🚀 Level 2 - Topic 5: Higher-Order Derivatives ⏫

1) Introduction: Differentiating Derivatives - Going Further

We've learned how to find the first derivative of a function \( f(x) \), which represents the instantaneous rate of change of \( f \) with respect to \( x \). But what if we want to know how the *rate of change itself* is changing? This leads us to the concept of higher-order derivatives.

Higher-order derivatives are simply derivatives of derivatives. We can differentiate the first derivative to get the second derivative, differentiate the second derivative to get the third derivative, and so on. These higher derivatives provide us with even more information about the behavior of a function, such as its concavity, rate of acceleration, and more.

In this topic, we'll explore how to find and interpret higher-order derivatives, focusing on second and third derivatives and their applications.

2) The Second Derivative: Rate of Change of the First Derivative

The second derivative of a function \( f(x) \) is the derivative of its first derivative \( f'(x) \). It's denoted by \( f''(x) \) or \( \frac{d^2y}{dx^2} \) (if \( y = f(x) \)).

Definition 5.1: Second Derivative

The second derivative of \( y = f(x) \) is defined as:

\( f''(x) = \frac{d}{dx}[f'(x)] = \frac{d}{dx}\left[\frac{dy}{dx}\right] \)

In Leibniz notation, we write:

\( \frac{d^2y}{dx^2} = \frac{d}{dx}\left(\frac{dy}{dx}\right) \)

In words: The second derivative measures the rate of change of the slope of the tangent line to the graph of \( f(x) \). It tells us how quickly the first derivative (the rate of change) is changing.

Examples of Finding Second Derivatives

Example 1: \( f(x) = x^4 - 3x^2 + 7x - 2 \)
First, find the first derivative \( f'(x) \):
\( f'(x) = \frac{d}{dx}[x^4 - 3x^2 + 7x - 2] = 4x^3 - 6x + 7 \)
Now, differentiate \( f'(x) \) to find the second derivative \( f''(x) \):
\( f''(x) = \frac{d}{dx}[f'(x)] = \frac{d}{dx}[4x^3 - 6x + 7] = 12x^2 - 6 \)
Thus, \( f''(x) = 12x^2 - 6 \).
Example 2: \( y = x^3 \sin(x) \)
First, find the first derivative \( \frac{dy}{dx} \) using the Product Rule:
\( \frac{dy}{dx} = \frac{d}{dx}[x^3 \sin(x)] = \frac{d}{dx}[x^3] \cdot \sin(x) + x^3 \cdot \frac{d}{dx}[\sin(x)] = 3x^2 \sin(x) + x^3 \cos(x) \)
Now, differentiate \( \frac{dy}{dx} \) to find the second derivative \( \frac{d^2y}{dx^2} \). We'll need the Product Rule again, applied to each term of \( \frac{dy}{dx} \):
\( \frac{d^2y}{dx^2} = \frac{d}{dx}[3x^2 \sin(x) + x^3 \cos(x)] = \frac{d}{dx}[3x^2 \sin(x)] + \frac{d}{dx}[x^3 \cos(x)] \)
For \( \frac{d}{dx}[3x^2 \sin(x)] \): \( = 3 \left( \frac{d}{dx}[x^2] \cdot \sin(x) + x^2 \cdot \frac{d}{dx}[\sin(x)] \right) = 3(2x \sin(x) + x^2 \cos(x)) = 6x \sin(x) + 3x^2 \cos(x) \)

For \( \frac{d}{dx}[x^3 \cos(x)] \): \( = \frac{d}{dx}[x^3] \cdot \cos(x) + x^3 \cdot \frac{d}{dx}[\cos(x)] = 3x^2 \cos(x) + x^3 (-\sin(x)) = 3x^2 \cos(x) - x^3 \sin(x) \)

Adding these together:
\( \frac{d^2y}{dx^2} = (6x \sin(x) + 3x^2 \cos(x)) + (3x^2 \cos(x) - x^3 \sin(x)) = 6x \sin(x) + 6x^2 \cos(x) - x^3 \sin(x) \)
So, \( \frac{d^2y}{dx^2} = 6x \sin(x) + 6x^2 \cos(x) - x^3 \sin(x) \).
Example 3: \( g(x) = \frac{x}{x + 1} \)
First, find the first derivative \( g'(x) \) using the Quotient Rule:
\( g'(x) = \frac{d}{dx}\left[\frac{x}{x + 1}\right] = \frac{\frac{d}{dx}[x] \cdot (x + 1) - x \cdot \frac{d}{dx}[x + 1]}{(x + 1)^2} = \frac{(1)(x + 1) - x(1)}{(x + 1)^2} = \frac{x + 1 - x}{(x + 1)^2} = \frac{1}{(x + 1)^2} = (x + 1)^{-2} \)
Now, differentiate \( g'(x) = (x + 1)^{-2} \) to find the second derivative \( g''(x) \) using the Chain Rule:
\( g''(x) = \frac{d}{dx}[(x + 1)^{-2}] = -2(x + 1)^{-3} \cdot \frac{d}{dx}[x + 1] = -2(x + 1)^{-3} \cdot (1) = -2(x + 1)^{-3} = -\frac{2}{(x + 1)^3} \)
Thus, \( g''(x) = -\frac{2}{(x + 1)^3} \).

3) Third and Higher-Order Derivatives

We can continue differentiating to find third derivatives, fourth derivatives, and so on. The third derivative is the derivative of the second derivative: \( f'''(x) = \frac{d}{dx}[f''(x)] = \frac{d^3y}{dx^3} \). For derivatives of order higher than three, we often use the notation \( f^{(n)}(x) \) or \( \frac{d^ny}{dx^n} \) for the \( n \)-th derivative.

Definition 5.2: Higher-Order Derivatives

The \( n \)-th derivative of \( y = f(x) \) is obtained by differentiating \( f(x) \) a total of \( n \) times. It is denoted by \( f^{(n)}(x) \) or \( \frac{d^ny}{dx^n} \).

\( f^{(n)}(x) = \frac{d}{dx}[f^{(n-1)}(x)] = \frac{d^n}{dx^n} [f(x)] \)

(where \( f^{(0)}(x) = f(x) \) and \( f^{(1)}(x) = f'(x) \), \( f^{(2)}(x) = f''(x) \), \( f^{(3)}(x) = f'''(x) \), etc.)

Examples of Third and Higher Derivatives

Example 1: \( f(x) = x^5 - 2x^3 + 6x \)
First derivative: \( f'(x) = 5x^4 - 6x^2 + 6 \)

Second derivative: \( f''(x) = 20x^3 - 12x \)

Third derivative: \( f'''(x) = \frac{d}{dx}[f''(x)] = \frac{d}{dx}[20x^3 - 12x] = 60x^2 - 12 \)

Fourth derivative: \( f^{(4)}(x) = \frac{d}{dx}[f'''(x)] = \frac{d}{dx}[60x^2 - 12] = 120x \)

Fifth derivative: \( f^{(5)}(x) = \frac{d}{dx}[f^{(4)}(x)] = \frac{d}{dx}[120x] = 120 \)

Sixth derivative: \( f^{(6)}(x) = \frac{d}{dx}[f^{(5)}(x)] = \frac{d}{dx}[120] = 0 \)

And all derivatives of order 6 and higher will also be zero for this polynomial.
Example 2: \( y = \sin(x) \) - Pattern in Derivatives
First derivative: \( \frac{dy}{dx} = \cos(x) \)

Second derivative: \( \frac{d^2y}{dx^2} = \frac{d}{dx}[\cos(x)] = -\sin(x) \)

Third derivative: \( \frac{d^3y}{dx^3} = \frac{d}{dx}[-\sin(x)] = -\cos(x) \)

Fourth derivative: \( \frac{d^4y}{dx^4} = \frac{d}{dx}[-\cos(x)] = -(-\sin(x)) = \sin(x) \)

Fifth derivative: \( \frac{d^5y}{dx^5} = \frac{d}{dx}[\sin(x)] = \cos(x) \) - and the pattern repeats!

The derivatives of \( \sin(x) \) cycle through \( \sin(x), \cos(x), -\sin(x), -\cos(x), \sin(x), \ldots \).

4) Applications of the Second Derivative: Concavity and Inflection Points

The second derivative has important applications in understanding the shape of a graph of a function. Two key applications are in determining the concavity of a graph and locating inflection points.

4.1) Concavity

The concavity of a function describes whether its graph curves upwards or downwards.

Definition 5.3: Concavity

A function \( f(x) \) is concave up on an interval if its graph curves upwards over that interval. Visually, the graph "holds water." More formally, the slopes of tangent lines are increasing as \( x \) increases.
A function \( f(x) \) is concave down on an interval if its graph curves downwards over that interval. Visually, the graph "spills water." More formally, the slopes of tangent lines are decreasing as \( x \) increases.

Test for Concavity Using the Second Derivative

Let \( f(x) \) be twice differentiable on an interval.

If \( f''(x) > 0 \) for all \( x \) in the interval, then \( f(x) \) is concave up on that interval.
If \( f''(x) < 0 \) for all \( x \) in the interval, then \( f(x) \) is concave down on that interval.
If \( f''(x) = 0 \) or \( f''(x) \) changes sign, we need to investigate further (possibly an inflection point).

Example: Determining Concavity

Determine the intervals where \( f(x) = x^3 - 3x^2 + 2x - 1 \) is concave up and concave down.

Find the second derivative \( f''(x) \).
First derivative: \( f'(x) = 3x^2 - 6x + 2 \)

Second derivative: \( f''(x) = 6x - 6 \)
Determine where \( f''(x) > 0 \) and \( f''(x) < 0 \).
Set \( f''(x) = 6x - 6 > 0 \Rightarrow 6x > 6 \Rightarrow x > 1 \). So, \( f''(x) > 0 \) when \( x > 1 \).

Set \( f''(x) = 6x - 6 < 0 \Rightarrow 6x < 6 \Rightarrow x < 1 \). So, \( f''(x) < 0 \) when \( x < 1 \).
State the concavity intervals.
\( f(x) \) is concave up on the interval \( (1, \infty) \) because \( f''(x) > 0 \) there.

\( f(x) \) is concave down on the interval \( (-\infty, 1) \) because \( f''(x) < 0 \) there.

4.2) Inflection Points

An inflection point is a point on the graph of a function where the concavity changes from concave up to concave down or vice versa.

Definition 5.4: Inflection Point

A point \( (c, f(c)) \) is an inflection point of the function \( f(x) \) if the concavity of \( f \) changes at \( x = c \). This typically occurs where \( f''(c) = 0 \) or \( f''(c) \) is undefined, provided that the concavity actually changes at \( x = c \).

Finding Inflection Points

Find the second derivative \( f''(x) \).
Find the critical points of \( f'(x) \) (i.e., solve for \( x \) where \( f''(x) = 0 \) or \( f''(x) \) is undefined). These are potential inflection points.
Check for concavity change at each potential inflection point. You can use a sign chart for \( f''(x) \) around these points. If the sign of \( f''(x) \) changes at \( x = c \), then \( (c, f(c)) \) is an inflection point. If the sign does not change, it's not an inflection point.

Example: Finding Inflection Points

Find the inflection points of \( f(x) = x^3 - 3x^2 + 2x - 1 \).

Find the second derivative \( f''(x) \).
We already found \( f''(x) = 6x - 6 \) in the previous example.
Find where \( f''(x) = 0 \) or is undefined.
Set \( f''(x) = 6x - 6 = 0 \Rightarrow 6x = 6 \Rightarrow x = 1 \). \( f''(x) \) is defined for all \( x \), so \( x = 1 \) is the only potential inflection point.
Check for concavity change at \( x = 1 \).
From the concavity example, we know \( f''(x) < 0 \) for \( x < 1 \) (concave down) and \( f''(x) > 0 \) for \( x > 1 \) (concave up). Since concavity changes at \( x = 1 \), there is an inflection point at \( x = 1 \).
Find the \( y \)-coordinate of the inflection point.
\( f(1) = (1)^3 - 3(1)^2 + 2(1) - 1 = 1 - 3 + 2 - 1 = -1 \). So the inflection point is at \( (1, -1) \).

5) Application of Third Derivative: Jerk (in Physics - Optional)

In physics, if \( s(t) \) represents the position of an object at time \( t \), then:

The first derivative \( v(t) = s'(t) \) represents the velocity (rate of change of position).
The second derivative \( a(t) = v'(t) = s''(t) \) represents the acceleration (rate of change of velocity).
The third derivative \( j(t) = a'(t) = v''(t) = s'''(t) \) represents the jerk (rate of change of acceleration). Jerk is related to the smoothness of motion and changes in forces.

While acceleration is more commonly encountered, the third derivative (jerk) and even higher derivatives have applications in fields like physics, engineering, and signal processing, especially when analyzing motion, vibrations, or rapid changes in systems.

Example: Position, Velocity, Acceleration, and Jerk

Suppose the position of a particle is given by \( s(t) = t^4 - 2t^3 + 3t^2 + 5t \).

Velocity: \( v(t) = s'(t) = 4t^3 - 6t^2 + 6t + 5 \)

Acceleration: \( a(t) = v'(t) = s''(t) = 12t^2 - 12t + 6 \)

Jerk: \( j(t) = a'(t) = s'''(t) = 24t - 12 \)

Fourth derivative (rate of change of jerk): \( s^{(4)}(t) = 24 \)

Fifth derivative and higher: \( s^{(n)}(t) = 0 \) for \( n \geq 5 \).

6) Practice Questions - Higher-Order Derivatives

Practice finding higher-order derivatives and applying them to concavity and inflection points.

Fundamental Practice Questions

Instructions: For each function, find the second derivative \( f''(x) \) or \( \frac{d^2y}{dx^2} \), and for some, the third derivative \( f'''(x) \) or \( \frac{d^3y}{dx^3} \).

Q1. \( f(x) = 5x^4 - 3x^3 + 2x^2 - 7x + 1 \) (Find \( f''(x) \) and \( f'''(x) \))
Q2. \( y = x^5 + \frac{1}{x} = x^5 + x^{-1} \) (Find \( \frac{d^2y}{dx^2} \))
Q3. \( g(x) = \sqrt{x} = x^{1/2} \) (Find \( g''(x) \) and \( g'''(x) \))
Q4. \( y = (2x + 1)^3 \) (Find \( \frac{d^2y}{dx^2} \))
Q5. \( f(x) = \sin(2x) \) (Find \( f''(x) \) and \( f'''(x) \))
Q6. \( y = x^2 \cos(x) \) (Find \( \frac{d^2y}{dx^2} \))
Q7. \( g(x) = \frac{x}{x - 2} \) (Find \( g''(x) \))
Q8. \( y = (x^2 + 3)^{1/2} \) (Find \( \frac{d^2y}{dx^2} \))
Q9. \( f(x) = \frac{1}{(x + 1)^2} = (x + 1)^{-2} \) (Find \( f''(x) \))
Q10. \( y = x \sqrt{x + 4} \) (Find \( \frac{d^2y}{dx^2} \))

Challenging Practice Questions

Instructions: These problems involve concavity, inflection points, or more complex higher-order derivatives.

Q1. For \( f(x) = x^4 - 6x^2 + 8x + 5 \), find the intervals where \( f(x) \) is concave up and concave down, and find all inflection points.
Q2. Find the inflection points of \( g(x) = \frac{1}{x^2 + 1} \). (You'll need to find \( g''(x) \) and solve \( g''(x) = 0 \)).
Q3. Determine the concavity of \( y = x\sqrt{x} \) for \( x > 0 \).
Q4. For what values of \( x \) is the graph of \( y = x^4 - 4x^3 + 6x^2 - 4x + 1 \) concave up?
Q5. The position of a particle is given by \( s(t) = t^3 - 6t^2 + 9t \). Find the times when the acceleration is zero. What is the jerk at these times?

7) Summary & Cheat Sheet - Higher-Order Derivatives

Let's summarize the key points about higher-order derivatives.

7.1) Notation for Higher-Order Derivatives

For \( y = f(x) \):

Order	Prime Notation	Leibniz Notation
First Derivative	\( f'(x) \)	\( \frac{dy}{dx} \)
Second Derivative	\( f''(x) \)	\( \frac{d^2y}{dx^2} \)
Third Derivative	\( f'''(x) \)	\( \frac{d^3y}{dx^3} \)
Fourth Derivative	\( f^{(4)}(x) \)	\( \frac{d^4y}{dx^4} \)
...	...	...
\( n \)-th Derivative	\( f^{(n)}(x) \)	\( \frac{d^ny}{dx^n} \)

7.2) Finding Higher-Order Derivatives - Process

To find the \( n \)-th derivative \( f^{(n)}(x) \), first find the first derivative \( f'(x) \).
Then, differentiate \( f'(x) \) to get the second derivative \( f''(x) \).
Continue differentiating the previous derivative to get the next higher derivative.
Repeat this process \( n \) times to obtain \( f^{(n)}(x) \).

7.3) Applications of Second Derivative

Concavity:
- \( f''(x) > 0 \) => Concave Up
- \( f''(x) < 0 \) => Concave Down
Inflection Points: Occur where concavity changes, typically when \( f''(x) = 0 \) or \( f''(x) \) is undefined, and concavity changes around that point.