Grade 12
CAPS Syllabus
Calculus
Advanced Differentiation

The Second Derivative

Learn what the second derivative tells us about how gradients change, concavity, and acceleration. Master this powerful tool for understanding the behavior of functions.

What You'll Learn
  • The second derivative is the derivative of the first derivative
  • It tells us how the gradient is changing
  • Notation: $f''(x)$ or $\frac{d^2y}{dx^2}$
  • How to calculate second derivatives using differentiation rules
  • Understanding concavity and what the sign of the second derivative means

What is the Second Derivative?

You already know that the first derivative of a function tells us the gradient at any point. But what if we want to know how that gradient is changing?

That's where the second derivative comes in!

Definition

The second derivative of a function is the derivative of the first derivative. It shows us the rate of change of the gradient.

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

In words: "Differentiate the first derivative"

Think About It:

If the first derivative tells us the velocity of a moving object, then the second derivative tells us the acceleration — how quickly the velocity is changing!

Notation for the Second Derivative

Just like the first derivative has multiple notations, the second derivative does too. Here are the main ways to write it:

Different Notations

If we have a function $f(x)$:

Prime notation:$f''(x)$
Leibniz notation:$\frac{d^2f}{dx^2}$

If we have $y = f(x)$:

Prime notation:$y''$
Leibniz notation:$\frac{d^2y}{dx^2}$
Expanded form:$\frac{d}{dx}\left[\frac{dy}{dx}\right]$

Important Note:

$\frac{d^2y}{dx^2}$ does NOT mean $\frac{d^2y}{(dx)^2}$. It's a special notation that means "the second derivative of y with respect to x".

What Does the Second Derivative Tell Us?

The second derivative gives us important information about how the function's gradient is changing:

$f''(x) > 0$

Gradient is increasing

The curve is concave up (shaped like a smile or cup)

$f''(x) = 0$

Gradient is constant

Possible point of inflection (where concavity changes)

$f''(x) < 0$

Gradient is decreasing

The curve is concave down (shaped like a frown or cap)

Real-World Example: Motion

Position: $s(t)$

Where the object is at time t

Velocity: $v(t) = s'(t)$

How fast the position is changing

Acceleration: $a(t) = v'(t) = s''(t)$

How fast the velocity is changing

Understanding Concavity Visually
Move the slider to explore how the second derivative relates to the curve's shape

At x = 0.0:

y = 0.00

First Derivative:

f'(0.0) = -1.50

Gradient = -1.50

Second Derivative:

f''(0.0) = 0.00

Inflection Point

😊
Concave Up (Smile)

f''(x) > 0

• Gradient is increasing

• Tangent lines get steeper from left to right

• Curve opens upward like a cup

Think: water would collect in it

☹️
Concave Down (Frown)

f''(x) < 0

• Gradient is decreasing

• Tangent lines get less steep from left to right

• Curve opens downward like a cap

Think: water would run off it

Key Observations:

  • • The green shaded region shows where f''(x) > 0 (concave up)
  • • The red shaded region shows where f''(x) < 0 (concave down)
  • • The orange dashed line shows the inflection point where f''(x) = 0
  • • When you check "Show multiple tangent lines", notice how the tangent slope changes differently in each region!

How to Find the Second Derivative

Finding the second derivative is straightforward: differentiate twice!

The Process

1Find the first derivative

Use the differentiation rules to find $f'(x)$ or $\frac{dy}{dx}$

2Differentiate again

Differentiate the first derivative to get $f''(x)$ or $\frac{d^2y}{dx^2}$

Key Insight:

The degree of the polynomial decreases by 1 each time you differentiate. For example:
$x^3$$3x^2$ $6x$
• Cubic → Quadratic → Linear

Worked Examples

Example 1: Polynomial Function
Calculate the second derivative of $k(x) = 2x^3 - 4x^2 + 9$

Step 1: Find the first derivative

$$\begin{aligned} k(x) &= 2x^3 - 4x^2 + 9 \\ k'(x) &= 2(3x^2) - 4(2x) + 0 \\ &= 6x^2 - 8x \end{aligned}$$

Step 2: Find the second derivative

$$\begin{aligned} k'(x) &= 6x^2 - 8x \\ k''(x) &= 6(2x) - 8 \\ &= 12x - 8 \end{aligned}$$

Answer:

$k''(x) = 12x - 8$

Example 2: Function with Fractions
Calculate the second derivative of $y = \frac{3}{x}$

Step 1: Rewrite using negative exponents

$$y = \frac{3}{x} = 3x^{-1}$$

Step 2: Find the first derivative

$$\begin{aligned} \frac{dy}{dx} &= 3(-1x^{-2}) \\ &= -3x^{-2} \\ &= -\frac{3}{x^2} \end{aligned}$$

Step 3: Find the second derivative

$$\begin{aligned} \frac{dy}{dx} &= -3x^{-2} \\ \frac{d^2y}{dx^2} &= -3(-2x^{-3}) \\ &= 6x^{-3} \\ &= \frac{6}{x^3} \end{aligned}$$

Answer:

$\frac{d^2y}{dx^2} = \frac{6}{x^3}$

Example 3: Expanding First
Find the first and second derivatives of $f(x) = x(x - 6) + 10$

Step 1: Expand the function

$$\begin{aligned} f(x) &= x(x - 6) + 10 \\ &= x^2 - 6x + 10 \end{aligned}$$

Step 2: Find the first derivative

$$f'(x) = 2x - 6$$

Step 3: Find the second derivative

$$f''(x) = 2$$

Answers:

$f'(x) = 2x - 6$

$f''(x) = 2$

Notice that the second derivative is a constant! This makes sense because the original function is a quadratic, so its gradient changes at a constant rate.

Interactive Visualization

Interactive Visualization: f(x), f'(x), and f''(x)
See how the function, first derivative, and second derivative relate to each other

Original Function f(x)

Shows the shape of the curve

First Derivative f'(x)

Shows the gradient at each point

Second Derivative f''(x)

Shows rate of change of gradient

Observe:

  • • Where $f''(x) > 0$, the graph of $f(x)$ is concave up (curves upward like a smile)
  • • Where $f''(x) < 0$, the graph of $f(x)$ is concave down (curves downward like a frown)
  • • Where $f''(x) = 0$, there may be a point of inflection
  • • Each derivative is one degree lower than the previous function

Practice Questions

Additional Practice Questions
Access more practice problems with PDF export functionality. Download questions-only or with complete solutions for offline study.
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Apply what you've learned with these practice problems.

Question 1
Question 2
Question 3

Quick Self-Check Quiz

Test your understanding with these questions.

Question 1
What is the second derivative?
Question 2
If $f(x) = x^3$, what is $f''(x)$?
Question 3
What does $f''(x) > 0$ tell us about the function?
Question 4
If the second derivative of position gives us acceleration, what does the first derivative give us?
Question 5
For $y = 5x^2 + 3x - 7$, what is $\\frac{d^2y}{dx^2}$?
Question 6
What happens to the degree of a polynomial each time you differentiate?
Question 7
If $g(x) = 2x^4$, what is $g''(0)$?

Summary

"The second derivative tells us how the rate of change is changing."

Key Formulas:

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

$\frac{d^2y}{dx^2} = \frac{d}{dx}\left[\frac{dy}{dx}\right]$

What It Tells Us:

  • $f''(x) > 0$ → Concave up (gradient increasing)
  • $f''(x) < 0$ → Concave down (gradient decreasing)
  • $f''(x) = 0$ → Possible inflection point

Remember:

Differentiate twice, using the same rules each time. The degree of the polynomial decreases with each derivative.