Second order derivatives

Differentiate once and you learn how fast $y$ is changing; differentiate again and you learn how that rate of change is itself changing. That second act of differentiation is the mathematical signature of curvature, of concavity, and — if $y$ tracks a moving particle's position — of acceleration. This is the final topic under Continuity, Differentiability and Differentiation in the Calculus unit, and the syllabus deliberately caps the ladder here: you differentiate an explicit, implicit, or parametric relation twice, never three times. The techniques are not new — chain rule, product rule, implicit differentiation, the two-step parametric trick — but applying them a second time, to an already-differentiated expression, is where the algebra gets messy and marks are lost. Master that repetition and you are set up perfectly for the second-derivative test on maxima and minima later in the same unit.

The core insight

The second derivative $\dfrac{d^2y}{dx^2}$ is found by differentiating $\dfrac{dy}{dx}$ a second time with respect to $x$ , and its sign tells you whether the curve bends upward or downward at that point.

The core idea

If $y = f(x)$ , the first derivative $\dfrac{dy}{dx}$ (also written $y'$ or $f'(x)$ ) is the instantaneous rate of change of $y$ . Differentiating that expression again, still with respect to $x$ , gives the second derivative:

\frac{d^2y}{dx^2} = \frac{d}{dx}\left(\frac{dy}{dx}\right), \quad \text{also written } y'' \text{ or } f''(x)

where: $y''$ and $f''(x)$ are alternative notations for the same quantity, and the process is simply "differentiate, then differentiate the result."

Explicit functions. Apply whatever rule produced $\dfrac{dy}{dx}$ — product, quotient, or chain rule — a second time to that new expression. Nothing conceptually new happens; the algebra is just longer.

Implicit relations. First find $\dfrac{dy}{dx}$ , which will typically contain both $x$ and $y$ . Differentiate that expression again with respect to $x$ , applying the chain rule to every $y$ -term, and substitute the already-known $\dfrac{dy}{dx}$ expression back in wherever it reappears, so the final answer is written purely in terms of $x$ and $y$ .

Parametric equations. For $x = x(t)$ , $y = y(t)$ , the second derivative is emphatically not $\dfrac{d^2y/dt^2}{d^2x/dt^2}$ . It is obtained by differentiating $\dfrac{dy}{dx}$ itself (a function of $t$ ) with respect to $t$ , then dividing by $\dfrac{dx}{dt}$ a second time:

\frac{d^2y}{dx^2} = \frac{\dfrac{d}{dt}\left(\dfrac{dy}{dx}\right)}{\dfrac{dx}{dt}}

where: the numerator is the $t$ -derivative of the already-found $\dfrac{dy}{dx}$ , and the denominator is the original $\dfrac{dx}{dt}$ — the division by $\dfrac{dx}{dt}$ is applied a second time, not the second $t$ -derivatives divided directly. Geometrically, $\dfrac{dy}{dx}$ gives the slope of the tangent at a point, while $\dfrac{d^2y}{dx^2}$ tells you how that slope is itself changing as $x$ increases — whether the curve is cupping upward (concave up, $y'' > 0$ ) or bending downward (concave down, $y'' < 0$ ). This is exactly the sign test used later for local maxima and minima: a positive second derivative at a critical point means the curve cups upward there, so the point is a local minimum, and a negative second derivative means a local maximum. Physically, if $y$ is the position of a particle at time $x$ , then $\dfrac{d^2y}{dx^2}$ is its acceleration, the rate of change of velocity itself — the first derivative of position gives velocity, and differentiating once more gives how quickly that velocity is itself changing. It helps to keep the three cases mentally separate before you start writing algebra. An explicit function needs nothing beyond repeating whatever rule built the first derivative. An implicit relation needs an extra bookkeeping step — carrying $\dfrac{dy}{dx}$ forward and re-substituting it. A parametric pair needs the two-stage "differentiate with respect to $t$ , then divide by $\dfrac{dx}{dt}$ " procedure, never a direct ratio of second $t$ -derivatives. Recognising which of the three you are looking at, before touching the algebra, saves most of the marks that otherwise get lost to a wrong method applied confidently.

Figure

Two curve arcs showing concave up with positive second derivative and concave down with negative second derivative

The parametric formula, derived

The parametric second-derivative rule is easy to misremember as a ratio of second $t$ -derivatives, so it is worth seeing exactly why that shortcut fails and what replaces it.

Derivation AWhy the parametric second derivative is not d two y over dt two divided by d two x over dt two

The goal is to show, using only the chain rule already established for the first parametric derivative, that $\dfrac{d^2y}{dx^2}$ must be built by differentiating $\dfrac{dy}{dx}$ with respect to $t$ and then dividing by $\dfrac{dx}{dt}$ a second time, and that this is generally a different quantity from the naive ratio $\dfrac{d^2y/dt^2}{d^2x/dt^2}$ .

Step 1. Start from the definition. By definition, $\dfrac{d^2y}{dx^2}$ means "differentiate $\dfrac{dy}{dx}$ with respect to $x$ ," where $\dfrac{dy}{dx}$ is itself the function $\dfrac{dy/dt}{dx/dt}$ built from the two given parametric derivatives. So the object we must differentiate with respect to $x$ is $g(t) = \dfrac{dy}{dx}$ , a function known only in terms of the parameter $t$ .

Step 2. Differentiate a $t$ -function with respect to $x$ using the chain rule. Since $g$ depends on $x$ only through $t$ , the chain rule gives $\dfrac{dg}{dx} = \dfrac{dg}{dt}\cdot\dfrac{dt}{dx}$ . This is the same chain-rule pattern used to get the first parametric derivative in the first place, just applied to $g$ instead of to $y$ .

Step 3. Rewrite $\dfrac{dt}{dx}$ as $1\big/\dfrac{dx}{dt}$ , since $t$ and $x$ are inverse functions of one another along the curve. Substituting gives $\dfrac{d^2y}{dx^2} = \dfrac{dg}{dt}\cdot\dfrac{1}{dx/dt} = \dfrac{\dfrac{d}{dt}\left(\dfrac{dy}{dx}\right)}{\dfrac{dx}{dt}}$ , exactly the formula stated above: differentiate $\dfrac{dy}{dx}$ with respect to $t$ , then divide once more by $\dfrac{dx}{dt}$ .

Step 4. Confirm the naive shortcut is a different quantity. The expression $\dfrac{d^2y}{dx^2}$ is built from three separate operations — form $\dfrac{dy}{dx}$ , differentiate it with respect to $t$ , divide by $\dfrac{dx}{dt}$ — none of which is "differentiate $y$ twice with respect to $t$ " or "differentiate $x$ twice with respect to $t$ " in isolation. So $\dfrac{d^2y/dt^2}{d^2x/dt^2}$ has no reason to equal $\dfrac{d^2y}{dx^2}$ , and in general it does not; the correct route always passes through $\dfrac{dy}{dx}$ first.

Worked example

Find $\dfrac{d^2y}{dx^2}$ if $y = e^{2x}\sin3x$ .

Step 1. Find the first derivative using the product rule (and chain rule on each factor). $\dfrac{dy}{dx} = 2e^{2x}\sin3x + e^{2x}\cdot3\cos3x = e^{2x}(2\sin3x+3\cos3x)$ .

Step 2. Differentiate again, applying the product rule to $e^{2x}$ and $(2\sin3x+3\cos3x)$ . $\dfrac{d^2y}{dx^2} = 2e^{2x}(2\sin3x+3\cos3x) + e^{2x}(6\cos3x-9\sin3x)$ .

Step 3. Expand and collect like terms. $\dfrac{d^2y}{dx^2} = e^{2x}\big[(4\sin3x+6\cos3x) + (6\cos3x-9\sin3x)\big] = e^{2x}(12\cos3x - 5\sin3x)$ .

Final answer: $\boxed{\dfrac{d^2y}{dx^2} = e^{2x}(12\cos3x-5\sin3x)}$ .

Competitive stretch

For the parabola $x = at^2$ , $y = 2at$ : Step 1. $\dfrac{dx}{dt} = 2at$ , $\dfrac{dy}{dt} = 2a$ , so $\dfrac{dy}{dx} = \dfrac{2a}{2at} = \dfrac{1}{t}$ . Step 2. Differentiate with respect to $t$ : $\dfrac{d}{dt}\left(\dfrac{1}{t}\right) = -\dfrac{1}{t^2}$ . Step 3. Divide by $\dfrac{dx}{dt}$ again: $\dfrac{d^2y}{dx^2} = \dfrac{-1/t^2}{2at} = -\dfrac{1}{2at^3}$ . This matches the derivation above exactly — no shortcut through $d^2y/dt^2$ was used or needed.

Common traps

Computing the parametric second derivative as $\dfrac{d^2y/dt^2}{d^2x/dt^2}$ — invalid, as the derivation above shows; you must go through $\dfrac{dy}{dx}$ first.
Leaving an implicit second derivative with a leftover $\dfrac{dy}{dx}$ term instead of substituting the first-derivative expression back in.
Sign errors from repeated product or chain rule applications — expand fully and collect terms rather than combining mentally.
Skipping straight to the second derivative of an implicit relation without first simplifying $\dfrac{dy}{dx}$ — the algebra becomes far harder to manage.

Ask the tutor: if the parametric formula or the implicit-substitution step still feels shaky, ask the AI tutor to walk through one more fully worked example side by side with you.

Why can't I just divide the second t-derivatives to get the parametric second derivative?

Because $\dfrac{d^2y}{dx^2}$ is defined as differentiating $\dfrac{dy}{dx}$ with respect to $x$ , not as differentiating $y$ and $x$ separately twice with respect to $t$ . The derivation shows the correct chain is: form $\dfrac{dy}{dx}$ first, differentiate that whole expression with respect to $t$ , then divide by $\dfrac{dx}{dt}$ once more. The naive ratio skips the crucial step of building $\dfrac{dy}{dx}$ first, so it generally gives a different, wrong number.

Why does an implicit second derivative still contain dy/dx sometimes?

When you differentiate the implicit first-derivative expression a second time, any $y$ -term produces a new $\dfrac{dy}{dx}$ through the chain rule, since $y$ is a function of $x$ . To get a fully simplified answer purely in terms of $x$ and $y$ , you must substitute the already-known expression for $\dfrac{dy}{dx}$ back in wherever it reappears — leaving it in place is usually marked as incomplete.

Will I ever need a third derivative in this syllabus?

No. The syllabus explicitly restricts successive differentiation to the second order only, so you will never be asked for $\dfrac{d^3y}{dx^3}$ in this course. What you do need is comfort applying the product, quotient, and chain rules a second time on an already-differentiated expression, since that repeated differentiation is usually algebraically messier than the first pass and is exactly where marks are lost.