What does the second derivative tell you?

The second derivative describes the concavity of the function: whether the curve bends upward or downward. Where f''(x) > 0, the function is concave up; where f''(x) < 0, it is concave down. Points where f''(x) = 0 often mark inflection points where concavity changes.

How many times do you differentiate to get the third derivative?

You differentiate the original function three times. The first derivative is f'(x), the second is f''(x) = \frac{d}{dx}f'(x), and the third is f'''(x) = \frac{d}{dx}f''(x). Each differentiation applies the power rule, chain rule, or other rules to the previous result.

Can a function have a second derivative if it is not differentiable everywhere?

No. If a function is not differentiable at a point, it cannot have a first derivative there, which means it cannot have a second derivative either. Higher order derivatives require all lower derivatives to exist, so a second derivative requires the function and its first derivative to both be differentiable.

Why do physicists use the second derivative?

In physics, if s(t) is position, then velocity is v(t) = s'(t) and acceleration is a(t) = s''(t). The second derivative directly gives acceleration, which is essential for applying Newton's second law (F = ma) and understanding motion and forces.

Higher Order Derivatives

Q: How many times do you differentiate to get the third derivative?

You differentiate the original function three times. The first derivative is f'(x), the second is f''(x) = \frac{d}{dx}f'(x), and the third is f'''(x) = \frac{d}{dx}f''(x). Each differentiation applies the power rule, chain rule, or other rules to the previous result.

Q: Can a function have a second derivative if it is not differentiable everywhere?

No. If a function is not differentiable at a point, it cannot have a first derivative there, which means it cannot have a second derivative either. Higher order derivatives require all lower derivatives to exist, so a second derivative requires the function and its first derivative to both be differentiable.

Q: Why do physicists use the second derivative?

In physics, if s(t) is position, then velocity is v(t) = s'(t) and acceleration is a(t) = s''(t). The second derivative directly gives acceleration, which is essential for applying Newton's second law (F = ma) and understanding motion and forces.

Higher order derivatives describe how the rate of change is changing; use them to find concavity and acceleration in functions.

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

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What each symbol means

What Higher Order Derivatives takes

Symbol	Meaning
$f$	The function being differentiated repeatedly; it represents the relationship between variables whose rate of change (and how that rate changes) you are analyzing.
$x$	The independent variable that you differentiate with respect to; confusing $x$ with constants or other variables causes errors when finding higher order derivatives.

When to use it

Reach for higher order derivatives when you need to understand concavity, acceleration, or how the rate of change itself is behaving.

Level

Usually taught in: Calculus I

Worked examples

1. Find the second derivative of a polynomial

Problem

Find

f''(x)

for

f(x) = x^3 + 2x^2 - 5x + 3

$f'(x) = \frac{d}{dx}(x^3 + 2x^2 - 5x + 3)$
Set up the first derivative using the power rule.
$f'(x) = 3x^2 + 4x - 5$
Apply the power rule term by term: the exponent moves down as a coefficient, and we subtract $1$ from each exponent.
$f''(x) = \frac{d}{dx}(3x^2 + 4x - 5)$
Now differentiate $f'(x)$ to find the second derivative.
$f''(x) = 6x + 4$
Apply the power rule again: $3x^2$ becomes $6x$ , $4x$ becomes $4$ , and the constant $-5$ becomes $0$ .

Answer: $f''(x) = 6x + 4$

Finding the second derivative involves applying the power rule twice—once to get $f'(x)$ , then again to get $f''(x)$ . Since this polynomial has simple powers of $x$ , we apply the power rule directly at each stage without needing other differentiation rules.

2. Find the second derivative with fractional and negative exponents

Problem

Find

f''(x)

for

f(x) = 4x^{3/2} - 2x^{-1}

$f'(x) = \frac{d}{dx}(4x^{3/2} - 2x^{-1})$
Set up the first derivative.
$f'(x) = 4 \cdot \frac{3}{2}x^{1/2} + 2x^{-2}$
Apply the power rule: for each term, multiply by the exponent and subtract $1$ from the exponent. For $-2x^{-1}$ , the exponent $-1$ becomes the coefficient: $-2 \cdot (-1) = +2$ .
$f'(x) = 6x^{1/2} + 2x^{-2}$
Simplify: $4 \cdot \frac{3}{2} = 6$ and the exponent becomes $\frac{1}{2}$ ; the second term has exponent $-1 - 1 = -2$ .
$f''(x) = \frac{d}{dx}(6x^{1/2} + 2x^{-2})$
Differentiate $f'(x)$ to find the second derivative.
$f''(x) = 6 \cdot \frac{1}{2}x^{-1/2} - 4x^{-3}$
Apply the power rule again: $6x^{1/2}$ gives $6 \cdot \frac{1}{2} = 3$ with exponent $-\frac{1}{2}$ ; for $2x^{-2}$ , we get $2 \cdot (-2) = -4$ with exponent $-3$ .
$f''(x) = 3x^{-1/2} - 4x^{-3}$
Final form: this can also be written as $f''(x) = \frac{3}{\sqrt{x}} - \frac{4}{x^3}$ .

Answer: $f''(x) = 3x^{-1/2} - 4x^{-3}$

This example shows that the power rule works with fractional and negative exponents, but you must carefully handle the algebra. Fractional exponents require careful simplification after each differentiation step. The power rule applies the same way: the exponent becomes a coefficient, and you subtract $1$ from the exponent.

3. Find the second derivative from a basketball scoring model

Problem

A basketball player's average points per game during a season is modeled by

P(t) = -0.1t^2 + 3t + 8

, where

t

is the number of games played. Find

P''(t)

$P'(t) = \frac{d}{dt}(-0.1t^2 + 3t + 8)$
Set up the first derivative, which represents the rate of change of average points.
$P'(t) = -0.2t + 3$
Apply the power rule: $-0.1 \cdot 2t = -0.2t$ , the constant $8$ becomes $0$ , and $3t$ becomes $3$ .
$P''(t) = \frac{d}{dt}(-0.2t + 3)$
Differentiate $P'(t)$ to find the second derivative, which represents how the rate of change is changing.
$P''(t) = -0.2$
Apply the power rule: $-0.2t$ becomes $-0.2$ , and the constant $3$ becomes $0$ .

Answer: $P''(t) = -0.2$

The second derivative is a negative constant, meaning the rate of change of scoring is constantly decreasing by the same amount each game. This tells us the player's average points per game is improving, but at a slower and slower rate throughout the season.

Common mistakes

Where Higher Order Derivatives usually goes wrong

The mistake	Why it is wrong	The fix
For $f(x) = x^3$ , writing $f''(x) = 3x^2$ .	This is only the first derivative; a second derivative requires differentiating twice, not once.	First, $f'(x) = 3x^2$ by the power rule. Then differentiate again: $f''(x) = 6x$ .
If $f'(x) = 2x + 5$ , writing $f''(x) = 2x + 5$ .	The constant term $5$ in the first derivative has no $x$ , so its derivative is $0$ , not itself.	Differentiate $f'(x) = 2x + 5$ to get $f''(x) = 2 + 0 = 2$ , not $2x + 5$ .
Confusing $f''(x)$ with $(f'(x))^2$ —for example, writing $f''(x) = (2x)^2 = 4x^2$ when $f(x) = x^2$ .	The notation $f''(x)$ means the derivative of the derivative, not the square of the first derivative; these are completely different operations.	Remember that $f''(x) = \frac{d}{dx}[f'(x)] = \frac{d}{dx}[2x] = 2$ , not $(2x)^2 = 4x^2$ .

Tips and when to use something else

The second derivative tells you concavity: if $f''(x) > 0$ , the curve bends upward (concave up); if $f''(x) < 0$ , it bends downward (concave down). Points where $f''(x) = 0$ are candidates for inflection points.
For piecewise or absolute value functions where the second derivative may not exist at certain points, use the First Derivative Test instead—it requires only that $f'(x)$ changes sign.
Many functions (especially trigonometric ones) have repeating patterns in their derivatives—for example, derivatives of $\sin(x)$ repeat every four steps—so recognize these cycles rather than computing every derivative.
To determine where a curve is concave up or down, you only need to test the sign of $f''(x)$ at sample points in each region; computing the exact value everywhere is unnecessary.

Related formulas

Power RuleThe Power Rule states that the derivative of x^n is nx^{n-1}, making it the essential tool for differentiating any polynomial expression.Critical PointsCritical points mark where a function's derivative is zero or undefined, revealing local extrema and where the graph changes direction.Second Derivative TestThe Second Derivative Test determines whether a critical point is a local minimum or maximum using the concavity at that point.Concavity and Inflection PointsPinpoint where a function's curve changes from bending upward to downward, revealing the transition point between concave up and concave down.Chain RuleThe Chain Rule tells you how to differentiate a composite function—when one function is plugged into another—by multiplying the outer and inner derivatives.

Frequently asked questions

What does the second derivative tell you?: The second derivative describes the concavity of the function: whether the curve bends upward or downward. Where $f''(x) > 0$ , the function is concave up; where $f''(x) < 0$ , it is concave down. Points where $f''(x) = 0$ often mark inflection points where concavity changes.
How many times do you differentiate to get the third derivative?: You differentiate the original function three times. The first derivative is $f'(x)$ , the second is $f''(x) = \frac{d}{dx}f'(x)$ , and the third is $f'''(x) = \frac{d}{dx}f''(x)$ . Each differentiation applies the power rule, chain rule, or other rules to the previous result.
Can a function have a second derivative if it is not differentiable everywhere?: No. If a function is not differentiable at a point, it cannot have a first derivative there, which means it cannot have a second derivative either. Higher order derivatives require all lower derivatives to exist, so a second derivative requires the function and its first derivative to both be differentiable.
Why do physicists use the second derivative?: In physics, if $s(t)$ is position, then velocity is $v(t) = s'(t)$ and acceleration is $a(t) = s''(t)$ . The second derivative directly gives acceleration, which is essential for applying Newton's second law ( $F = ma$ ) and understanding motion and forces.

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Reviewed 2026-09-18