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Calculus Solver

Second Derivative Calculator

Enter a function to get its second derivative with every step shown: first f′(x), then f″(x). Use the result to see where a graph is concave up or down, locate inflection points, or classify critical points.

Your expression will be shown here as it will be read.

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What is the second derivative?

The second derivative is the derivative of the derivative. If f′(x) tells you the slope of a graph, f″(x) tells you how fast that slope is changing. A positive second derivative means the slope is increasing, so the graph bends upward; a negative one means it bends downward.

To find it, differentiate once, simplify, and differentiate again. Simplifying in between is not optional in practice: a tidy f′(x) can turn a long second differentiation into a short one.

f′′(x)=ddx[f′(x)]=d2ydx2f''(x) = \frac{d}{dx}\big[f'(x)\big] = \frac{d^{2}y}{dx^{2}}

What the second derivative tells you

The second derivative is used far more often than higher derivatives, because of these interpretations.

  • Concave up

    f′′(x)>0f''(x) > 0

    The graph bends upward like a cup, and its slope is increasing.

  • Concave down

    f′′(x)<0f''(x) < 0

    The graph bends downward like a cap, and its slope is decreasing.

  • Inflection point

    f′′ changes sign at x=cf''\ \text{changes sign at}\ x = c

    Where concavity switches. f″(c) = 0 alone is not enough: x⁴ has f″(0) = 0 but no inflection point.

  • Second derivative test

    f′(c)=0:f′′(c)>0⇒min⁡,f′′(c)<0⇒max⁡f'(c) = 0:\quad f''(c) > 0 \Rightarrow \min,\quad f''(c) < 0 \Rightarrow \max

    A fast way to classify critical points. If f″(c) = 0, the test is inconclusive.

  • Acceleration

    s(t)→s′(t)=v(t)→s′′(t)=a(t)s(t) \to s'(t) = v(t) \to s''(t) = a(t)

    For a position function, the second derivative is acceleration.

  • Notation

    f′′(x),y′′,d2ydx2,D2ff''(x),\quad y'',\quad \frac{d^{2}y}{dx^{2}},\quad D^{2}f

    All four mean the same thing. The exponent in d²y/dx² marks the order, not a power.

How this calculator works

The calculator differentiates your function once, showing every rule it uses, and simplifies the result exactly. It then differentiates that simplified first derivative again, with its own list of steps, so you can follow both passes separately.

The answer is checked by comparing it with a numerical finite-difference estimate of the first derivative's slope at several points. The Order menu goes up to the 10th derivative if you need third or higher derivatives.

For the full details, including how answers are verified, see how the solver works.

Worked examples

Work through these by hand first, then enter them above to compare each step.

Example 1

Polynomial and inflection points

f(x)=x4−4x3+2f(x) = x^{4} - 4x^{3} + 2

  1. First derivative by the power rule:

    f′(x)=4x3−12x2f'(x) = 4x^{3} - 12x^{2}

  2. Differentiate again:

    f′′(x)=12x2−24x=12x(x−2)f''(x) = 12x^{2} - 24x = 12x(x - 2)

  3. f″ changes sign at x = 0 and x = 2, so both are inflection points. The graph is concave up for x < 0 and x > 2, and concave down between.

Answer

f′′(x)=12x2−24xf''(x) = 12x^{2} - 24x

Example 2

Product rule twice

f(x)=x exf(x) = x\,e^{x}

  1. Product rule:

    f′(x)=ex+xex=(x+1)exf'(x) = e^{x} + x e^{x} = (x + 1)e^{x}

  2. Product rule again on (x + 1)eˣ:

    f′′(x)=ex+(x+1)exf''(x) = e^{x} + (x + 1)e^{x}

Answer

f′′(x)=(x+2) exf''(x) = (x + 2)\,e^{x}

Example 3

Chain rule, then the quotient rule

f(x)=ln⁡(x2+1)f(x) = \ln\left(x^{2} + 1\right)

  1. Chain rule:

    f′(x)=2xx2+1f'(x) = \frac{2x}{x^{2} + 1}

  2. Quotient rule on f′:

    f′′(x)=2(x2+1)−2x⋅2x(x2+1)2f''(x) = \frac{2\left(x^{2} + 1\right) - 2x\cdot 2x}{\left(x^{2} + 1\right)^{2}}

  3. Simplify the numerator: 2x² + 2 − 4x² = 2 − 2x².

Answer

f′′(x)=2−2x2(x2+1)2f''(x) = \frac{2 - 2x^{2}}{\left(x^{2} + 1\right)^{2}}

Example 4

Trigonometric function

f(x)=sin⁡(2x)f(x) = \sin(2x)

  1. Chain rule:

    f′(x)=2cos⁡(2x)f'(x) = 2\cos(2x)

  2. Chain rule again:

    f′′(x)=2⋅(−sin⁡(2x))⋅2f''(x) = 2\cdot\big(-\sin(2x)\big)\cdot 2

Answer

f′′(x)=−4sin⁡(2x)f''(x) = -4\sin(2x)

Example 5

Acceleration from position

s(t)=t3−6t2+9ts(t) = t^{3} - 6t^{2} + 9t

  1. Velocity is the first derivative:

    v(t)=s′(t)=3t2−12t+9v(t) = s'(t) = 3t^{2} - 12t + 9

  2. Acceleration is the second derivative:

    a(t)=s′′(t)=6t−12a(t) = s''(t) = 6t - 12

  3. The acceleration is zero at t = 2, negative before and positive after.

Answer

s′′(t)=6t−12s''(t) = 6t - 12

Common mistakes

  • Not simplifying f′ first

    Differentiating an unsimplified first derivative multiplies the work, and the chance of error, for the second pass.

  • Assuming f″ = 0 means an inflection point

    For x⁴, f″(0) = 0, yet the graph is concave up on both sides. Check that f″ actually changes sign.

  • Confusing concavity with increasing

    A function can be decreasing and concave up at the same time, like e^(−x). Concavity is about the slope's change, not its sign.

  • Squaring the first derivative

    f″(x) is not (f′(x))². The notation d²y/dx² records the order, not a power.

Limitations

  • The calculator returns f″(x); it does not list intervals of concavity automatically. Find where f″ is zero or undefined, then test its sign between those points.
  • Second derivatives of implicit equations (d²y/dx²) are not supported; the implicit differentiation calculator finds dy/dx.
  • Answers may be in an equivalent form to your textbook's, for example factored versus expanded.

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Frequently asked questions

How do you find the second derivative?

Differentiate the function once to get f′(x), simplify it, then differentiate f′(x) again. The calculator shows both passes with every rule used.

What does a negative second derivative mean?

The graph is concave down there: its slope is decreasing. At a critical point, a negative second derivative means a local maximum.

Can I find the third derivative or higher?

Yes. Choose a higher order from the Order menu; it goes up to the 10th derivative.

Is the second derivative the same as the derivative squared?

No. The second derivative differentiates twice. For f(x) = x³, f″(x) = 6x, while (f′(x))² = 9x⁴.

Next step

Pick one of the worked examples, solve it on paper, then enter it in the calculator and compare your steps with each rule shown. Spotting where your work differs is the fastest way to improve.

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