💡 Note: After the third derivative, we use numbers in parentheses: $f^{(4)}(x)$, $f^{(5)}(x)$, etc.

Notation for Higher-Order Derivatives

Prime Notation

• $f'(x)$ — first derivative
• $f''(x)$ — second derivative (two primes)
• $f'''(x)$ — third derivative (three primes)
• $f^{(4)}(x)$ — fourth derivative (number in parentheses)

Leibniz Notation

• $\frac{dy}{dx}$ — first derivative
• $\frac{d^2y}{dx^2}$ — second derivative
• $\frac{d^3y}{dx^3}$ — third derivative
• $\frac{d^ny}{dx^n}$ — nth derivative

Important: In $\frac{d^2y}{dx^2}$, the "2" is NOT an exponent! It indicates the second derivative.

Finding Higher-Order Derivatives

Simply differentiate repeatedly!

Example 1: Polynomial

Find all derivatives of $f(x) = x^4 - 3x^3 + 2x^2 - 5x + 7$

$f(x) = x^4 - 3x^3 + 2x^2 - 5x + 7$

$f'(x) = 4x^3 - 9x^2 + 4x - 5$

$f''(x) = 12x^2 - 18x + 4$

$f'''(x) = 24x - 18$

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

$f^{(5)}(x) = 0$ (and all higher derivatives are 0)

💡 Pattern: For a polynomial of degree $n$, the $(n+1)$-th derivative and all higher derivatives are zero!

Example 2: Exponential

Find the first four derivatives of $f(x) = e^{2x}$

$f(x) = e^{2x}$

$f'(x) = 2e^{2x}$

$f''(x) = 4e^{2x}$

$f'''(x) = 8e^{2x}$

$f^{(4)}(x) = 16e^{2x}$

Pattern: $f^{(n)}(x) = 2^n e^{2x}$

Example 3: Trigonometric

Find the pattern for $f(x) = \sin x$

$f(x) = \sin x$

$f'(x) = \cos x$

$f''(x) = -\sin x$

$f'''(x) = -\cos x$

$f^{(4)}(x) = \sin x$ (back to the start!)

Pattern: The derivatives of sine cycle every 4 derivatives!

What Does the Second Derivative Tell Us?

The second derivative has special meaning:

Physical Interpretation

If $s(t)$ represents position:

• $s'(t) = v(t)$ is velocity (rate of change of position)
• $s''(t) = v'(t) = a(t)$ is acceleration (rate of change of velocity)

Example: If $s(t) = 16t^2$ (position in feet at time $t$ seconds)

• $v(t) = s'(t) = 32t$ ft/sec (velocity)
• $a(t) = s''(t) = 32$ ft/sec² (acceleration)

Concavity

The second derivative tells us about the concavity (curvature) of a function:

• If $f''(x) > 0$, the graph is concave up (curves upward, like ∪)
• If $f''(x) < 0$, the graph is concave down (curves downward, like ∩)
• If $f''(x) = 0$, there might be an inflection point

💡 Memory Trick: Concave up looks like a cup that holds water ∪

Inflection Points

An inflection point is where the concavity changes (from up to down, or vice versa).

How to Find Inflection Points

1. Find $f''(x)$
2. Set $f''(x) = 0$ and solve
3. Check that $f''(x)$ changes sign on either side

Example: Find inflection points of $f(x) = x^3$

$f'(x) = 3x^2$

$f''(x) = 6x$

Set $f''(x) = 0$: $6x = 0$, so $x = 0$

Check signs:

• When $x < 0$: $f''(x) < 0$ (concave down)
• When $x > 0$: $f''(x) > 0$ (concave up)

Since concavity changes, $(0, 0)$ is an inflection point! ✓

The Second Derivative Test

The second derivative can help classify critical points!

Second Derivative Test for Extrema

If $f'(c) = 0$ (critical point):

• If $f''(c) > 0$ → local minimum at $x = c$ (concave up)
• If $f''(c) < 0$ → local maximum at $x = c$ (concave down)
• If $f''(c) = 0$ → test is inconclusive (use first derivative test)

Why it works: If the graph is curving upward at a critical point, it must be a minimum!

Higher-Order Derivatives with Chain Rule

When using Chain Rule multiple times, things get complicated!

Example: $y = e^{3x}$

$y' = 3e^{3x}$

$y'' = 9e^{3x}$

$y''' = 27e^{3x}$

Pattern: $y^{(n)} = 3^n e^{3x}$

Example: $y = \sin(2x)$

$y' = 2\cos(2x)$

$y'' = -4\sin(2x)$

$y''' = -8\cos(2x)$

$y^{(4)} = 16\sin(2x)$

Special Patterns

Pattern 1: $f(x) = x^n$

$f^{(k)}(x) = n(n-1)(n-2)\cdots(n-k+1)x^{n-k}$

For $k = n$: $f^{(n)}(x) = n!$ (factorial)

For $k > n$: $f^{(k)}(x) = 0$

Pattern 2: $f(x) = e^{kx}$

$f^{(n)}(x) = k^n e^{kx}$

Pattern 3: $f(x) = \sin(kx)$

The derivatives cycle with period 4:

• $f^{(4m)}(x) = k^{4m}\sin(kx)$
• $f^{(4m+1)}(x) = k^{4m+1}\cos(kx)$
• $f^{(4m+2)}(x) = -k^{4m+2}\sin(kx)$
• $f^{(4m+3)}(x) = -k^{4m+3}\cos(kx)$

Pattern 4: $f(x) = \ln x$

$f'(x) = \frac{1}{x} = x^{-1}$

$f''(x) = -x^{-2}$

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

$f^{(4)}(x) = -6x^{-4}$

Pattern: $f^{(n)}(x) = \frac{(-1)^{n-1}(n-1)!}{x^n}$

⚠️ Common Mistakes

Mistake 1: Wrong Notation

❌ $\frac{d^2y}{dx^2} = \left(\frac{dy}{dx}\right)^2$ (WRONG!) ✅ $\frac{d^2y}{dx^2} = \frac{d}{dx}\left[\frac{dy}{dx}\right]$ (second derivative)

Mistake 2: Sign Errors

When differentiating $f''(x)$ to get $f'''(x)$, be careful with negative signs!

Mistake 3: Forgetting Chain Rule

When taking second derivatives of composite functions, don't forget the Chain Rule!

Mistake 4: Inflection Point Errors

Just because $f''(c) = 0$ doesn't guarantee an inflection point - concavity must change!

Applications

Physics - Motion

• Position: $s(t)$
• Velocity: $v(t) = s'(t)$
• Acceleration: $a(t) = s''(t) = v'(t)$
• Jerk: $j(t) = s'''(t) = a'(t)$ (rate of change of acceleration)

Economics

• Total Cost: $C(x)$
• Marginal Cost: $C'(x)$
• Rate of change of Marginal Cost: $C''(x)$

Curve Sketching

Use $f''(x)$ to determine:

• Where graph is concave up or down
• Location of inflection points
• Classification of critical points (max or min)

📝 Practice Tips

1. Differentiate carefully - each derivative builds on the previous one
2. Look for patterns - especially with trig, exponential, and polynomial functions
3. Check your work - a simple error early on compounds with each derivative
4. Remember: Second derivative relates to concavity and acceleration
5. Don't confuse $\frac{d^2y}{dx^2}$ with $\left(\frac{dy}{dx}\right)^2$ - they're completely different!

$v(t) = s'(t) = 3t^2 - 12t + 9$

Step 2: Evaluate velocity at $t = 2$

$v(2) = 3(2)^2 - 12(2) + 9$

$= 3(4) - 24 + 9$

$= 12 - 24 + 9$

$= -3$ meters/second

Step 3: Find acceleration (second derivative of position)

$a(t) = s''(t) = v'(t) = 6t - 12$

Step 4: Evaluate acceleration at $t = 2$

$a(2) = 6(2) - 12 = 12 - 12 = 0$ meters/second²

Interpretation:

• At $t = 2$ seconds, the particle is moving at $-3$ m/s (moving backward/left)
• The acceleration is $0$ m/s² (velocity is neither increasing nor decreasing at this instant)

Answer: Velocity = $-3$ m/s, Acceleration = $0$ m/s²

Step 2: Find the second derivative

$f''(x) = 12x^2 - 24x$

Step 3: Set $f''(x) = 0$ and solve

$12x^2 - 24x = 0$

$12x(x - 2) = 0$

So $x = 0$ or $x = 2$

Step 4: Test for sign changes in $f''(x)$

We need to check intervals: $(-\infty, 0)$, $(0, 2)$, $(2, \infty)$

Test $x = -1$ (in $(-\infty, 0)$): $f''(-1) = 12(-1)^2 - 24(-1) = 12 + 24 = 36 > 0$ ✓ concave up

Test $x = 1$ (in $(0, 2)$): $f''(1) = 12(1)^2 - 24(1) = 12 - 24 = -12 < 0$ ✓ concave down

Test $x = 3$ (in $(2, \infty)$): $f''(3) = 12(3)^2 - 24(3) = 108 - 72 = 36 > 0$ ✓ concave up

Step 5: Identify inflection points

At $x = 0$: concavity changes from up to down → inflection point ✓

At $x = 2$: concavity changes from down to up → inflection point ✓

Step 6: Find y-coordinates

$f(0) = 0^4 - 4(0)^3 = 0$

$f(2) = 2^4 - 4(2)^3 = 16 - 32 = -16$

Answer: Inflection points at $(0, 0)$ and $(2, -16)$