Differentiation

Limits and Continuity

Intuitive Notion of a Limit

The limit of a function $f(x)$ as $x$ approaches $a$ is the value that $f(x)$ approaches, regardless of whether $f(a)$ is defined:

\lim_\{x \to a\} f(x) = L

This means that as $x$ gets arbitrarily close to $a$ , $f(x)$ gets arbitrarily close to $L$ .

Left-Hand and Right-Hand Limits

A two-sided limit exists if and only if both one-sided limits exist and are equal:

\lim_\{x \to a\} f(x) = L \iff \lim_\{x \to a^-\} f(x) = \lim_\{x \to a^+\} f(x) = L

Find $\displaystyle\lim_{x \to 0} \frac{|x|}{x}$ .

\lim_\{x \to 0^-\} \frac\{|x|\}\{x\} = \lim_\{x \to 0^-\} \frac\{-x\}\{x\} = -1

\lim_\{x \to 0^+\} \frac\{|x|\}\{x\} = \lim_\{x \to 0^+\} \frac\{x\}\{x\} = 1

Since the one-sided limits are not equal, the limit does not exist.

Limit	Value
$\displaystyle\lim_{x \to 0} \frac{\sin x}{x}$	$1$
$\displaystyle\lim_{x \to 0} \frac{1 - \cos x}{x}$	$0$
$\displaystyle\lim_{x \to \infty} \frac{1}{x}$	$0$
$\displaystyle\lim_{x \to \infty} \left(1 + \frac{1}{x}\right)^x$	$e$
$\displaystyle\lim_{x \to 0} \frac{e^x - 1}{x}$	$1$
$\displaystyle\lim_{x \to 0} \frac{\ln(1+x)}{x}$	$1$

The Squeeze Theorem

If $g(x) \le f(x) \le h(x)$ for all $x$ near $a$ (except possibly at $a$ ), and:

\lim_\{x \to a\} g(x) = \lim_\{x \to a\} h(x) = L

Then $\displaystyle\lim_{x \to a} f(x) = L$ .

Show that $\displaystyle\lim_{x \to 0} x^2 \sin\!\left(\frac{1}{x}\right) = 0$ .

Since $-1 \le \sin\!\left(\frac{1}{x}\right) \le 1$ We have $-x^2 \le x^2 \sin\!\left(\frac{1}{x}\right) \le x^2$ .

Both $\displaystyle\lim_{x \to 0}(-x^2) = 0$ and $\displaystyle\lim_{x \to 0} x^2 = 0$ .

By the squeeze theorem, $\displaystyle\lim_{x \to 0} x^2 \sin\!\left(\frac{1}{x}\right) = 0$ .

Continuity

A function $f$ is continuous at $x = a$ if all three conditions hold:

$f(a)$ is defined
$\displaystyle\lim_{x \to a} f(x)$ exists
$\displaystyle\lim_{x \to a} f(x) = f(a)$

When asked whether a piecewise function is continuous at a boundary point, always check that the Left-hand limit, right-hand limit, and function value all agree.

The Derivative

Definition from First Principles

The derivative of $f$ at $x = a$ is defined as:

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

This limit, when it exists, gives the instantaneous rate of change of $f$ at $x = a$ And equals the Gradient of the tangent line to the curve $y = f(x)$ at that point.

Alternative Form

F'(x) = \lim_\{\Delta x \to 0\} \frac\{f(x + \Delta x) - f(x)\}\{\Delta x\}

Differentiability Implies Continuity

If $f$ is differentiable at $x = a$ Then $f$ is continuous at $x = a$ . The converse is not true: a Function can be continuous but not differentiable (e.g., $f(x) = |x|$ at $x = 0$ ).

Differentiate $f(x) = x^2$ from first principles.

\begin`\{aligned}` F'(x) &= \lim_\{h \to 0\} \frac\{(x+h)^2 - x^2\}\{h\} \\[6pt] &= \lim_\{h \to 0\} \frac\{x^2 + 2xh + h^2 - x^2\}\{h\} \\[6pt] &= \lim_\{h \to 0\} \frac\{2xh + h^2\}\{h\} \\[6pt] &= \lim_\{h \to 0\}(2x + h) \\[6pt] &= 2x \end`\{aligned}`

Notation for Derivatives

Several notations are equivalent:

$f'(x)$ , Lagrange notation (prime notation)
$\dfrac{dy}{dx}$ , Leibniz notation
$\dfrac{d}{dx}[f(x)]$ , operator notation
$\dot{y}$ , Newton notation (time derivatives in physics)

Differentiation Rules

The Power Rule

For $n \in \mathbb{R}$ :

\frac\{d\}`\{dx}`[x^n] = nx^\{n-1\}

Constant Multiple Rule

\frac{d\}\{dx}[cf(x)] = c \cdot f'(x)

Sum and Difference Rule

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

The Product Rule

If $u = f(x)$ and $v = g(x)$ Then:

\frac\{d\}`\{dx}`[uv] = u\frac`\{dv}, \{dx}` + v\frac`\{du}, \{dx}`

Differentiate $\displaystyle f(x) = \sin(2x^2 + 1)$ .

Let $u = 2x^2 + 1$ So $y = \sin u$ .

\frac`\{dy}, \{dx}` = \cos u \cdot 4x = 4x\cos(2x^2+1)

Derivatives of Standard Functions

$f(x)$	$f'(x)$
$x^n$	$nx^{n-1}$
$e^x$	$e^x$
$a^x$	$a^x \ln a$
$\ln x$	$\dfrac{1}{x}$
$\log_a x$	$\dfrac{1}{x \ln a}$
$\sin x$	$\cos x$
$\cos x$	$-\sin x$
$\tan x$	$\sec^2 x$
$\csc x$	$-\csc x \cot x$
$\sec x$	$\sec x \tan x$
$\cot x$	$-\csc^2 x$
$\arcsin x$	$\dfrac{1}{\sqrt{1-x^2}}$
$\arccos x$	$\dfrac{-1}{\sqrt{1-x^2}}$
$\arctan x$	$\dfrac{1}{1+x^2}$

Find $\dfrac{d^2y}{dx^2}$ for $x^2 + y^2 = 25$ .

We have $\dfrac{dy}{dx} = -\dfrac{x}{y}$ .

Differentiate with respect to $x$ :

\frac{d^2y\}\{dx^2\} = -\frac{y \cdot 1 - x \cdot \\\{dy}, \{dx}\}\{y^2\}

Substitute $\dfrac{dy}{dx} = -\dfrac{x}{y}$ :

\frac\{d^2y\}\{dx^2\} = -\frac\{y - x\left(-\frac\{x\}\{y\}\right)\}\{y^2\} = -\frac\{y + \frac\{x^2\}\{y\}\}\{y^2\} = -\frac\{y^2 + x^2\}\{y^3\} = -\frac\{25\}\{y^3\}

When two or more quantities are related by an equation, their rates of change are also related.

Strategy

Identify the quantities that are changing and the rate(s) given.
Write an equation relating the quantities.
Differentiate both sides with respect to time ( $t$ ).
Substitute known values and solve for the unknown rate.

Find the equation of the tangent and normal to $y = x^3 - 3x + 2$ at $x = 1$ .

At $x = 1$ : $y = 1 - 3 + 2 = 0$ and $y' = 3x^2 - 3 = 0$ .

Since $y' = 0$ The tangent is horizontal: $y = 0$ .

The normal is vertical: $x = 1$ .

Increasing and Decreasing Functions

$f$ is increasing on an interval if $f'(x) \gt 0$ for all $x$ in that interval.
$f$ is decreasing on an interval if $f'(x) \lt 0$ for all $x$ in that interval.
$f$ is stationary at $x = a$ if $f'(a) = 0$ .

Stationary Points

A stationary point occurs where $f'(x) = 0$ . There are three types:

Type	First Derivative Test	Second Derivative Test
Local maximum	$f'$ changes from $+$ to $-$	$f''(x) \lt 0$
Local minimum	$f'$ changes from $-$ to $+$	$f''(x) \gt 0$
Point of inflection	$f'$ does not change sign	$f''(x) = 0$ (inconclusive alone)

Find and classify the stationary points of $f(x) = x^3 - 6x^2 + 9x + 1$ .

F'(x) = 3x^2 - 12x + 9 = 3(x^2 - 4x + 3) = 3(x-1)(x-3)

Setting $f'(x) = 0$ : $x = 1$ or $x = 3$ .

F''(x) = 6x - 12

At $x = 1$ : $f''(1) = -6 \lt 0$ So local maximum. $f(1) = 1 - 6 + 9 + 1 = 5$ . Point: $(1, 5)$ .

At $x = 3$ : $f''(3) = 6 \gt 0$ So local minimum. $f(3) = 27 - 54 + 27 + 1 = 1$ . Point: $(3, 1)$ .

Concavity and Points of Inflection

{/prettier-ignore/}

Use the sliders to adjust coefficients of a polynomial and observe how $f'(x)$ and $f''(x)$ Determine stationary points and concavity.

$f$ is concave up on an interval if $f''(x) \gt 0$ (the graph curves upward).
$f$ is concave down on an interval if $f''(x) \lt 0$ (the graph curves downward).
A point of inflection occurs where the concavity changes, i.e., $f''(x)$ changes sign.

Find the points of inflection of $f(x) = x^4 - 4x^3$ .

F'(x) = 4x^3 - 12x^2

F''(x) = 12x^2 - 24x = 12x(x - 2)

$f''(x) = 0$ when $x = 0$ or $x = 2$ .

Interval	Sign of $f''$	Concavity
$x \lt 0$	$+$	Concave up
$0 \lt x \lt 2$	$-$	Concave down
$x \gt 2$	$+$	Concave up

Concavity changes at both $x = 0$ and $x = 2$ So both are points of inflection.

At $x = 0$ : $f(0) = 0$ . Point: $(0, 0)$ .

At $x = 2$ : $f(2) = 16 - 32 = -16$ . Point: $(2, -16)$ .

Optimization

Optimization problems involve finding the maximum or minimum value of a quantity subject to Constraints.

Strategy

Define variables and identify the quantity to be optimised.
Write an expression for the quantity in terms of a single variable.
Differentiate and set the derivative equal to zero.
Verify that the critical point gives a maximum or minimum.
Answer the question in context.

An open-top cylindrical can is to hold $500\mathrm{ cm}^3$ of liquid. Find the dimensions that Minimise the surface area.

Volume: $V = \pi r^2 h = 500$ So $h = \dfrac{500}{\pi r^2}$ .

Surface area (no top): $A = \pi r^2 + 2\pi r h = \pi r^2 + \dfrac{1000}{r}$ .

\frac`\{dA}, \{dr}` = 2\pi r - \frac\{1000\}\{r^2\} = 0

2\pi r = \frac\{1000\}\{r^2\} \implies 2\pi r^3 = 1000 \implies r^3 = \frac\{500\}\{\pi\}

R = \left(\frac\{500\}\{\pi\}\right)^\{1/3\} \approx 5.42 \mathrm\{ cm\}

H = \frac\{500\}\{\pi \cdot \left(\frac\{500\}\{\pi\}\right)^\{2/3\}\} = \left(\frac\{500\}\{\pi\}\right)^\{1/3\} \approx 5.42 \mathrm\{ cm\}

The optimal open-top can has $h = r$ I.e., height equals radius.

L’Hopital’s Rule

If $\displaystyle\lim_{x \to a}\frac{f(x)}{g(x)}$ gives an indeterminate form $\dfrac{0}{0}$ or $\dfrac{\pm\infty}{\pm\infty}$ Then:

\lim_\{x \to a\}\frac\{f(x)\}\{g(x)\} = \lim_\{x \to a\}\frac\{f'(x)\}\{g'(x)\}

Provided the limit on the right exists.

Evaluate $\displaystyle\lim_{x \to 0}\frac{e^x - 1 - x}{x^2}$ .

Direct substitution gives $\dfrac{0}{0}$ So apply L’Hopital’s rule:

\lim_\{x \to 0\}\frac\{e^x - 1\}\{2x\}

Still $\dfrac{0}{0}$ Apply again:

\lim_\{x \to 0\}\frac\{e^x\}\{2\} = \frac\{1\}\{2\}

IB Exam-Style Questions

Question 1 (Paper 1 style)

Let $f(x) = 2x^3 - 3x^2 - 36x + 5$ .

(a) Find $f'(x)$ .

F'(x) = 6x^2 - 6x - 36

(b) Find the $x$ -coordinates of the stationary points.

6x^2 - 6x - 36 = 0 \implies x^2 - x - 6 = 0 \implies (x-3)(x+2) = 0

$x = 3$ or $x = -2$ .

(c) Determine the nature of each stationary point.

F''(x) = 12x - 6

At $x = -2$ : $f''(-2) = -30 \lt 0$ So local maximum.

At $x = 3$ : $f''(3) = 30 \gt 0$ So local minimum.

(d) Find the equation of the tangent to the curve at $x = 0$ .

$f(0) = 5$ and $f'(0) = -36$ .

Y - 5 = -36(x - 0) \implies y = -36x + 5

Question 2 (Paper 2 style)

A rectangular garden is to be constructed adjacent to a house, with the house forming one side of the rectangle. Fencing is required for the other three sides. $60\mathrm{ m}$ of fencing is Available.

(a) If the side parallel to the house has length $x\mathrm{ m}$ Show that the area is:

A = x\left(30 - \frac\{x\}\{2\}\right) = 30x - \frac\{x^2\}\{2\}

The two perpendicular sides have total length $60 - x$ So each is $\dfrac{60-x}{2} = 30 - \dfrac{x}{2}$ .

A = x\left(30 - \frac\{x\}\{2\}\right)

(b) Find the value of $x$ that maximises the area.

\frac`\{dA}, \{dx}` = 30 - x = 0 \implies x = 30

\frac\{d^2A\}\{dx^2\} = -1 \lt 0 \implies \mathrm\{maximum\}

Maximum area $= 30 \times 15 = 450\mathrm{ m}^2$ .

Question 3 (Paper 1 style)

Given $x^2 + xy + y^2 = 7$ :

(a) Find $\dfrac{dy}{dx}$ in terms of $x$ and $y$ .

2x + y + x\frac`\{dy}, \{dx}` + 2y\frac`\{dy}, \{dx}` = 0

(x + 2y)\frac`\{dy}, \{dx}` = -(2x + y)

\frac`\{dy}, \{dx}` = -\frac\{2x+y\}\{x+2y\}

(b) Verify that the point $(1, 2)$ lies on the curve and find the gradient at that point.

$1 + 2 + 4 = 7$ . Yes.

\frac`\{dy}, \{dx}`\bigg|_\{(1,2)\} = -\frac\{2+2\}\{1+4\} = -\frac\{4\}\{5\}

Question 4 (Paper 2 style)

The curve $C$ has equation $y = x e^{-x}$ .

(a) Find $\dfrac{dy}{dx}$ .

Using the product rule with $u = x$ and $v = e^{-x}$ :

\frac`\{dy}, \{dx}` = 1 \cdot e^\{-x\} + x \cdot (-e^\{-x\}) = e^\{-x\}(1 - x)

(b) Find the coordinates of the stationary point and determine its nature.

$\dfrac{dy}{dx} = 0 \implies 1 - x = 0 \implies x = 1$ .

$y = e^{-1} = \dfrac{1}{e}$ .

$\dfrac{d^2y}{dx^2} = -e^{-x}(1-x) + e^{-x}(-1) = e^{-x}(x - 2)$ .

At $x = 1$ : $\dfrac{d^2y}{dx^2} = e^{-1}(-1) \lt 0$ So local maximum at $\left(1, \dfrac{1}{e}\right)$ .

(c) Find the point of inflection.

$\dfrac{d^2y}{dx^2} = 0 \implies x = 2$ .

At $x = 2$ : $y = 2e^{-2} = \dfrac{2}{e^2}$ .

Since $\dfrac{d^2y}{dx^2}$ changes sign at $x = 2$ This is a point of inflection at $\left(2, \dfrac{2}{e^2}\right)$ .

(d) Find the equation of the tangent at $x = 0$ .

$y(0) = 0$ and $y'(0) = e^0(1) = 1$ .

Y = x

Question 5 (Paper 1 style)

The radius of a circle is increasing at a rate of $0.5\mathrm{ cm/s}$ . Find the rate of change of the area when the radius is $4\mathrm{ cm}$ .

A = \pi r^2

\frac`\{dA}, \{dt}` = 2\pi r \frac`\{dr}, \{dt}` = 2\pi(4)(0.5) = 4\pi \mathrm\{ cm\}^2\mathrm\{/s\}

Question 6 (Paper 2 style)

Let $f(x) = \dfrac{x}{x^2 + 1}$ .

(a) Find $f'(x)$ .

Using the quotient rule with $u = x$$v = x^2 + 1$ :

F'(x) = \frac\{(x^2+1)(1) - x(2x)\}\{(x^2+1)^2\} = \frac\{1 - x^2\}\{(x^2+1)^2\}

(b) Find the range of values of $x$ for which $f$ is increasing.

$f'(x) \gt 0 \implies 1 - x^2 \gt 0 \implies x^2 \lt 1 \implies -1 \lt x \lt 1$ .

(c) Find the stationary points.

$f'(x) = 0 \implies x = \pm 1$ .

At $x = 1$ : $f(1) = \dfrac{1}{2}$ .

At $x = -1$ : $f(-1) = -\dfrac{1}{2}$ .

Since $f'$ changes from $+$ to $-$ at $x = 1$ : local maximum at $\left(1, \dfrac{1}{2}\right)$ .

Since $f'$ changes from $-$ to $+$ at $x = -1$ : local minimum at $\left(-1, -\dfrac{1}{2}\right)$ .

(d) Find the equations of the asymptotes.

As $x \to \pm\infty$ : $f(x) \to 0$ So $y = 0$ is a horizontal asymptote.

The denominator $x^2 + 1 \neq 0$ for all real $x$ So there are no vertical asymptotes.

Summary of Key Results

Concept	Formula
First principles	$\displaystyle f'(x) = \lim_{h \to 0}\frac{f(x+h)-f(x)}{h}$
Power rule	$\dfrac{d}{dx}[x^n] = nx^{n-1}$
Product rule	$(uv)' = u'v + uv'$
Quotient rule	$\left(\dfrac{u}{v}\right)' = \dfrac{u'v - uv'}{v^2}$
Chain rule	$\dfrac{dy}{dx} = \dfrac{dy}{du} \cdot \dfrac{du}{dx}$
Tangent at $(a, f(a))$	$y - f(a) = f'(a)(x - a)$
Normal gradient	$-\dfrac{1}{f'(a)}$
Increasing	$f'(x) \gt 0$
Decreasing	$f'(x) \lt 0$
Local max	$f'(a) = 0$$f''(a) \lt 0$
Local min	$f'(a) = 0$$f''(a) \gt 0$

Diagnostic Test Ready to test your understanding of Differentiation? The contains the hardest questions

within the IB specification for this topic, each with a full worked solution.

Unit tests probe edge cases and common misconceptions. Integration tests combine Differentiation with other IB mathematics topics to test synthesis under exam conditions.

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Intuition

Differentiation asks a deceptively simple question: how fast is something changing right now? The derivative is the slope of the tangent line, which approximates the function locally as a straight line. The chain rule is the composition of rates, if a small change in x produces a change in u, and a change in u produces a change in y, then the total rate is the product of these two rates. The product rule captures how two growing quantities interact: when both x and y increase, the area of their product grows from two directions simultaneously. Implicit differentiation extends these ideas to curves defined implicitly, where y is entangled with x in ways that resist explicit solution.

Common Pitfalls

Forgetting to use the chain rule for composite functions. Identify the inner function first.
Confusing the derivative of $\ln(x)$ with the derivative of $\log_a(x)$ . The latter requires the change of base formula.
Misreading the question, particularly with ‘hence’ vs ‘hence or otherwise’. The former requires using previous work.

Summary

Chain rule: $\frac{d}{dx}[f(g(x))] = f'(g(x)) \cdot g'(x)$
Product rule: $(uv)' = u'v + uv'$ ; quotient rule: $\left(\frac{u}{v}\right)' = \frac{u'v - uv'}{v^2}$
Implicit differentiation: apply chain rule to terms with $y$
Related rates: differentiate with respect to $t$

Cross-References

Topic	Site	Link
[Differentiation]	A-Level	View
[Differentiation]	IB	View

Worked Examples

Worked examples demonstrating the application of key concepts are covered in the detailed sub-pages linked above.