 2 — (ii) Same. $2\sec^2 x - 7\sec x\tan x$.

Takeaway: Watch the ONE sign flip in the family — $(\cos x)' = -\sin x$ turns the $-6\cos x$ term positive; the constant dies silently.