Calculus Differentiation

Higher-order derivatives

Q: What is a second derivative?

The derivative of the first derivative, written f double prime of x. It measures how the slope is changing and controls concavity.

Q: What does the second derivative tell you?

Concavity: where the second derivative is positive the graph is concave up, where negative it is concave down. In motion it is acceleration.

Q: How is acceleration a derivative?

Acceleration is the derivative of velocity, and velocity is the derivative of position, so acceleration is the second derivative of position, s double prime of t.

Q: Do the derivatives of sin x repeat?

Yes. Sine goes to cosine to negative sine to negative cosine back to sine, a cycle of length four. And e to the x stays e to the x at every order.

20 practice questions 0 video lessons Theory + worked examples

Higher-Order Derivatives

California Calculus • Standard 7.0 • Differentiation

Higher-Order Derivatives is a topic in Differentiation in the California Calculus Standards. It is aligned to Standard 7.0, which requires students to compute derivatives of higher orders.

A higher-order derivative is the derivative of a derivative. The second derivative \(f''(x)\) measures how the slope is changing — it gives concavity, and for motion it is acceleration.

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Theory

A higher-order derivative is the derivative of a derivative. The second derivative \(f''(x)\) measures how the slope itself is changing (concavity), and for motion it is acceleration. You can keep differentiating to get \(f'''\), \(f''''\), and beyond.

Differentiating a function gives \(f'(x)\); differentiating that gives the second derivative \(f''(x)\), and so on:

\[f^{\prime\prime}(x)=\dfrac{d}{dx}f^{\prime}(x),\qquad f^{\prime\prime\prime}(x)=\dfrac{d}{dx}f^{\prime\prime}(x)\]

The second derivative describes the rate of change of the slope — it controls concavity (up or down).

In motion, if \(s(t)\) is position then \(s'(t)\) is velocity and \(s''(t)\) is acceleration.

Key idea: the derivatives of \(\sin x\), \(\cos x\), and \(e^{x}\) repeat in a pattern — sine returns to itself every four derivatives, and \(e^{x}\) never changes.

A cubic \(f\) and its second derivative \(f''\) (a line).

Position \(s(t)\) and velocity \(v=s'\); acceleration is \(s''\).

The notation for higher derivatives:

\[f^{\prime\prime}(x),\ f^{\prime\prime\prime}(x),\ f^{(4)}(x),\ \dots,\ f^{(n)}(x)\]

second derivative, third derivative, and the n-th derivative

Motion (US units):

\[v(t)=s^{\prime}(t)\ \text{(ft/s)},\qquad a(t)=s^{\prime\prime}(t)\ \text{(ft/s}^2)\]

velocity is the first derivative and acceleration the second derivative of position

Repeating patterns: \(\dfrac{d}{dx}e^{x}=e^{x}\) forever, and the sine derivatives cycle \(\sin\to\cos\to-\sin\to-\cos\to\sin\).

How to find a higher-order derivative

Differentiate to get \(f'(x)\), simplifying fully.
Differentiate again for \(f''(x)\); repeat for each higher order.
Substitute a value only at the end if you need \(f''(a)\) or an acceleration.

Example 1 — Second and third derivatives

For \(f(x)=x^4\), find \(f''(x)\) and \(f'''(x)\).

Solution

Differentiate repeatedly with the power rule.

\(f'(x)\)	\(=\)	\(4x^3\)
\(f''(x)\)	\(=\)	\(12x^2\)
\(f'''(x)\)	\(=\)	\(24x\)

second derivative 12 x squared, third derivative 24x

Example 2 — Evaluate a second derivative

For \(f(x)=x^3-2x^2+5x\), find \(f''(x)\).

Solution

Differentiate twice.

\(f'(x)\)	\(=\)	\(3x^2-4x+5\)
\(f''(x)\)	\(=\)	\(6x-4\)

second derivative is 6x minus 4

Example 3 — Acceleration

An object has position \(s(t)=t^3\) ft. Find its acceleration at \(t=2\) s.

Solution

Acceleration is the second derivative, \(a(t)=s''(t)\).

\(v(t)=s'(t)\)	\(=\)	\(3t^2\)
\(a(t)=s''(t)\)	\(=\)	\(6t\)
\(a(2)\)	\(=\)	\(12\)

The acceleration is \(12\) ft/s\(^2\).

acceleration equals 12 feet per second squared

Example 4 — A repeating pattern

Find the fourth derivative of \(f(x)=\sin x\).

Solution

The derivatives of sine cycle every four steps.

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

fourth derivative of sine is sine again

Common pitfalls

Simplify before differentiating again. Carrying an unsimplified \(f'\) into the next step multiplies the chance of an algebra slip.

Acceleration is \(s''\), not \(s'\). Velocity is the first derivative; acceleration is the second.

Powers drop fast. Each derivative of a polynomial lowers the degree by one, so a high enough derivative of a polynomial is eventually \(0\).

Frequently asked questions

What is a second derivative?

The derivative of the first derivative, written \(f''(x)\). It measures how the slope is changing and controls concavity.

What does the second derivative tell you?

Concavity: where \(f''>0\) the graph is concave up, where \(f''<0\) it is concave down. In motion it is acceleration.

How is acceleration a derivative?

Acceleration is the derivative of velocity, and velocity is the derivative of position, so acceleration is the second derivative of position, \(s''(t)\).

Do the derivatives of sin x repeat?

Yes. \(\sin x\to\cos x\to-\sin x\to-\cos x\to\sin x\), a cycle of length four. And \(e^{x}\) stays \(e^{x}\) at every order.

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Higher-order derivatives

Higher-Order Derivatives

📖 Theory

How to find a higher-order derivative

Common pitfalls

Frequently asked questions

Theory