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.
Meet the destination, bring back what you already know, and gather the terms and formulas this lesson leans on.
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Work through the core explanation before applying it.
Meet the destination, bring back what you already know, and gather the terms and formulas this lesson leans on.
Work through the core explanation before applying it.
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
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Quick-fire practice, 5 problems +2 XP per reveal
Run the quick drill and copy the summary into your book.
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.
Answer the drill bank and rate your confidence.
Pick your answer, then rate your confidence, that tells the system what to drill next. Each retry pulls a fresh mix from the bank.
Write full responses, then check them against the model answers.
For $f(x) = 2x^3 - 9x^2 + 12x$, find $f''(x)$ and determine where the curve is concave up. [3 marks]
$f'(x) = 6x^2 - 18x + 12$ [0.5]
$f''(x) = 12x - 18$ [0.5]
Concave up when $f''(x) > 0$, so $12x - 18 > 0$, giving $x > \frac{3}{2}$ [2]
Show that $y = x^4 - 4x^3$ has a point of inflection at $x = 2$. [3 marks]
$\frac{dy}{dx} = 4x^3 - 12x^2$ 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]
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]
$v(t) = 3t^2 - 12t + 9$ [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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