Second Derivative Calculator

Q: What does the second derivative tell you?

How the slope is changing. f''(x) > 0 means the slope is increasing (concave up); f''(x) < 0 means it is decreasing (concave down). In physics, if f is position, f'' is acceleration.

Q: How do you find the second derivative?

Differentiate once to get f'(x), then differentiate that result again. The calculator shows both rounds of steps.

Q: What is the notation?

f''(x), \frac{d^2y}{dx^2}, y'' and D^2 f all mean the same thing.

Q: How is it used to classify maxima and minima?

At a critical point c (where f'(c) = 0): f''(c) > 0 is a local minimum, f''(c) < 0 is a local maximum. See the critical points calculator.

Differentiate twice (or up to five times) with every step, evaluate at a point, and read concavity straight off the graph.

Function f(x)

Reads as

x^{3} - 3 x + \ln\left(x\right)

Try:

OrderEvaluate at x =

Answer

f''(x) = 6 x - \frac{1}{x^{2}}

$f''(1) = 5$ · concave up here

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Step by step

Derivative #1
$\frac{d}{dx}\left[x^{3} - 3 x + \ln\left(x\right)\right]$
Sum rule
Differentiate term by term.
$\frac{d}{dx}\left[x^{3} - 3 x + \ln\left(x\right)\right] = \frac{d}{dx}\left[-3 x\right] + \frac{d}{dx}\left[x^{3}\right] + \frac{d}{dx}\left[\ln\left(x\right)\right]$
Constant multiple rule
Pull the constant factor out front.
$\frac{d}{dx}\left[-3 x\right] = -3 \cdot \frac{d}{dx}\left[x\right]$
Power rule
d/dx[xⁿ] = n·xⁿ⁻¹.
$\frac{d}{dx}\left[x^{3}\right] = 3 x^{2}$
Logarithm rule
$\frac{d}{dx}\left[\ln\left(x\right)\right] = \frac{1}{x}$
Result
$f^{(1)}(x) = 3 x^{2} + \frac{1}{x} - 3$
Derivative #2
$\frac{d}{dx}\left[3 x^{2} + \frac{1}{x} - 3\right]$
Sum rule
Differentiate term by term.
$\frac{d}{dx}\left[3 x^{2} + \frac{1}{x} - 3\right] = \frac{d}{dx}\left[3 x^{2}\right] + \frac{d}{dx}\left[\frac{1}{x}\right] + \frac{d}{dx}\left[-3\right]$
Constant multiple rule
Pull the constant factor out front.
$\frac{d}{dx}\left[3 x^{2}\right] = 3 \cdot \frac{d}{dx}\left[x^{2}\right]$
Power rule
d/dx[xⁿ] = n·xⁿ⁻¹.
$\frac{d}{dx}\left[x^{2}\right] = 2 x$
Power rule
d/dx[xⁿ] = n·xⁿ⁻¹.
$\frac{d}{dx}\left[\frac{1}{x}\right] = -\frac{1}{x^{2}}$
Result
$f^{(2)}(x) = 6 x - \frac{1}{x^{2}}$

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An AI tutor reads the problem and steps above and explains them in plain words. It can make mistakes, so check it against the worked steps.

Reading the three curves together

Where $f''$ (amber) is above the axis, $f'$ (grey) is rising and $f$ bends upward. Where $f''$ crosses zero, $f'$ peaks or bottoms out and $f$ has an inflection point. In motion problems: position $s(t)$ , velocity $s'(t)$ , acceleration $s''(t)$ .

Questions students ask

What does the second derivative tell you?

How the slope is changing. $f''(x) > 0$ means the slope is increasing (concave up); $f''(x) < 0$ means it is decreasing (concave down). In physics, if $f$ is position, $f''$ is acceleration.

How do you find the second derivative?

Differentiate once to get $f'(x)$ , then differentiate that result again. The calculator shows both rounds of steps.

What is the notation?

$f''(x)$ , $\frac{d^2y}{dx^2}$ , $y''$ and $D^2 f$ all mean the same thing.

How is it used to classify maxima and minima?

At a critical point $c$ (where $f'(c) = 0$ ): $f''(c) > 0$ is a local minimum, $f''(c) < 0$ is a local maximum. See the critical points calculator.