CalculusConcept reference

Second derivatives

The second derivative is the derivative of the first derivative.

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Overview
The first derivative is itself a function
For f(x)=2x^3-3x^2, the first derivative is 6x^2-6x
A nonzero first derivative can have zero derivative
Key takeaway
Sources & further reading
Concept connections

01The first derivative is itself a function#

The first derivative is itself a function. Its derivative is the second derivative: it measures how the first rate changes with the input.

f\prime\prime=(f\prime)\prime

For $f(x)=x^3+2x^2$ , differentiate every term once, then differentiate the resulting function.

Figure 1For f=x³+2x², f′=3x²+4x and f″=6x+4. At x=0 the tangent slope is 0; at x=1 it is 7. On [0,1], f″ is positive, so the slope increases.

Check your reasoning

$f(x)=x^{4}$ . Find $f\prime\prime(x)$ .

A $4x^{3}$
B $12x^{2}$
C $16x^{6}$

Show answer and explanation

$12x^{2}$

Differentiate once to get $4x^{3}$ , then again to get $12x^{2}$ .

02For f(x)=2x^3-3x^2, the first derivative is 6x^2-6x#

For $f(x)=2x^3-3x^2$ , the first derivative is $6x^2-6x$ . Keep those coefficients in the second step.

f\prime\prime(x)=12x-6

At $x=2$ , the second derivative is $18$ . Differentiate first, then evaluate.

If $s(x)$ is position and $x$ is time, $s\prime(x)$ is velocity and $s\prime\prime(x)$ is acceleration. With meters and seconds, their units are meters per second and meters per second squared.

Check your reasoning

Position in meters is $s(x)=x^{3}+2x$ , where $x$ is time in seconds. Acceleration is $s\prime\prime(x)$ . Find acceleration at $x=2$ , in $\mathrm{m/s^2}$ .

Show answer and explanation

Differentiate position twice: $s\prime\prime(x)=6x$ . At $2$ , this is $12$ .

03A nonzero first derivative can have zero derivative#

A nonzero first derivative can have zero derivative. For $f(x)=4x-7$ , the first derivative is the constant $4$ , so the second derivative is $0$ .

Two primes mean differentiate twice. They do not mean $(f\prime)^2$ .

Check your reasoning

$f(x)=5x+3$ . A learner squares the first derivative. Repair $f\prime\prime(x)$ .

A $5$
B $25$
C $0$

Show answer and explanation

$0$

Differentiate once to get $5$ , then again to get $0$ .

Key takeaway

Find the first derivative, then differentiate that function again.

Compute the second derivative of a scalar function.

Sources & further reading

[1]
OpenStax Calculus Volume 1, 3.2: Higher-order derivatives ↗OpenStax Calculus Volume 1 · Book

01The first derivative is itself a function#

02For f(x)=2x^3-3x^2, the first derivative is 6x^2-6x#

03A nonzero first derivative can have zero derivative#

Find the first derivative, then differentiate that function again.

Sources & further reading

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