MathsISC Class 11 · Limits and Derivatives

Derivative from First Principles, the Product and Quotient Rules and Rates of Change

Updated 17 October 2026 · Fact-checked

The derivative f′(x) = lim_{h→0} [f(x + h) − f(x)]/h is the slope of the tangent and the instantaneous rate of change. Standard results: d(xⁿ)/dx = nxⁿ⁻¹, d(sin x)/dx = cos x, d(cos x)/dx = −sin x. Rules: sum, product (uv)′ = u′v + uv′ and quotient (u/v)′ = (u′v − uv′)/v².

Understand Differentiation from First Principles and Rules of Differentiation

Definition. $f' (x) = lim_{h \to 0} \frac{f ( x + h ) - f ( x )}{h}$ , if the limit exists. Geometrically it is the slope of the tangent to y = f(x) at x; physically it is the rate of change of y with respect to x. The slope of the tangent at x = a is f′(a), and the equation of the tangent is $y - f (a) = f' (a) (x - a)$ .

From first principles. For $f (x) = x^{2}$ : $\frac{( x + h ) ^{2} - x ^{2}}{h} = 2 x + h \to 2 x$ . For $f (x) = sin x$ : $\frac{s i n ( x + h ) - s i n x}{h} = cos x \cdot \frac{s i n h}{h} - sin x \cdot \frac{1 - c o s h}{h} \to cos x$ .

Standard results. $\frac{d}{d x} (x^{n}) = n x^{n - 1}$ ; $\frac{d}{d x} (c) = 0$ ; $\frac{d}{d x} (sin x) = cos x$ ; $\frac{d}{d x} (cos x) = - sin x$ ; $\frac{d}{d x} (tan x) = sec^{2} x$ ; $\frac{d}{d x} (e^{x}) = e^{x}$ .

Rules. (cu)′ = cu′; (u ± v)′ = u′ ± v′; product (uv)′ = u′v + uv′; quotient (u/v)′ = (u′v − uv′)/v². Differentiate each term separately in a polynomial.

Uses. Horizontal tangent: set f′(x) = 0. Increasing function: f′(x) > 0. Rate of change: if the radius of a circle changes at rate $\frac{d r}{d t}$ then $\frac{d A}{d t} = 2 π r \frac{d r}{d t}$ .

Differentiation from First Principles and Rules of Differentiation: formulas to remember

First principles: $f′(x) = \lim_{h→0} \frac{f(x + h) - f(x)}{h}$; Slope of the tangent.
Standard results: $(x^n)′ = nx^{n-1}$ | $(\sin x)′ = \cos x$ | $(\cos x)′ = -\sin x$; x in radians.
Product and quotient: $(uv)′ = u′v + uv′$ | $(\frac{u}{v})′ = \frac{u′v - uv′}{v^2}$; Do not differentiate term by term.
Tangent: $$y - f(a) = f'(a)(x - a)$$; Slope f′(a).

Differentiation from First Principles and Rules of Differentiation in real life

Each concept, met in an everyday situation. Read the situation, try the problem, then check the solution. The one-line hook is what to remember.

Growth of an oil spill

Suppose an oil spill is a circle whose radius grows at 2 cm per second.

Find the rate at which its area increases when the radius is 5 cm.

Try it first, then show the solutionSolution

$A = π r^{2}$ , so $\frac{d A}{d t} = 2 π r \frac{d r}{d t}$ .
With $r = 5$ and $\frac{d r}{d t} = 2$ : $\frac{d A}{d t} = 2 π \times 5 \times 2 = 20 π$ .
$\approx 62.8$ cm² per second.

Answer: 20π ≈ 62.8 cm² s⁻¹.

Memory hook: Multiply by dr/dt (chain rule).

Tricks to remember Differentiation from First Principles and Rules of Differentiation

Short memory aids. Say each one aloud once, then use it on a question the same day.

Bring the power down
Multiply by the power and reduce the power by 1.
Use it when: Differentiating xⁿ.
Product: first dash, second plus second dash first
Keep the pattern u′v + uv′.
Use it when: Product rule.

Differentiation from First Principles and Rules of Differentiation: solved examples

Ordered from board level to HOTS. Try each one before you open the solution.

Example 1: Derivative of a power · 1 mark · MCQBoard level

The derivative of $x^{5}$ with respect to x is:
(a) $x^{4}$
(b) $5 x^{4}$
(c) $5 x^{5}$
(d) $4 x^{5}$

Try it first, then show the solutionSolution

$n x^{n - 1}$ with n = 5.

Answer: (b) $5 x^{4}$

Example 2: First principles · 3 marks · Short answerBoard level

Find the derivative of $x^{2}$ from first principles and its value at x = 3.

Try it first, then show the solutionSolution

$f' (x) = lim_{h \to 0} \frac{( x + h ) ^{2} - x ^{2}}{h}$ . (1 mark)
$= lim (2 x + h) = 2 x$ . (1 mark)
At x = 3, $f' (3) = 6$ . (1 mark)

Answer: 2x; 6.

Example 3: Product and quotient rules · 4 marks · Short answerHarder board level

Differentiate (i) $y = (x^{2} + 1) (x^{3} - 2 x)$ and (ii) $y = \frac{x + 1}{x - 1}$ .

Try it first, then show the solutionSolution

(i) $y' = 2 x (x^{3} - 2 x) + (x^{2} + 1) (3 x^{2} - 2)$ . (1 mark)
$= 2 x^{4} - 4 x^{2} + 3 x^{4} - 2 x^{2} + 3 x^{2} - 2 = 5 x^{4} - 3 x^{2} - 2$ . (1.5 marks)
(ii) $y' = \frac{( x - 1 ) - ( x + 1 )}{( x - 1 ) ^{2}} = - \frac{2}{( x - 1 ) ^{2}}$ . (1.5 marks)

Answer: 5x⁴ − 3x² − 2; −2/(x − 1)².

Example 4: Tangent to a curve · 4 marks · Short answerHarder board level

For $y = x^{3} - 3 x + 2$ find the slope of the tangent at x = 2, the equation of that tangent and the points where the tangent is horizontal.

Try it first, then show the solutionSolution

$y' = 3 x^{2} - 3$ ; at x = 2, slope $= 9$ . (1 mark)
At x = 2, $y = 8 - 6 + 2 = 4$ ; tangent $y - 4 = 9 (x - 2)$ , i.e. $9 x - y - 14 = 0$ . (1.5 marks)
Horizontal: $3 x^{2} - 3 = 0$ , so $x = \pm 1$ ; points (1, 0) and (−1, 4). (1.5 marks)

Answer: Slope 9; 9x − y − 14 = 0; (1, 0) and (−1, 4).

Example 5: Derivative of sin x and a rate of change · 5 marks · Long answerHOTS / competency

Using first principles, derive the derivative of $sin x$ . A circular ripple has radius growing at 3 cm s⁻¹; find the rate of increase of its circumference and of its area when r = 10 cm.

Try it first, then show the solutionSolution

$\frac{s i n ( x + h ) - s i n x}{h} = cos x \cdot \frac{s i n h}{h} - sin x \cdot \frac{1 - c o s h}{h}$ . (1.5 marks)
As h → 0, $\frac{s i n h}{h} \to 1$ and $\frac{1 - c o s h}{h} \to 0$ , so $(sin x)' = cos x$ . (1 mark)
Circumference $C = 2 π r$ : $\frac{d C}{d t} = 2 π \times 3 = 6 π \approx 18.85$ cm s⁻¹. (1 mark)
Area: $\frac{d A}{d t} = 2 π r \frac{d r}{d t} = 2 π \times 10 \times 3 = 60 π \approx 188.5$ cm² s⁻¹. (1.5 marks)

Answer: cos x; 6π cm s⁻¹ and 60π cm² s⁻¹.

Where students lose marks in Differentiation from First Principles and Rules of Differentiation

The slips examiners see most often, and what to write instead.

Mistake: Differentiating a product term by term
Writing (uv)′ = u′v′.
Do this: Use (uv)′ = u′v + uv′.
Usually costs: 2 marks.
Mistake: Wrong sign in the quotient rule
Writing uv′ − u′v.
Do this: Numerator is u′v − uv′.
Usually costs: 1 to 2 marks.
Mistake: Forgetting the h in first principles
Not dividing by h or not taking the limit.
Do this: State f′(x) = lim [f(x+h) − f(x)]/h and simplify.
Usually costs: 1 to 2 marks.

Exam tips

Write the definition for questions asking first principles.
Label u and v clearly.
Check the sign of the slope with a sketch.
Give units in rate-of-change problems.

Practice questions on Differentiation from First Principles and Rules of Differentiation

Write each answer on paper first. Open the hint only if you are stuck, and the answer only when you have finished.

Q1 · 1 mark · MCQBoard level
The derivative of $cos x$ is:
(a) $sin x$
(b) $- sin x$
(c) $cos x$
(d) $- cos x$
Hint
Standard result.
Answer
(b) $- sin x$
Q2 · 2 marks · Short answerBoard level
Differentiate $y = 3 x^{4} - 2 x^{2} + 5 x - 7$ .
Hint
Term by term.
Answer
$y' = 12 x^{3} - 4 x + 5$ .
Q3 · 3 marks · Short answerHarder board level
Find the derivative of $y = x^{2} sin x$ .
Hint
Product rule.
Answer
$y' = 2 x sin x + x^{2} cos x$ .
Q4 · 3 marks · Short answerHarder board level
Differentiate $y = \frac{s i n x}{x}$ and evaluate at $x = π$ .
Hint
Quotient rule.
Answer
$y' = \frac{x c o s x - s i n x}{x ^{2}}$ ; at $x = π$ : $\frac{- π - 0}{π ^{2}} = - \frac{1}{π}$ .
Q5 · 3 marks · Short answerHarder board level
Find the equation of the tangent to $y = x^{2} + 2 x$ at the point (1, 3).
Hint
Slope = y′ at x = 1.
Answer
$y' = 2 x + 2 = 4$ ; $y - 3 = 4 (x - 1)$ , i.e. $4 x - y - 1 = 0$ .
Q6 · 4 marks · Case-basedHOTS / competency
Case study: Suppose the distance of a car is $s = 2 t^{3} - 9 t^{2} + 12 t$ metres at time t seconds.
(i) Find the velocity. (1 mark)
(ii) Find the velocity at t = 3. (1 mark)
(iii) Find the times when the car is momentarily at rest. (1 mark)
(iv) Find the acceleration at t = 3. (1 mark)
Hint
v = ds/dt, a = dv/dt.
Answer
(i) $v = 6 t^{2} - 18 t + 12$ . (ii) $54 - 54 + 12 = 12$ m s⁻¹. (iii) $6 (t - 1) (t - 2) = 0$ : t = 1 s and t = 2 s. (iv) $a = 12 t - 18 = 18$ m s⁻².

Test your recall

Answer in your head or on paper first, then open the card. Pulling an answer out of memory makes it stick far better than re-reading.

1.Derivative of xⁿ?Show answer
n xⁿ⁻¹.
2.Derivative of cos x?Show answer
−sin x.
3.Quotient rule?Show answer
(u′v − uv′)/v².
4.Slope of the tangent?Show answer
f′(a).

Differentiation from First Principles and Rules of Differentiation: frequently asked questions

What does the derivative measure?

The instantaneous rate of change, and the slope of the tangent line.

Why is the derivative of a constant zero?

A constant does not change, so the rate of change is zero.

Can a function be continuous but not differentiable?

Yes, for example |x| at x = 0 has a corner, so there is no unique tangent.