Module 40 of 46Lesson 1 of 8 · HL · AA 5.12 — First principles; continuity and differentiability; higher derivatives

Course map · Topic 5: Calculus

5.12 First principles; continuity and differentiability; higher derivatives

HL only11 worked questions

HL asks where derivatives come from: the limit of a difference quotient. You must differentiate from first principles, decide whether a function is continuous and differentiable at a point (especially where a piecewise rule switches), and work with higher derivatives and nth-derivative patterns.

✏️ Start with the questions. Try each one on paper, then open the solution and compare.

Stuck? Open the notes ↓

✏️ Practice questions

Try each question on paper first. Then open the solution and compare every step.

Question 1●○○ Warm-up✍️ By hand · Paper 1 · no calculator[5 marks]

Let

f(x)=x^{2}+3x

. Use differentiation from first principles to show that

f'(x)=2x+3

Show the full step-by-step solution

Step 1🧠 ThinkGradient of a tiny chord

\dfrac{f(x+h)-f(x)}{h}

is the gradient of the chord from

x

x+h

. As

h

shrinks to 0 the chord becomes the tangent. Work out

f(x+h)

carefully first.

Step 2✍️ By handSubstitute, expand, cancel

f'(x)=\displaystyle\lim_{h\to0}\dfrac{(x+h)^{2}+3(x+h)-(x^{2}+3x)}{h}

=\displaystyle\lim_{h\to0}\dfrac{x^{2}+2xh+h^{2}+3x+3h-x^{2}-3x}{h}

=\displaystyle\lim_{h\to0}\dfrac{2xh+h^{2}+3h}{h}=\lim_{h\to0}(2x+h+3)

M1A1

=2x+3

Final answer

f'(x)=2x+3

(shown).

Question 2●○○ Warm-up✍️ By hand · Paper 1 · no calculator[5 marks]

Differentiate

f(x)=x^{3}

from first principles.

Show the full step-by-step solution

Step 1🧠 ThinkExpand (x + h)³ with the binomial theorem

(x+h)^{3}=x^{3}+3x^{2}h+3xh^{2}+h^{3}

. The

x^{3}

cancels, and every term left has an

h

to cancel.

Step 2✍️ By handFirst principles

f'(x)=\displaystyle\lim_{h\to0}\dfrac{(x+h)^{3}-x^{3}}{h}

=\displaystyle\lim_{h\to0}\dfrac{3x^{2}h+3xh^{2}+h^{3}}{h}

=\displaystyle\lim_{h\to0}\left(3x^{2}+3xh+h^{2}\right)

=3x^{2}

A1A1

Final answer

f'(x)=3x^{2}

Question 3●○○ Warm-up✍️ By hand · Paper 1 · no calculator[4 marks]

Let

f(x)=x^{5}-2x^{3}+\mathrm{e}^{2x}

. Find

f'''(x)

and the value of

f^{(4)}(0)

Show the full step-by-step solution

Step 1🧠 ThinkDifferentiate again and again

Each dash is one more differentiation. Note

\mathrm{e}^{2x}

gains a factor 2 every time.

Step 2✍️ By handFour derivatives

f'(x)=5x^{4}-6x^{2}+2\mathrm{e}^{2x}

f''(x)=20x^{3}-12x+4\mathrm{e}^{2x}

f'''(x)=60x^{2}-12+8\mathrm{e}^{2x}

f^{(4)}(x)=120x+16\mathrm{e}^{2x}\Rightarrow f^{(4)}(0)=16

M1A1

Final answer

f'''(x)=60x^{2}-12+8\mathrm{e}^{2x}

;

f^{(4)}(0)=16

Question 4●●○ Standard✍️ By hand · Paper 1 · no calculator[5 marks]

Use first principles to find the derivative of

f(x)=\dfrac{1}{x}

x\neq0

Show the full step-by-step solution

Step 1🧠 ThinkCombine the two fractions first

The top

\dfrac{1}{x+h}-\dfrac{1}{x}

needs a common denominator

x(x+h)

. After that, an

h

appears on top and cancels.

Step 2✍️ By handFirst principles with fractions

f'(x)=\displaystyle\lim_{h\to0}\dfrac{\dfrac{1}{x+h}-\dfrac{1}{x}}{h}

=\displaystyle\lim_{h\to0}\dfrac{x-(x+h)}{h\,x(x+h)}=\lim_{h\to0}\dfrac{-h}{h\,x(x+h)}

A1A1

=\displaystyle\lim_{h\to0}\dfrac{-1}{x(x+h)}

=-\dfrac{1}{x^{2}}

Final answer

f'(x)=-\dfrac{1}{x^{2}}

Question 5●●○ Standard✍️ By hand · Paper 1 · no calculator[5 marks]

The function

f

is defined by

f(x)=\begin{cases}x^{2}+k, & x\lt 1\\ 3x-1, & x\ge1\end{cases}

(a) Find the value of $k$ for which $f$ is continuous at $x=1$ .[2]
(b) For this value of $k$ , determine whether $f$ is differentiable at $x=1$ .[3]

Show the full step-by-step solution

Part (a)

Step 1🧠 ThinkThe two pieces must meet

Approach

x=1

from the left using

x^{2}+k

and from the right using

3x-1

. Continuity needs both to give the same value.

Step 2✍️ By handLeft limit = right limit

\displaystyle\lim_{x\to1^-}f(x)=1+k,\quad\lim_{x\to1^+}f(x)=f(1)=2

1+k=2\Rightarrow k=1

Part (b)

Step 1✍️ By handCompare the gradients at the join

left:

\dfrac{\mathrm{d}}{\mathrm{d}x}(x^{2}+1)=2x=2

x=1

right:

\dfrac{\mathrm{d}}{\mathrm{d}x}(3x-1)=3

2\neq3

: the graph has a corner, so

f

is not differentiable at

x=1

Final answer (a)

k=1

(b) not differentiable (left gradient 2, right gradient 3).

Question 6●●○ Standard✍️ By hand · Paper 1 · no calculator[5 marks]

The function

g(x)=\begin{cases}ax^{2}+b, & x\le2\\ 4x-1, & x\gt 2\end{cases}

is differentiable for all

x

. Find

a

and

b

Show the full step-by-step solution

Step 1🧠 ThinkTwo conditions at the join

Differentiable means smooth: the pieces must meet (continuity) and have the same gradient at

x=2

. That gives two equations for two unknowns.

Step 2✍️ By handSame gradient

2ax=4

x=2\Rightarrow4a=4\Rightarrow a=1

M1A1

Step 3✍️ By handSame value

4a+b=4(2)-1=7

4+b=7\Rightarrow b=3

A1A1

Final answer

a=1,\ b=3

Question 7●●○ Standard✍️ + 🖩 · Paper 2[6 marks]

Let

f(x)=x^{3}

(a) Use your GDC to find the value of $\dfrac{f(2+h)-f(2)}{h}$ for $h=0.1$ , $0.01$ and $0.001$ .[2]
(b) Show that $\dfrac{f(2+h)-f(2)}{h}=12+6h+h^{2}$ , and hence write down $f'(2)$ .[4]

Show the full step-by-step solution

Step 1🧠 ThinkThe calculator can only estimate

A chord gradient with a small

h

is close to the tangent gradient, but it is never exact. The algebra in (b) proves what the numbers suggest.

Part (a)

Step 1🖩 On the GDCType the difference quotient three times

((2+0.1)^3−8)/0.1›enter💻 Enter

Press ▲ to copy the line and change only

h

the values approach 12

Part (b)

Step 1✍️ By handExpand

(2+h)^{3}=8+12h+6h^{2}+h^{3}

M1A1

\dfrac{12h+6h^{2}+h^{3}}{h}=12+6h+h^{2}

f'(2)=\displaystyle\lim_{h\to0}(12+6h+h^{2})=12

✅ Check it

h=0.1

12+0.6+0.01=12.61

✔ — matches the screen.

Final answer (a)

12.61,\ 12.0601,\ 12.006

(b) shown;

f'(2)=12

Question 8●●○ Standard✍️ + 🖩 · Paper 2[7 marks]

Let

f(x)=\mathrm{e}^{x}\sin x

(a) Show that $f''(x)=2\mathrm{e}^{x}\cos x$ .[3]
(b) Use your GDC to verify the value of $f''(1)$ .[1]
(c) Show that $f^{(4)}(x)=-4f(x)$ .[3]

Show the full step-by-step solution

Part (a)

Step 1✍️ By handProduct rule twice

f'(x)=\mathrm{e}^{x}\sin x+\mathrm{e}^{x}\cos x

M1A1

f''(x)=(\mathrm{e}^{x}\sin x+\mathrm{e}^{x}\cos x)+(\mathrm{e}^{x}\cos x-\mathrm{e}^{x}\sin x)=2\mathrm{e}^{x}\cos x

Part (b)

Step 1🖩 On the GDC2nd Derivative option (RAD mode)

menu💻 click menu on keypad›4: Calculus›1: Numerical Derivative at a Point…›x›tab💻 Tab›1›tab💻 Tab›▼💻 ↓ key›enter💻 Enter

Choose 2nd Derivative in the dialog, then type the function in the template.

both 2.93739 ✔

Part (c)

Step 1✍️ By handTwo more derivatives

f'''(x)=2\mathrm{e}^{x}\cos x-2\mathrm{e}^{x}\sin x

f^{(4)}(x)=(2\mathrm{e}^{x}\cos x-2\mathrm{e}^{x}\sin x)-(2\mathrm{e}^{x}\sin x+2\mathrm{e}^{x}\cos x)

=-4\mathrm{e}^{x}\sin x=-4f(x)

Final answer (a) shown (b)

f''(1)=2\mathrm{e}\cos1\approx2.94

Question 9●●● Exam-level✍️ + 🖩 · Paper 2[7 marks]

A cycle path in Gaborone is designed along the curve

y=\begin{cases}ax^{2}+bx, & 0\le x\le2\\ \ln(x-1)+c, & x\gt 2\end{cases}

(units in hundreds of metres). The path passes through

(1,\ 0.5)

, and the two sections must join smoothly (the function is differentiable at

x=2

(a) Write down two equations in $a$ and $b$ .[3]
(b) Solve them to find $a$ and $b$ .[2]
(c) Find the exact value of $c$ .[2]

Show the full step-by-step solution

Part (a)

Step 1🧠 ThinkPoint + equal gradients

The point

(1,\ 0.5)

is on the first section. “Joins smoothly” at

x=2

means the gradients from both sides are equal. The gradient of

\ln(x-1)

x=2

\dfrac{1}{2-1}=1

Step 2✍️ By handThe two equations

a+b=0.5

2ax+b=\dfrac{1}{x-1}

x=2

4a+b=1

M1A1

Part (b)

Step 1🖩 On the GDCsimult(

simult([1,1;4,1],[0.5;1])›enter💻 Enter

📍 Where to find it on the TI-Nspire

simult(is inmenu💻 click menu on keypad›7: Matrix & Vector›6: Simultaneous

a =

\dfrac{1}{6}

, b =

\dfrac{1}{3}

Step 2✍️ By handExact values

3a=0.5\Rightarrow a=\dfrac16,\quad b=\dfrac13

A1A1

Part (c)

Step 1✍️ By handContinuity at x = 2

\ln(1)+c=4a+2b=\dfrac46+\dfrac23

c=\dfrac43

Step 2🖩 On the GDCSee the smooth join

no jump and no corner at x = 2

Final answer (a)

a+b=0.5,\ 4a+b=1

(b)

a=\dfrac16,\ b=\dfrac13

(c)

c=\dfrac43

Question 10●●● Exam-level✍️ By hand · Paper 1 · no calculator[8 marks]

Let

y=x\mathrm{e}^{x}

(a) Find $\dfrac{\mathrm{d}y}{\mathrm{d}x}$ , $\dfrac{\mathrm{d}^{2}y}{\mathrm{d}x^{2}}$ and $\dfrac{\mathrm{d}^{3}y}{\mathrm{d}x^{3}}$ .[3]
(b) Use mathematical induction to prove that $\dfrac{\mathrm{d}^{n}y}{\mathrm{d}x^{n}}=(x+n)\mathrm{e}^{x}$ for all $n\in\mathbb{Z}^{+}$ .[5]

Show the full step-by-step solution

Part (a)

Step 1✍️ By handProduct rule each time

y'=\mathrm{e}^{x}+x\mathrm{e}^{x}=(x+1)\mathrm{e}^{x}

y''=\mathrm{e}^{x}+(x+1)\mathrm{e}^{x}=(x+2)\mathrm{e}^{x}

y'''=(x+3)\mathrm{e}^{x}

Part (b)

Step 1🧠 ThinkInduction: start, assume, step, conclude

Show it is true for

n=1

. Assume it is true for

n=k

. Differentiate once more to show it is then true for

n=k+1

. Finish with the standard concluding sentence.

Step 2✍️ By handProof

n=1

y'=(x+1)\mathrm{e}^{x}

✔ trueR1

Assume

\dfrac{\mathrm{d}^{k}y}{\mathrm{d}x^{k}}=(x+k)\mathrm{e}^{x}

\dfrac{\mathrm{d}^{k+1}y}{\mathrm{d}x^{k+1}}=\dfrac{\mathrm{d}}{\mathrm{d}x}\big[(x+k)\mathrm{e}^{x}\big]=\mathrm{e}^{x}+(x+k)\mathrm{e}^{x}

=\big(x+(k+1)\big)\mathrm{e}^{x}

True for

n=1

, and true for

n=k

implies true for

n=k+1

; so true for all

n\in\mathbb{Z}^{+}

by induction.R1

Final answer (a)

(x+1)\mathrm{e}^{x},\ (x+2)\mathrm{e}^{x},\ (x+3)\mathrm{e}^{x}

(b) proved by induction.

Question 11●●● Exam-level✍️ + 🖩 · Paper 3[14 marks]

Investigation: the derivative of $a^{x}$ . Let

f(x)=a^{x}

, where

a\gt 0

(a) Show from first principles that $f'(x)=a^{x}\times\displaystyle\lim_{h\to0}\dfrac{a^{h}-1}{h}$ .[3]
(b) Let $k_a=\displaystyle\lim_{h\to0}\dfrac{a^{h}-1}{h}$ . Use your GDC to evaluate $\dfrac{a^{h}-1}{h}$ for $h=0.1,\ 0.01,\ 0.001,\ 0.0001$ when $a=2$ and when $a=3$ . Hence estimate $k_2$ and $k_3$ to 3 significant figures.[3]
(c) There is one value of $a$ for which $k_a=1$ . By solving $\dfrac{a^{0.0001}-1}{0.0001}=1$ , estimate this value of $a$ .[2]
(d) Compare $k_2$ and $k_3$ with $\ln2$ and $\ln3$ . Write down a conjecture for $k_a$ .[2]
(e) By writing $a^{h}=\mathrm{e}^{h\ln a}$ , and given that $\displaystyle\lim_{t\to0}\dfrac{\mathrm{e}^{t}-1}{t}=1$ , prove your conjecture and hence state $\dfrac{\mathrm{d}}{\mathrm{d}x}(a^{x})$ .[4]

Show the full step-by-step solution

Part (a)

Step 1✍️ By handFactor out a^x

f'(x)=\displaystyle\lim_{h\to0}\dfrac{a^{x+h}-a^{x}}{h}

=\displaystyle\lim_{h\to0}\dfrac{a^{x}(a^{h}-1)}{h}

a^{x}

does not depend on

h

, so

f'(x)=a^{x}\displaystyle\lim_{h\to0}\dfrac{a^{h}-1}{h}

Part (b)

Step 1🖩 On the GDCTable of difference quotients

(2^0.1−1)/0.1›enter💻 Enter

a = 2

a = 3

Step 2✍️ By handEstimates

k_2\approx0.693,\quad k_3\approx1.10

A1A1A1

Part (c)

Step 1🖩 On the GDCnSolve for a

menu💻 click menu on keypad›3: Algebra›1: Numerical Solve›(x^0.0001−1)/0.0001=1,x›enter💻 Enter

a ≈ 2.718 — very close to e

Step 2✍️ By handRecognise the number

a\approx2.718\approx\mathrm{e}

A1A1

Part (d)

Step 1🖩 On the GDCCompare with ln

k₂ ≈ ln 2 and k₃ ≈ ln 3

Step 2✍️ By handConjecture

k_2\approx\ln2,\ k_3\approx\ln3

Conjecture:

k_a=\ln a

Part (e)

Step 1✍️ By handSubstitute t = h ln a

\dfrac{a^{h}-1}{h}=\dfrac{\mathrm{e}^{h\ln a}-1}{h}=\ln a\times\dfrac{\mathrm{e}^{h\ln a}-1}{h\ln a}

M1A1

let

t=h\ln a

; as

h\to0

t\to0

, so

k_a=\ln a\times\displaystyle\lim_{t\to0}\dfrac{\mathrm{e}^{t}-1}{t}=\ln a

\dfrac{\mathrm{d}}{\mathrm{d}x}(a^{x})=a^{x}\ln a

💡 Remember (For

a=1

\ln a=0

and the step dividing by

\ln a

is not needed:

1^{x}

is constant, derivative 0 ✔.) With

a=\mathrm{e}

k_{\mathrm{e}}=1

, which is why

\dfrac{\mathrm{d}}{\mathrm{d}x}\mathrm{e}^{x}=\mathrm{e}^{x}

Final answer (a) shown (b)

k_2\approx0.693,\ k_3\approx1.10

(c)

a\approx2.718\ (\mathrm{e})

(d)

k_a=\ln a

(e) proved;

\dfrac{\mathrm{d}}{\mathrm{d}x}a^{x}=a^{x}\ln a

🆘 Stuck on a question? Open the notes, GDC toolkit and common mistakes

💻 Using the TI-Nspire CX Student Software on a laptop

The screens in these notes are drawn exactly as the Student Software shows them. The keypad on the left of the software works like the handheld — click its keys with the mouse.
menu: click menu on the on-screen keypad (or Documents Toolbox ▸ 🔧). The menu numbers and letters are the same as in these notes.
enter = Enter · tab = Tab (next box) · esc = Esc · del = Backspace.
Powers: type ^ (Shift+6), the power, then press → to come back down before typing more. x² = ^2 then →.
Fraction: type / · √: type sqrt( · π: type pi · eˣ: type exp( · store: type :=. For EE (×10ⁿ) and i, click them on the keypad.
New page: Ctrl+I (or the Insert menu) · change page: Ctrl+← / Ctrl+→ · graph entry line: Ctrl+G · var: Ctrl+L.
Radian / Degree: click Settings in the status bar at the bottom of the window. The mode (RAD or DEG) shows in the top bar.
IB exam mode (Press-to-Test): everything in these notes works on the non-CAS TI-Nspire with the IB exam settings. In exam mode there are no dot-product, cross-product, unit-vector or eigenvalue keys and no Diff Eq graphs — the notes show the allowed way (type the components, use polyRoots on the characteristic equation, do Euler steps on a Calculator page).
If you type nDeriv(, the software shows it as centralDiff( — it is the same command.

🎯 What you must know

First principles: $f'(x)=\displaystyle\lim_{h\to0}\dfrac{f(x+h)-f(x)}{h}$ . Expand, cancel $f(x)$ , divide by $h$ , then let $h\to0$ .
Continuous at $x=a$ : $\displaystyle\lim_{x\to a^-}f(x)=\lim_{x\to a^+}f(x)=f(a)$ — no jump, no hole.
Differentiable at $x=a$ : continuous and the left and right gradients are equal — no corner (e.g. $|x|$ at 0 is continuous but not differentiable).
Differentiable ⇒ continuous, but continuous does not imply differentiable.
Piecewise function smooth at the join: (1) the two pieces give the same value; (2) the two derivatives give the same value at the join.
Higher derivatives: $f''(x),\ f'''(x),\ f^{(4)}(x),\dots,\ f^{(n)}(x)=\dfrac{\mathrm{d}^{n}y}{\mathrm{d}x^{n}}$ . Spot a pattern for $f^{(n)}$ , then prove it by induction.
GDC (Paper 2): a difference quotient $\dfrac{f(a+h)-f(a)}{h}$ for small $h$ estimates $f'(a)$ ; the d/dx and $\dfrac{d^{2}}{dx^{2}}$ templates give numerical values. No limit command in exam mode — limits are proved by hand.

Derivative from first principles

f'(x)=\displaystyle\lim_{h\to0}\dfrac{f(x+h)-f(x)}{h}

📘 In the formula booklet

Continuity at x = a

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

🧠 Learn this one

nth derivative notation

f^{(n)}(x)=\dfrac{\mathrm{d}^{n}y}{\mathrm{d}x^{n}}

🧠 Learn this one

🖩 Your GDC toolkit for this subtopic

Follow the steps on your own TI-Nspire CX II. The screens show what you should see.

Estimate f′(a): type the difference quotient (f(a+h)−f(a))/h for h = 0.1, 0.01, 0.001

((2+0.01)^3−8)/0.01›enter💻 Enter

Type ((2+0.01)^3−8)/0.01.
Press enter.

the values approach 12

Worked example: Question 7 — Type the difference quotient three times

Press ▲ to copy the line and change only

h

f′(a) or f″(a): Numerical Derivative at a Point (choose 1st or 2nd Derivative)

menu💻 click menu on keypad›4: Calculus›1: Numerical Derivative at a Point…

Start on a Calculator page (ctrl doc adds one).
Press menu.
Choose 4: Calculus.
Choose 1: Numerical Derivative at a Point….
A box opens. Fill in each field (press tab to move between fields), then press enter. The results appear on the screen.

The menu path

both 2.93739 ✔

Worked example: Question 8 — 2nd Derivative option (RAD mode)

Choose 2nd Derivative in the dialog, then type the function in the template.

Solve the two join conditions of a piecewise function

menu💻 click menu on keypad›3: Algebra›2: Solve System of Linear Equations…

Start on a Calculator page (ctrl doc adds one).
Press menu.
Choose 3: Algebra.
Choose 2: Solve System of Linear Equations….
A box opens. Fill in each field (press tab to move between fields), then press enter. The results appear on the screen.

The menu path

⚠️ Where students lose marks

Letting $h\to0$ before dividing by $h$ — you get $\dfrac00$ . Simplify first, cancel $h$ , then take the limit.
Writing “ $\lim$ ” on some lines and not others, or dropping it before the end. Keep $\displaystyle\lim_{h\to0}$ on every line until $h$ is gone.
Checking only that the pieces join (continuity) and claiming differentiability. You also need equal gradients.
Guessing an nth-derivative pattern from $f'$ and $f''$ only. Work out at least three derivatives before conjecturing.

IB DP Mathematics: Analysis and Approaches (AA) — SL and HL HL · AA 5.12 — First principles; continuity and differentiability; higher derivatives

IB DP Mathematics: Analysis and Approaches (AA) — SL and HL

5.12 First principles; continuity and differentiability; higher derivatives

✏️ Practice questions

🎯 What you must know

🖩 Your GDC toolkit for this subtopic

⚠️ Where students lose marks