Higher-Order Derivatives

Second derivatives, third derivatives, and beyond

🔢 Higher-Order Derivatives

What are Higher-Order Derivatives?

The first derivative f(x)f'(x) tells us the rate of change of f(x)f(x).

But we can also take the derivative of the derivative! This gives us the second derivative, third derivative, and so on.

The First Few Derivatives

  • First derivative: f(x)f'(x) or dydx\frac{dy}{dx}
  • Second derivative: f(x)f''(x) or d2ydx2\frac{d^2y}{dx^2}
  • Third derivative: f(x)f'''(x) or d3ydx3\frac{d^3y}{dx^3}
  • Fourth derivative: f(4)(x)f^{(4)}(x) or d4ydx4\frac{d^4y}{dx^4}
  • nth derivative: f(n)(x)f^{(n)}(x) or dnydxn\frac{d^ny}{dx^n}

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


Notation for Higher-Order Derivatives

Prime Notation

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

Leibniz Notation

  • dydx\frac{dy}{dx} — first derivative
  • d2ydx2\frac{d^2y}{dx^2} — second derivative
  • d3ydx3\frac{d^3y}{dx^3} — third derivative
  • dnydxn\frac{d^ny}{dx^n} — nth derivative

Important: In d2ydx2\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)=x43x3+2x25x+7f(x) = x^4 - 3x^3 + 2x^2 - 5x + 7

f(x)=x43x3+2x25x+7f(x) = x^4 - 3x^3 + 2x^2 - 5x + 7

f(x)=4x39x2+4x5f'(x) = 4x^3 - 9x^2 + 4x - 5

f(x)=12x218x+4f''(x) = 12x^2 - 18x + 4

f(x)=24x18f'''(x) = 24x - 18

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

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

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

Example 2: Exponential

Find the first four derivatives of f(x)=e2xf(x) = e^{2x}

f(x)=e2xf(x) = e^{2x}

f(x)=2e2xf'(x) = 2e^{2x}

f(x)=4e2xf''(x) = 4e^{2x}

f(x)=8e2xf'''(x) = 8e^{2x}

f(4)(x)=16e2xf^{(4)}(x) = 16e^{2x}

Pattern: f(n)(x)=2ne2xf^{(n)}(x) = 2^n e^{2x}

Example 3: Trigonometric

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

f(x)=sinxf(x) = \sin x

f(x)=cosxf'(x) = \cos x

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

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

f(4)(x)=sinxf^{(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)s(t) represents position:

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

Example: If s(t)=16t2s(t) = 16t^2 (position in feet at time tt seconds)

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

Concavity

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

  • If f(x)>0f''(x) > 0, the graph is concave up (curves upward, like ∪)
  • If f(x)<0f''(x) < 0, the graph is concave down (curves downward, like ∩)
  • If f(x)=0f''(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)f''(x)
  2. Set f(x)=0f''(x) = 0 and solve
  3. Check that f(x)f''(x) changes sign on either side

Example: Find inflection points of f(x)=x3f(x) = x^3

f(x)=3x2f'(x) = 3x^2

f(x)=6xf''(x) = 6x

Set f(x)=0f''(x) = 0: 6x=06x = 0, so x=0x = 0

Check signs:

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

Since concavity changes, (0,0)(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)=0f'(c) = 0 (critical point):

  • If f(c)>0f''(c) > 0local minimum at x=cx = c (concave up)
  • If f(c)<0f''(c) < 0local maximum at x=cx = c (concave down)
  • If f(c)=0f''(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=e3xy = e^{3x}

y=3e3xy' = 3e^{3x}

y=9e3xy'' = 9e^{3x}

y=27e3xy''' = 27e^{3x}

Pattern: y(n)=3ne3xy^{(n)} = 3^n e^{3x}

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

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

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

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

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


Special Patterns

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

f(k)(x)=n(n1)(n2)(nk+1)xnkf^{(k)}(x) = n(n-1)(n-2)\cdots(n-k+1)x^{n-k}

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

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

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

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

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

The derivatives cycle with period 4:

  • f(4m)(x)=k4msin(kx)f^{(4m)}(x) = k^{4m}\sin(kx)
  • f(4m+1)(x)=k4m+1cos(kx)f^{(4m+1)}(x) = k^{4m+1}\cos(kx)
  • f(4m+2)(x)=k4m+2sin(kx)f^{(4m+2)}(x) = -k^{4m+2}\sin(kx)
  • f(4m+3)(x)=k4m+3cos(kx)f^{(4m+3)}(x) = -k^{4m+3}\cos(kx)

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

f(x)=1x=x1f'(x) = \frac{1}{x} = x^{-1}

f(x)=x2f''(x) = -x^{-2}

f(x)=2x3f'''(x) = 2x^{-3}

f(4)(x)=6x4f^{(4)}(x) = -6x^{-4}

Pattern: f(n)(x)=(1)n1(n1)!xnf^{(n)}(x) = \frac{(-1)^{n-1}(n-1)!}{x^n}


⚠️ Common Mistakes

Mistake 1: Wrong Notation

d2ydx2=(dydx)2\frac{d^2y}{dx^2} = \left(\frac{dy}{dx}\right)^2 (WRONG!) ✅ d2ydx2=ddx[dydx]\frac{d^2y}{dx^2} = \frac{d}{dx}\left[\frac{dy}{dx}\right] (second derivative)

Mistake 2: Sign Errors

When differentiating f(x)f''(x) to get f(x)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)=0f''(c) = 0 doesn't guarantee an inflection point - concavity must change!


Applications

Physics - Motion

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

Economics

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

Curve Sketching

Use f(x)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 d2ydx2\frac{d^2y}{dx^2} with (dydx)2\left(\frac{dy}{dx}\right)^2 - they're completely different!

📚 Practice Problems

1Problem 1easy

Question:

Find the second derivative of f(x)=1x2f(x) = \frac{1}{x^2}.

💡 Show Solution

Step 1: Rewrite using negative exponents

f(x)=1x2=x2f(x) = \frac{1}{x^2} = x^{-2}


Step 2: Find the first derivative

Using the Power Rule:

f(x)=2x3=2x3f'(x) = -2x^{-3} = -\frac{2}{x^3}


Step 3: Find the second derivative

Differentiate f(x)=2x3f'(x) = -2x^{-3}:

f(x)=2(3)x4=6x4f''(x) = -2 \cdot (-3)x^{-4} = 6x^{-4}


Step 4: Rewrite in fraction form

f(x)=6x4f''(x) = \frac{6}{x^4}

Answer: f(x)=6x4f''(x) = \frac{6}{x^4} or 6x46x^{-4}

2Problem 2medium

Question:

A particle moves along a line with position function s(t)=t36t2+9ts(t) = t^3 - 6t^2 + 9t meters at time tt seconds. Find the velocity and acceleration at t=2t = 2 seconds.

💡 Show Solution

Step 1: Find velocity (first derivative of position)

s(t)=t36t2+9ts(t) = t^3 - 6t^2 + 9t

v(t)=s(t)=3t212t+9v(t) = s'(t) = 3t^2 - 12t + 9


Step 2: Evaluate velocity at t=2t = 2

v(2)=3(2)212(2)+9v(2) = 3(2)^2 - 12(2) + 9

=3(4)24+9= 3(4) - 24 + 9

=1224+9= 12 - 24 + 9

=3= -3 meters/second


Step 3: Find acceleration (second derivative of position)

a(t)=s(t)=v(t)=6t12a(t) = s''(t) = v'(t) = 6t - 12


Step 4: Evaluate acceleration at t=2t = 2

a(2)=6(2)12=1212=0a(2) = 6(2) - 12 = 12 - 12 = 0 meters/second²


Interpretation:

  • At t=2t = 2 seconds, the particle is moving at 3-3 m/s (moving backward/left)
  • The acceleration is 00 m/s² (velocity is neither increasing nor decreasing at this instant)

Answer: Velocity = 3-3 m/s, Acceleration = 00 m/s²

3Problem 3hard

Question:

Find all inflection points of f(x)=x44x3f(x) = x^4 - 4x^3.

💡 Show Solution

Step 1: Find the first derivative

f(x)=x44x3f(x) = x^4 - 4x^3

f(x)=4x312x2f'(x) = 4x^3 - 12x^2


Step 2: Find the second derivative

f(x)=12x224xf''(x) = 12x^2 - 24x


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

12x224x=012x^2 - 24x = 0

12x(x2)=012x(x - 2) = 0

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


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

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

Test x=1x = -1 (in (,0)(-\infty, 0)): f(1)=12(1)224(1)=12+24=36>0f''(-1) = 12(-1)^2 - 24(-1) = 12 + 24 = 36 > 0 ✓ concave up

Test x=1x = 1 (in (0,2)(0, 2)): f(1)=12(1)224(1)=1224=12<0f''(1) = 12(1)^2 - 24(1) = 12 - 24 = -12 < 0 ✓ concave down

Test x=3x = 3 (in (2,)(2, \infty)): f(3)=12(3)224(3)=10872=36>0f''(3) = 12(3)^2 - 24(3) = 108 - 72 = 36 > 0 ✓ concave up


Step 5: Identify inflection points

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

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


Step 6: Find y-coordinates

f(0)=044(0)3=0f(0) = 0^4 - 4(0)^3 = 0

f(2)=244(2)3=1632=16f(2) = 2^4 - 4(2)^3 = 16 - 32 = -16

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