HL · Further calculus 1 · Continuity, differentiability & first principles

Module 37 of 46Lesson 1 of 8 · HL · Further calculus 1 · Continuity, differentiability & first principles

First principles, limits, implicit differentiation & further derivatives · Essential understanding 1

Continuity, differentiability & differentiation from first principles

IB AA AHL 5.12with Prof. Kitso

💻 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.

The big idea: Every derivative rule you know comes from one idea: the gradient of a chord that shrinks to the gradient of a tangent. Written as a limit:

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

.
In this lesson you learn to use that limit directly, and to decide when a function is continuous and when it is differentiable. They are not the same thing.

By the end of this lesson you can:

decide whether a function is continuous at a point by comparing left limit, right limit and function value
decide whether a function is differentiable at a point, and explain why differentiable means continuous but not the other way round
differentiate from first principles for polynomials, $\dfrac{1}{x}$ and $\sqrt{x}$ type functions, showing every line
use the notation $f''(x)$ , $f'''(x)$ , $\dfrac{d^2y}{dx^2}$ and $f^{(n)}(x)$ for higher derivatives
use the word converges correctly for a limit that settles on one finite value

🧑🏾 Imagine this…

Stand on a smooth hill and look at the ground just in front of your feet. It looks flat. The closer you look, the flatter it gets. That is differentiability: zoom in far enough and the curve looks like a straight line, and the gradient of that line is the derivative. Now stand at the tip of a roof. However far you zoom, you still see a sharp corner. The roof is continuous (no gap in the surface) but not differentiable at the tip.

Limit

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

means

f(x)

gets as close as you like to

L

when

x

is close enough to

a

(from either side). The function does not even have to be defined at

a

Converges

A limit converges when it settles on one finite number. If the values grow without bound, or jump between two numbers, the limit diverges (it does not exist).

Continuous at x = a

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

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

f(a)

. Left limit, right limit and the actual value all agree: you can draw through

x = a

without lifting your pen.

Differentiable at x = a

The gradient limit

\lim_{h \to 0} \dfrac{f(a+h) - f(a)}{h}

exists, so the limit from the left (

h < 0

) equals the limit from the right (

h > 0

Higher derivatives

Differentiate again:

f''(x) = \dfrac{d^2 y}{dx^2}

, then

f'''(x) = \dfrac{d^3 y}{dx^3}

, and in general

f^{(n)}(x)

for the

n

th derivative.

★ Differentiation from first principles

$f'(x) = \lim_{h \to 0} \dfrac{f(x+h) - f(x)}{h}$ . The fraction is the gradient of the chord from $x$ to $x + h$ .
Method: (1) write $f(x+h)$ and expand; (2) form $f(x+h) - f(x)$ , where the old terms cancel; (3) divide by $h$ , so every remaining term has a factor $h$ cancelled; (4) let $h \to 0$ .
For $\sqrt{\ }$ functions, multiply top and bottom by the conjugate. For fractions, combine them over a common denominator first.

★ Continuity and differentiability

Differentiable at a point ⇒ continuous at that point.
Continuous does not mean differentiable. The graph of $y = |x|$ has no break at $x = 0$ but has a corner, so the gradient from the left is $-1$ and from the right is $+1$ .
A function can also fail to be differentiable where the tangent is vertical (gradient infinite) or where there is a jump or hole.
For a piecewise function that must be smooth at the join $x = a$ : the two pieces must have the same value at $a$ and the same gradient at $a$ .

⚠️ Watch out

In first principles you must write the limit and keep $\lim_{h \to 0}$ in front until the $h$ terms have gone. Writing $h = 0$ too early gives $\dfrac{0}{0}$ and loses the marks.
Expand $(x+h)^2$ as $x^2 + 2xh + h^2$ , never $x^2 + h^2$ . For $(x+h)^3$ and $(x+h)^4$ use the binomial coefficients 1, 3, 3, 1 and 1, 4, 6, 4, 1.
Do not cancel an $h$ from just one term of the numerator. Take out the common factor $h$ from the whole numerator first.

💡 Command terms

Differentiate from first principles and use the definition of the derivative both demand the limit method. Quoting the power rule earns nothing. Hence means use the result you just found, and show that needs every line written out.

🧪 Try it yourselfChords closing in on a tangent

The blue curve is the gradient of the chord from x to x + h for f(x) = x³, namely

\dfrac{(x + h)^3 - x^3}{h}

. The red curve is the true derivative 3x². Drag h down towards 0 and watch the blue curve fall onto the red one. This is convergence.

Worked examples

Example 1.1First principles for a quadratic

Use the definition

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

to differentiate

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

🧠 Think it through

Replace $x$ by $x + h$ everywhere and expand.
Subtract the original $f(x)$ : the terms without $h$ must cancel.
Divide every term by $h$ , then let $h \to 0$ .

✍️ By hand
   (M1)
   (A1)
   (M1)
   (A1)

🔢 On your TI-Nspire CX II

Check at $x = 2$ : the formula gives $6(2) + 2 = 14$ . On a Calculator page use the derivative template (press menu💻 click menu on keypad, then Calculus, then Numerical Derivative at a Point).

✅ Answer

f'(x) = 6x + 2

Example 1.2First principles for a cubic, then higher derivatives

(a) Differentiate

f(x) = 2x^3 - x

from first principles. (b) Hence find

f''(x)

and

f'''(x)

, and the gradient of the curve at

x = -2

🧠 Think it through

Use $(x+h)^3 = x^3 + 3x^2h + 3xh^2 + h^3$ . The $2x^3$ means all four terms are doubled.
Once you have f′(x) you may use ordinary rules for the higher derivatives.

✍️ By hand
(a) 
   (M1)
   (A1)
   (M1)
As , the terms with  vanish:
   (A1)
(b)  and .
At : .

✅ Answer (a)

f'(x) = 6x^2 - 1

. (b)

f''(x) = 12x

f'''(x) = 12

; the gradient at

x = -2

is 23.

Example 1.3First principles with a square root

Use the definition of the derivative to show that the derivative of

f(x) = \sqrt{x}

\dfrac{1}{2\sqrt{x}}

, for

x > 0

. Hence find the gradient of

y = \sqrt{x}

x = 9

🧠 Think it through

The top is $\sqrt{x+h} - \sqrt{x}$ . Multiply top and bottom by the conjugate $\sqrt{x+h} + \sqrt{x}$ so the top becomes a difference of squares.
After that, the h on the top cancels the h on the bottom.

✍️ By hand
Multiply top and bottom by the conjugate :   (M1)
   (A1)
As :      (AG)
At : .

🔢 On your TI-Nspire CX II

Check with the derivative template at $x = 9$ .

✅ Answer

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

; the gradient at

x = 9

\dfrac{1}{6}

Example 1.4A piecewise function that must be smooth

The function

f

is defined by

f(x) = 2x^2 + 1

for

x \leq 1

, and

f(x) = ax + b

for

x > 1

. Find the values of

a

and

b

so that

f

is continuous and differentiable at

x = 1

🧠 Think it through

Continuity: the left value at $x = 1$ must equal the right value.
Differentiability: the gradients must also match at $x = 1$ . The gradient of $2x^2 + 1$ is $4x$ .
That gives two equations for two unknowns.

✍️ By hand
Left value at : . Right value: .
Continuous: .   (M1)(A1)
Left gradient:  at . Right gradient: .
Differentiable: .   (M1)(A1)
So .   (A1)

✅ Answer

a = 4

and

b = -1

. The right-hand piece is the tangent line

y = 4x - 1

to the curve at

x = 1

The line y = 4x − 1 is exactly the tangent to y = 2x² + 1 at x = 1. A smooth join means the second piece starts as the tangent to the first.

Example 1.5Continuous but not differentiable

Show that (a)

f(x) = |x - 2|

is continuous but not differentiable at

x = 2

; (b)

g(x) = \sqrt[3]{x}

is continuous at

x = 0

but not differentiable there.

🧠 Think it through

For (a) work out the gradient limit from the left ( $h < 0$ , so $|h| = -h$ ) and from the right ( $h > 0$ , so $|h| = h$ ).
For (b) form $\dfrac{g(h) - g(0)}{h}$ and see what happens as $h$ gets small.

✍️ By hand
(a) . As  from either side, , so  is continuous at 2.   (R1)
.   (M1)
For  this is ; for  it is . The two one-sided limits differ, so  does not exist.   (R1)
(b) , and  as , so  is continuous.   (R1)
   (M1)
This grows without bound as , so the limit does not converge. The tangent at the origin is vertical.   (R1)

🔢 On your TI-Nspire CX II

See it on the graph page: plot abs(x-2) and zoom in on $x = 2$ . The corner never smooths out.
📍 Where to find it on the TI-Nspire
⌨️ type abs( on the keyboard; the screen shows it as | |

✅ Answer (a) Left gradient

-1

, right gradient

+1

: not differentiable at

x = 2

. (b) The gradient limit is infinite (vertical tangent): not differentiable at

x = 0

Challenge questions

⭐ Challenge 1.6First principles for x to the power 4Paper 1 style

(a) Use the binomial theorem to expand

(x+h)^4

. (b) Hence differentiate

f(x) = x^4

from first principles. (c) Find

f''(2)

and state

f^{(4)}(x)

. (d) Show from first principles that the derivative of

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

-\dfrac{1}{x^2}

✏️ Try it on paper and on your calculator first. Then check.

Show the full solution

🧠 Think it through

Binomial coefficients for power 4: 1, 4, 6, 4, 1.
After subtracting $x^4$ , every remaining term contains $h$ .

✍️ By hand
(a) 
   (A1)
(b) 
   (M1)(A1)
As  the terms with  vanish:      (A1)
(c) , so .   (A1)
 and .   (A1)
(d) 
   (M1)(A1)
As :    (AG)

✅ Answer (b)

f'(x) = 4x^3

. (c)

f''(2) = 48

;

f^{(4)}(x) = 24

. (d)

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

⭐ Challenge 1.7The function x|x|: how smooth is it?Paper 3 style

In this investigation

f(x) = x|x|

, that is

f(x) = x^2

for

x \geq 0

and

f(x) = -x^2

for

x < 0

(a) Show that $f$ is continuous at $x = 0$ .
(b) Use first principles to show that $f'(0)$ exists and find its value.
(c) Find $f'(x)$ for $x > 0$ and for $x < 0$ , and write $f'(x)$ as a single formula.
(d) Show that $f''(0)$ does not exist.
(e) Graph $f$ and $f'$ on your GDC. Describe, in words, how the two graphs differ at $x = 0$ .

✏️ Try it on paper and on your calculator first. Then check.

Show the full solution

🧠 Think it through

(a) Check the left limit, right limit and the value at 0.
(b) $\dfrac{f(h) - f(0)}{h} = \dfrac{h|h|}{h} = |h|$ .
(c) Differentiate each piece separately. Then notice that 2x for x > 0 and −2x for x < 0 is 2|x|.
(d) Repeat the first-principles idea, but on the function f′.

✍️ By hand
(a) , , and . All three agree, so  is continuous at 0.   (R1)
(b) 
   (M1)
As  from either side, . So .   (A1)
(c) For : . For : .   (A1)
Together with :  for all .   (A1)
(d) , which is  for  and  for .   (M1)
The one-sided limits differ, so  does not exist.   (R1)

🔢 On your TI-Nspire CX II

Graph x*abs(x) as f1 and 2*abs(x) as f2. The first passes smoothly through the origin with a horizontal tangent. The second has a sharp corner at the origin.
📍 Where to find it on the TI-Nspire
⌨️ type abs( on the keyboard; the screen shows it as | |

✅ Answer

f'(x) = 2|x|

, so

f

is differentiable everywhere, but

f'

has a corner at 0 so

f''(0)

does not exist.

Lesson idea: differentiable once does not mean differentiable twice. Each derivative can lose some smoothness.

Next: Lesson 2 · Limits and l'Hôpital's rule

IB DP Mathematics: Analysis and Approaches (AA) — SL and HL HL · Further calculus 1 · Continuity, differentiability & first principles

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

Continuity, differentiability & differentiation from first principles

Worked examples

Challenge questions