Second-order derivatives

Differentiate a function once and you learn its rate of change. Differentiate it again and you learn how that rate of change is itself changing — the second-order derivative, written $\dfrac{d^2y}{dx^2}$ or $y''$ . It closes out the Continuity and Differentiability chapter by asking you to reuse every tool built earlier — the chain rule, implicit differentiation, logarithmic differentiation, parametric differentiation — one extra time. The computation is mechanically simple: differentiate, then differentiate the result. What takes practice is doing this correctly when $y$ is not written as a bare formula in $x$ — when it is tangled up in an implicit relation, or split across a parameter $t$ . This article walks through all three settings: explicit, implicit, and parametric, with one fully-solved example in each style that you can use as a template.

The core insight

The second-order derivative $\dfrac{d^2y}{dx^2} = \dfrac{d}{dx}\left(\dfrac{dy}{dx}\right)$ is found by differentiating a function's first derivative a second time, with the exact technique — direct, implicit, or via a parameter — matching however $y$ was originally given.

The core idea

Given $y = f(x)$ , call the first derivative $y' = \dfrac{dy}{dx}$ . The second-order derivative is what you get by differentiating $y'$ once more with respect to $x$ :

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

where: $y$ is the original function, $\dfrac{dy}{dx}$ is its first derivative (itself a function of $x$ ), and $\dfrac{d^2y}{dx^2}$ is the derivative of that derivative. The syllabus asks you to compute this in three settings, and the technique changes with the setting.

Explicit functions. If $y = f(x)$ directly, there is nothing new: differentiate $f(x)$ to get $f'(x)$ , then differentiate $f'(x)$ again to get $f''(x)$ . Every differentiation rule you already know — power, product, quotient, chain — applies at each stage independently.

Implicit relations. When $x$ and $y$ are tangled together in one equation, the first derivative $\dfrac{dy}{dx}$ itself usually contains $y$ (and sometimes $\dfrac{dy}{dx}$ appears on both sides after simplification). To get the second derivative, differentiate that first-derivative expression with respect to $x$ once more — applying the chain rule to every $y$ -term as $\dfrac{d}{dx}(y) = \dfrac{dy}{dx}$ — and then substitute back the already-known expression for $\dfrac{dy}{dx}$ to simplify the result purely in terms of $x$ and $y$ .

Parametric curves. If $x = f(t)$ and $y = g(t)$ , first find $\dfrac{dy}{dx}$ as a function of $t$ in the usual parametric way. Then treat that expression as a new " $y$ -like" quantity depending on $x$ only through $t$ , and apply the same chain-rule move used to get the first derivative:

\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 $\dfrac{dx}{dt}$ again — the same one used in step one. Differentiating $\dfrac{dy}{dx}$ with respect to $t$ and stopping there, without dividing by $\dfrac{dx}{dt}$ , is the single most common parametric mistake.

Figure

Three small graphs stacked showing a cubic curve its slope curve and its slope of slope curve for the same function

The core idea, continued: why the parametric formula is not a new rule

Once $\dfrac{dy}{dx}$ is written as a function of $t$ , it depends on $x$ only indirectly, through $t$ . Finding its derivative with respect to $x$ therefore needs exactly the same conversion used for the first derivative: differentiate with respect to $t$ (the rate "per unit $t$ "), then divide by $\dfrac{dx}{dt}$ to convert it into a rate "per unit $x$ ." Seen this way, the second-order parametric formula is the same idea applied twice, not a separate rule to memorise.

Derivation AThe parametric second-derivative formula

The goal of this derivation is to show, starting only from the definition $\dfrac{d^2y}{dx^2} = \dfrac{d}{dx}\left(\dfrac{dy}{dx}\right)$ and the ordinary chain rule for a function of a parameter, exactly why $\dfrac{d^2y}{dx^2}$ for a parametric curve $x = f(t)$ , $y = g(t)$ works out to be the $t$ -derivative of $\dfrac{dy}{dx}$ divided once more by $\dfrac{dx}{dt}$ , rather than simply the $t$ -derivative of $\dfrac{dy}{dx}$ on its own.

Step 1. Start from the first-derivative stage, which you already know: for a parametric curve, $\dfrac{dy}{dx} = \dfrac{dy/dt}{dx/dt}$ , obtained by treating $y$ and $x$ as both varying with $t$ and using the chain rule $\dfrac{dy}{dt} = \dfrac{dy}{dx}\cdot\dfrac{dx}{dt}$ , then solving for $\dfrac{dy}{dx}$ . Call this ratio $g(t)$ ; it is a specific function of $t$ , found once and reused below.

Step 2. By definition, the second derivative is $\dfrac{d^2y}{dx^2} = \dfrac{d}{dx}\big(g(t)\big)$ — the derivative of $g(t)$ with respect to $x$ . But $g(t)$ is written as a function of $t$ , not $x$ , so this derivative cannot be taken directly; it must go through $t$ using the chain rule, exactly as any function of $t$ is differentiated with respect to $x$ .

Step 3. Apply the chain rule the same way it was applied in Step 1: $\dfrac{d}{dx}\big(g(t)\big) = \dfrac{d}{dt}\big(g(t)\big)\cdot\dfrac{dt}{dx}$ . Since $\dfrac{dt}{dx} = 1\big/\dfrac{dx}{dt}$ , this becomes $\dfrac{d}{dx}\big(g(t)\big) = \dfrac{\dfrac{d}{dt}\big(g(t)\big)}{\dfrac{dx}{dt}}$ .

Step 4. Substitute back $g(t) = \dfrac{dy}{dx}$ to read off the final formula: $\dfrac{d^2y}{dx^2} = \dfrac{\dfrac{d}{dt}\left(\dfrac{dy}{dx}\right)}{\dfrac{dx}{dt}}$ . The division by $\dfrac{dx}{dt}$ is not optional bookkeeping — it is the same conversion factor from "per unit $t$ " to "per unit $x$ " used in Step 1, applied a second time because $\dfrac{dy}{dx}$ is itself a quantity that only depends on $x$ through $t$ .

Worked example

Find $\dfrac{d^2y}{dx^2}$ if $y = x^3 - 6x^2 + 9x$ .

Step 1. Find the first derivative by differentiating term by term:

\frac{dy}{dx} = 3x^2 - 12x + 9

Step 2. Differentiate the first derivative once more to get the second derivative:

\frac{d^2y}{dx^2} = \frac{d}{dx}(3x^2 - 12x + 9) = 6x - 12

Step 3. State and check the result. So $\dfrac{d^2y}{dx^2} = 6x - 12$ . At $x = 2$ , this evaluates to $6(2) - 12 = 0$ , consistent with $x = 2$ being the midpoint of the cubic's turning behaviour (a fact examined further, using concavity, in the next chapter — not required here). Final answer: $\dfrac{d^2y}{dx^2} = 6x - 12$ .

Competitive stretch

Find $\dfrac{d^2y}{dx^2}$ if $x^2 + y^2 = 1$ (an implicit relation).

Step 1. Differentiate both sides implicitly with respect to $x$ : $2x + 2y\dfrac{dy}{dx} = 0$ , which gives $\dfrac{dy}{dx} = -\dfrac{x}{y}$ .

Step 2. Differentiate $\dfrac{dy}{dx} = -\dfrac{x}{y}$ again using the quotient rule, treating $y$ as a function of $x$ so that $\dfrac{d}{dx}(y) = \dfrac{dy}{dx}$ :

\frac{d^2y}{dx^2} = -\frac{y\cdot 1 - x\cdot\dfrac{dy}{dx}}{y^2}

Step 3. Substitute $\dfrac{dy}{dx} = -\dfrac{x}{y}$ from Step 1 and simplify: $\dfrac{d^2y}{dx^2} = -\dfrac{y + x^2/y}{y^2} = -\dfrac{x^2+y^2}{y^3}$ .

Step 4. Use the original relation $x^2 + y^2 = 1$ to finish: $\dfrac{d^2y}{dx^2} = -\dfrac{1}{y^3}$ . Final answer: $\dfrac{d^2y}{dx^2} = -\dfrac{1}{y^3}$ .

Common traps

Not substituting the first derivative back in for an implicit relation — leaving $\dfrac{d^2y}{dx^2}$ expressed in terms of $\dfrac{dy}{dx}$ instead of purely $x$ and $y$ .
Differentiating $\dfrac{dy}{dx}$ directly with respect to $x$ for a parametric curve, instead of with respect to $t$ followed by division by $\dfrac{dx}{dt}$ — the single most common parametric error.
Sign and arithmetic slips carried over from the first derivative into the second, since any early error compounds through the second round of differentiation.
Assuming the answer must simplify to something "neat." Many correct second derivatives remain moderately complex expressions in $x$ , $y$ , or $t$ ; do not over-simplify or suspect an error just because the algebra stays busy.

Ask the tutor: if the parametric "differentiate then divide again" pattern still feels slippery, ask the AI tutor to work through one more parametric example with you, step by step.

Why does the second derivative of an implicit relation need substitution at the end, but an explicit one does not?

For an explicit function $y = f(x)$ , differentiating twice gives an answer purely in $x$ automatically, since $y$ never appears on the right-hand side. For an implicit relation, the first differentiation step produces $\dfrac{dy}{dx}$ in terms of both $x$ and $y$ , and differentiating that expression again naturally reintroduces $\dfrac{dy}{dx}$ inside the working (via the product or quotient rule). Substituting the known formula for $\dfrac{dy}{dx}$ back in is what removes it, leaving a final answer in $x$ and $y$ only — skipping this step leaves the answer only partially simplified.

Why can't I just differentiate dy/dx with respect to t for a parametric curve and stop there?

Because $\dfrac{dy}{dx}$ , once written as a function of $t$ , is still a rate measured "per unit $x$ ," not "per unit $t$ ." Differentiating it with respect to $t$ gives you how fast that ratio changes as $t$ increases, which is a different quantity from how fast it changes as $x$ increases. Dividing by $\dfrac{dx}{dt}$ converts the rate back into "per unit $x$ " terms, exactly as it did when you first built $\dfrac{dy}{dx}$ from $\dfrac{dy}{dt}$ and $\dfrac{dx}{dt}$ . Skipping this division is the most common parametric second-derivative mistake.

Do I need to know about concavity or curve sketching to answer second-order derivative questions?

No. In this chapter, computing $\dfrac{d^2y}{dx^2}$ is a purely computational exercise in reapplying differentiation rules — explicit, implicit, or parametric — one extra time. The geometric meaning of the second derivative, such as concavity (whether a curve bends upward or downward) and its use in locating maxima and minima, is developed in the next chapter, Applications of Derivatives. You are not expected to interpret the sign of $\dfrac{d^2y}{dx^2}$ geometrically here, only to compute it correctly.