The Second Derivative
The first derivative tells you the gradient. The second derivative tells you how the gradient itself is changing: whether the curve is bending upward or downward, like a road that transitions from uphill to steeper uphill.
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Three printable worksheets that build from foundations to mastery — or build your own from any module’s questions.
Before we start, what do you already know about this topic?
Key facts
- Find the second derivative $f''(x)$ or $\frac{d^2y}{dx^2}$
- The notation for first and second derivatives
Concepts
- Interpret the second derivative as the rate of change of the gradient
- Use the second derivative to determine concavity
Skills
- Relate the second derivative to acceleration in kinematics
- Find and verify points of inflection
If the first derivative measures how a function is changing, the second derivative measures how that rate of change is itself changing. In motion, if position is $s(t)$, velocity is $s'(t)$ and acceleration is $s''(t)$. On a graph, $f''(x) > 0$ means the curve bends upward like a cup; $f''(x) < 0$ means it bends downward like a cap.
The second derivative $f''(x)$ is found by differentiating $f'(x)$ again; Concave up: $f''(x) > 0$ — curve bends like a cup (U-shape)
Pause — copy the second derivative definition ($f''(x)$ = differentiate $f'(x)$ again), the concavity rules ($f''(x) > 0$ = concave up/cup, $f''(x) < 0$ = concave down), and the point-of-inflection test into your book.
Did you get this? True or false: if $f''(x) > 0$ at a point, the curve is concave down at that point.
Quick check: Which notation represents the second derivative?
We just saw that $f''(x)$ is found by differentiating $f'(x)$ again, and $f''(x) = 0$ locates possible inflection points. That raises a question: how does this work in practice on a polynomial? This card answers it → differentiating $3x^4 - 2x^3 + x$ twice and factorising to locate where concavity may change.
Step 1: Find $f'(x)$ using the power rule on each term; Step 2: Differentiate $f'(x)$ again to get $f''(x)$
Pause — copy the two-step second derivative procedure (differentiate once, then differentiate again) with the worked solution for $f(x) = 3x^4 - 2x^3 + x$ into your book.
Quick check: If $f(x) = x^4$, what is $f''(x)$?
We just saw how to find $f''(x)$ from a polynomial. That raises a question: once we have $f''(x) = 0$, how do we confirm whether that point is actually an inflection point (and find its coordinates)? This card answers it → solving $f''(x) = 0$, verifying the sign of $f''$ changes, then substituting into $f(x)$ for the $y$-coordinate.
To find a point of inflection: solve $f''(x) = 0$, then verify the sign of $f''(x)$ changes; Always find the $y$-coordinate by substituting back into $f(x)$ (not $f'$ or $f''$)
Pause — copy the inflection-point procedure: solve $f''(x) = 0$, verify sign change, then substitute into $f(x)$ (not $f'$ or $f''$) for the $y$-coordinate into your book.
Did you get this? True or false: to confirm a point of inflection, it is enough to just show $f''(x) = 0$.
We just saw how to locate and confirm inflection points using the second derivative sign test. That raises a question: does the second derivative have applications beyond curve sketching — does it connect to real-world quantities? This card answers it → in kinematics, $s''(t)$ is acceleration: differentiating position twice gives the rate at which velocity is changing.
Kinematics chain: $s(t)$ (position) → $s'(t) = v(t)$ (velocity) → $s''(t) = a(t)$ (acceleration); Each differentiation step uses the same power rule
Pause — copy the kinematics differentiation chain: $s(t) \to s'(t) = v(t) \to s''(t) = a(t)$, and note that each step uses the power rule into your book.
Quick check: For $s(t) = t^2 + 4t$, what is the acceleration $a(t)$?
Common errors — the 3 traps that cost marks
Quick-fire practice — 5 problems +2 XP per reveal
Find $f''(x)$ for $f(x) = x^5 - 4x^3 + 2x$.
Find $\frac{d^2y}{dx^2}$ for $y = \sin x$.
Determine the concavity of $y = x^4 - 2x^2$ at $x = 1$.
Find the point of inflection of $y = x^3 - 3x^2$.
If $s(t) = 2t^3 - t^2$, find the acceleration at $t = 1$.
Earlier you were asked what you know about rates of change of rates of change. The second derivative captures exactly this: how the slope itself is changing. A positive second derivative means the curve is getting steeper in the positive direction.
Pick your answer, then rate your confidence — that tells the system what to drill next. Each retry pulls a fresh mix from the bank.
Find and classify the second derivative
For $f(x) = 2x^3 - 9x^2 + 12x$, find $f''(x)$ and determine where the curve is concave up. [3 marks]
View comprehensive answer
$f'(x) = 6x^2 - 18x + 12$ [0.5]
$f''(x) = 12x - 18$ [0.5]
Concave up when $f''(x) > 0$, so $12x - 18 > 0$, giving $x > \frac{3}{2}$ [2]
Point of inflection with verification
Show that $y = x^4 - 4x^3$ has a point of inflection at $x = 2$. [3 marks]
View comprehensive answer
$\frac{dy}{dx} = 4x^3 - 12x^2$ and $\frac{d^2y}{dx^2} = 12x^2 - 24x$ [0.5]
Set $\frac{d^2y}{dx^2} = 0$: $12x(x-2) = 0$, so $x = 0$ or $x = 2$ [0.5]
At $x = 2$: for $x = 1$, $f''(1) = -12 < 0$; for $x = 3$, $f''(3) = 36 > 0$. Sign changes, so inflection at $x = 2$ [2]
Kinematics application
A particle moves with displacement $s(t) = t^3 - 6t^2 + 9t$ metres. Find the time when the acceleration is zero and the velocity at that time. [4 marks]
View comprehensive answer
$v(t) = 3t^2 - 12t + 9$ [0.5]
$a(t) = 6t - 12$ [0.5]
Set $a(t) = 0$: $6t - 12 = 0$, so $t = 2$ s [1]
$v(2) = 12 - 24 + 9 = -3$ m/s [2]
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