Skip to main content
Mathovia

Calculus / Calculus 1 · Derivatives

Higher-Order Derivatives

A higher-order derivative is what you get by differentiating again. The second derivative measures how fast the slope is changing, which is why it tells you how a graph bends and why it's acceleration in physics.

Practice Problems
Written by
MPhil in Mathematics, Founder of Mathovia
Published

Mathematics verification: Mathematical formulas, calculations, worked examples and solutions are reviewed for accuracy before publication. How we check our work

Chapter 3 · Lesson 12 of 13Calculus 1 · Derivatives
  1. 1.Limit Definition of the Derivative
  2. 2.Differentiability
  3. 3.Power Rule for Derivatives
  4. 4.Product Rule for Derivatives
  5. 5.Quotient Rule for Derivatives
  6. 6.Chain Rule
  7. 7.Derivatives of Trig Functions
  8. 8.Derivatives of Exponential and Logarithmic Functions
  9. 9.Derivatives of Inverse Trig Functions
  10. 10.Implicit Differentiation
  11. 11.Logarithmic Differentiation
  12. 12.Higher-Order Derivatives
  13. 13.Tangent Lines and Normal Lines
See all calculus chapters →

The derivative of a function is another function, so you can differentiate it again. The result is the second derivative, and you can keep going to the third, fourth and beyond. These higher-order derivatives aren’t just repetition. The second derivative measures how the slope itself is changing, which tells you how a graph bends and gives you acceleration in physics. Later it powers the second derivative test, concavity and Taylor series.

The second derivative f double prime of x is the derivative of f prime of x, written d squared y over d x squared. The nth derivative f superscript n of x is written d to the n y over d x to the n
The second derivative is the derivative of the derivative

What Is the Second Derivative?

The second derivative of is the derivative of . You find it in two passes: differentiate once with the rules you already know, then differentiate the answer.

Take . The power rule gives

and differentiating again gives

Keep going and you get the third derivative , then the fourth derivative . Every derivative after that is 0 as well.

Notation for Higher-Order Derivatives

OrderPrime notationLeibniz notationAlso written
First
Second
Third
Fourth
nth

Two details trip people up:

  • Brackets after three primes. Writing is hard to read, so from the fourth derivative on you put the order in brackets: . The brackets matter, because means to the fourth power.
  • Where the 2 goes in . It’s applied twice, . The “d” on top is used twice, and on the bottom it’s twice, written . It is not , and it is not .

What the Second Derivative Tells You

Graphs of f of x equals x cubed minus 3x in blue, its derivative 3x squared minus 3 in orange, and its second derivative 6x in green. The orange curve crosses zero at x equals minus 1 and 1, exactly below the peak and valley of the blue curve. The green line crosses zero at x equals 0, where the blue curve switches from bending down to bending up
f′ is zero at the peak and valley; f″ is zero where the bend flips

The first derivative is the slope. The second derivative is the rate of change of the slope: whether the slope is increasing or decreasing as you move right.

  • : the slope is increasing. The graph bends upward like a cup. This is called concave up.
  • : the slope is decreasing. The graph bends downward like a cap. This is concave down.
  • changes sign: the bend flips. That point is an inflection point.

Look at in the graph. For , is negative and the blue curve is shaped like a cap, with its peak at . For , is positive and the curve is a cup, with its valley at . At the bend switches.

The first derivative finds where the curve turns around ( at ). The second derivative tells you which way it bends there: , so is a peak; , so is a valley. That’s the idea behind the second derivative test, which you’ll use for optimization later in the course.

Position, Velocity and Acceleration

Position s of t equals t cubed minus 6 t squared plus 9t in blue, velocity in orange and acceleration in green, for t from 0 to about 4. The velocity is zero at t equals 1 and t equals 3, where the position curve turns around. The acceleration is zero at t equals 2
Velocity is the first derivative of position; acceleration is the second

If is the position of an object at time , then

Velocity is how fast the position changes. Acceleration is how fast the velocity changes. So yes, the second derivative of position is acceleration.

A particle moves along a line with meters after seconds.

Step 1 — velocity, the first derivative:

Step 2 — acceleration, the second derivative:

Step 3 — read the motion off the signs. The particle stops () at and . At the acceleration is m/s², so it’s being pushed backward, and it reverses. At it’s moving forward at 9 m/s but m/s², so it’s slowing down.

A useful rule: an object speeds up when and have the same sign and slows down when they have opposite signs. Here, between and , both are negative, so it’s moving backward and speeding up.

Example 1: A Polynomial All the Way Down

Find every derivative of .

DerivativeResult

Each pass lowers the degree by one, so a polynomial of degree has a constant th derivative, and every derivative after that is zero. The constant is times the leading coefficient: here .

Example 2: The Derivatives of sin x Repeat

Four cards in a row: f of x equals sine x, f prime equals cosine x, f double prime equals minus sine x, f triple prime equals minus cosine x. An arrow loops from the last card back to the first: the fourth derivative is sine x again
Every fourth derivative of sin x brings you back to sin x

Using the derivatives of the trig functions:

The cycle has length 4. To find a high derivative, divide the order by 4 and look at the remainder. For the 99th derivative: , so it matches the third derivative, .

Notice that . Sine is a function whose second derivative is its own negative, and that’s why it describes springs and waves.

Example 3: The nth Derivative of e^(2x)

Find a formula for the th derivative of .

The chain rule brings out a factor of 2 every time:

The pattern is clear:

Example 4: Using the Product Rule Twice

Find for .

Step 1 — first derivative, by the product rule:

Step 2 — differentiate , again a product:

Simplifying before step 2 kept the second pass short. The pattern continues: .

Example 5: A Pattern With Factorials

Find the first four derivatives of , and the th.

Rewrite as so the power rule applies:

DerivativeResult

The signs alternate and the numbers are factorials, . So

Example 6: An Implicit Second Derivative

For the hyperbola , implicit differentiation gives , so . Find .

Step 1 — differentiate with the quotient rule, remembering depends on :

Step 2 — substitute :

Step 3 — use the original equation, , so :

Above the x-axis , so : those parts of the hyperbola bend down. Below the axis they bend up. The implicit differentiation lesson does the same calculation for a circle.

Common Mistakes

  • Squaring the first derivative. is not . For , , but .
  • Mixing up and . Brackets mean the fourth derivative; no brackets means the fourth power.
  • Forgetting the chain-rule factor on every pass. Each derivative of brings out another 2, so the third derivative is , not .
  • Treating as a constant in an implicit second derivative. It depends on , so substitute its formula in before simplifying.

Practice Problems

Try each one before opening the answer. The practice set has 12 more with instant checking.

Problem 1. Find for .

Show answer

Step 1 — first derivative:

Step 2 — differentiate again:

Answer:

Problem 2. Find for .

Show answer

Step 1 — write it as a power and differentiate:

Step 2 — differentiate again:

Answer:

Problem 3. Find the third derivative of .

Show answer

Each pass multiplies by :

Answer:

Problem 4. Find for .

Show answer

Step 1 — product rule:

Step 2 — product rule on the first term again:

Answer:

Problem 5. A particle has position . Find its acceleration at each time its velocity is zero.

Show answer

Step 1 — velocity:

So at and .

Step 2 — acceleration:

Step 3 — evaluate:

Answer: and

Problem 6. Find the 50th derivative of .

Show answer

The derivatives of cycle every 4: . Since , the 50th derivative matches the second.

Answer:

You can now differentiate a function as many times as you like. Next, the first derivative goes to work on geometry: finding the equation of the tangent line to any curve.

Frequently Asked Questions

What is the second derivative?+

The second derivative is the derivative of the derivative. If is the slope of , then is the rate at which that slope changes. For , and .

How do you find the second derivative?+

Find the first derivative with the usual rules, simplify it, then differentiate that result again. There is no separate formula: the second derivative of is just the first derivative of .

What does the second derivative tell you?+

It tells you how the graph bends. Where the slope is increasing and the graph is concave up, like a cup. Where the slope is decreasing and the graph is concave down, like a cap.

Is the second derivative acceleration?+

Yes, when the function is a position. If is position, then is velocity and is acceleration, the rate at which the velocity changes.

How do you write the second derivative?+

Common notations are , and . In the Leibniz form the 2 goes on top next to the and on the bottom after the , because it is applied twice to .

What is the nth derivative?+

The nth derivative, written or , is what you get after differentiating times. For example, every derivative of is , and the nth derivative of is .

Can the second derivative be zero when the first derivative isn't?+

Yes. For , but . The graph is still sloping down at ; it is just switching from bending down to bending up.

Related lessons

Power Rule for DerivativesChain RuleDerivatives of Trig FunctionsImplicit Differentiation