[Figure: FiveStudy logo]
FiveStudy
Calculus AB · Unit 3 · 3.6
AP Calculus · Differentiation: Composite, Implicit, and Inverse Functions
Calculating Higher-Order Derivatives
Differentiate again. The technique is unchanged — the notation and the meaning are what shift.
key idea$f''=\left(f'\right)';\qquad \frac{d^{2}y}{dx^{2}}=\frac{d}{dx}\left[\frac{dy}{dx}\right]$
The rule
A higher-order derivative is nothing new — just differentiate again. What is new is the notation, which is inconsistent between the two systems, and the interpretation of the second derivative as a rate of change of a rate.
| Order | Lagrange | Leibniz | Means |
|---|
| 1st | $f'(x)$ | $\dfrac{dy}{dx}$ | rate of change of $f$ |
| 2nd | $f''(x)$ | $\dfrac{d^{2}y}{dx^{2}}$ | rate of change of $f'$ — concavity |
| 3rd | $f'''(x)$ | $\dfrac{d^{3}y}{dx^{3}}$ | rate of change of $f''$ |
| 4th and beyond | $f^{(4)}(x)$ | $\dfrac{d^{4}y}{dx^{4}}$ | primes are replaced by a bracketed number |
| $n$-th | $f^{(n)}(x)$ | $\dfrac{d^{n}y}{dx^{n}}$ | — |
From the 4th onward, $f''''$ is not written — use $f^{(4)}$. The brackets matter: $f^{4}$ would mean the function raised to the fourth power.
$\dfrac{d^{2}y}{dx^{2}}$Read ‘d two y d x squared’. The $2$ is not an exponent — it counts how many times you differentiated.
$f^{(n)}(x)$The $n$-th derivative. The brackets distinguish it from a power.
Velocity and accelerationIf $s(t)$ is position, $s'(t)$ is velocity and $s''(t)$ is acceleration — the most common physical reading of a second derivative.
ConcavityThe sign of $f''$: positive means the graph bends upward, negative downward.
Pattern recognition
what to expect
- Each order is one more differentiation — no separate technique exists.
- A degree-$n$ polynomial has zero $(n+1)$-th derivative.
- $e^{x}$ is unchanged at every order; $\sin$ and $\cos$ cycle with period 4.
- Simplify at each stage — unsimplified expressions compound quickly.
Worked examples
EX 1
Straight repetition
$f(x)=3x^{4}-2x^{2}+7$Find the first four derivatives of $f(x)=3x^{4}-2x^{2}+7$.
2
Second
$f''(x)=36x^{2}-4$.
4
Fourth
$f^{(4)}(x)=72$ — note the bracketed notation.
5
Predict the fifth
$f^{(5)}(x)=0$. A degree-$4$ polynomial dies at the $5$th derivative, and stays zero forever after.
EX 2
Second derivative needs the chain rule twice
$y=e^{2x}$Find $\dfrac{d^{2}y}{dx^{2}}$ for $y=e^{2x}$.
1
First derivative
Chain rule: $\dfrac{dy}{dx}=e^{2x}\cdot2=2e^{2x}$.
2
Differentiate again
$\dfrac{d^{2}y}{dx^{2}}=2\cdot\dfrac{d}{dx}\left[e^{2x}\right]=2\cdot2e^{2x}$.
4
Spot the pattern
Each differentiation contributes another factor of $2$, so $\dfrac{d^{n}y}{dx^{n}}=2^{n}e^{2x}$.
5
The common error
Writing $2e^{2x}$ for the second derivative — forgetting that the inner factor reappears every time.
EX 3
Implicit second derivative
$x^{2}+y^{2}=25$Find $\dfrac{d^{2}y}{dx^{2}}$ for $x^{2}+y^{2}=25$.
1
Start from the first derivative
Implicit differentiation gives $\dfrac{dy}{dx}=-\dfrac{x}{y}$.
2
Differentiate again with the quotient rule
$\dfrac{d^{2}y}{dx^{2}}=-\dfrac{(1)(y)-(x)\dfrac{dy}{dx}}{y^{2}}$.
3
Substitute the first derivative
$=-\dfrac{y-x\left(-\dfrac{x}{y}\right)}{y^{2}}=-\dfrac{y+\dfrac{x^{2}}{y}}{y^{2}}$.
4
Clear the inner fraction
$=-\dfrac{y^{2}+x^{2}}{y^{3}}$.
5
Use the original relation
Since $x^{2}+y^{2}=25$: $\dfrac{d^{2}y}{dx^{2}}=-\dfrac{25}{y^{3}}$. Substituting the relation back is what turns a messy expression into a clean one.
EX 4
Physical reading
motionA particle's position is $s(t)=t^{3}-6t^{2}+9t$ metres. Find its velocity and acceleration at $t=2$, and describe the motion.
1
Velocity is the first derivative
$v(t)=s'(t)=3t^{2}-12t+9$, so $v(2)=12-24+9=-3$ m/s.
2
Acceleration is the second
$a(t)=s''(t)=6t-12$, so $a(2)=0$ m/s$^{2}$.
3
Read the velocity
Negative, so the particle is moving in the negative direction at $t=2$.
4
Read the acceleration
Zero, so at that instant the velocity is momentarily neither increasing nor decreasing.
5
Note the units
Position in metres, velocity in m/s, acceleration in m/s$^{2}$ — each derivative divides by time once more.
Watch out
$\dfrac{d^{2}y}{dx^{2}}$ contains no squares
The superscripts count differentiations. It is not $\left(\dfrac{dy}{dx}\right)^{2}$, which is a completely different quantity.
$f^{(4)}$ needs its brackets
$f^{4}$ means $\left[f(x)\right]^{4}$. Dropping the brackets changes a derivative into a power.
The chain factor reappears at every order
For $e^{2x}$ each differentiation brings another $2$. Applying it only once is the standard slip in Example 2.
Simplify as you go
An unsimplified first derivative makes the second far harder than it needs to be — and in the implicit case, substituting the original relation collapses the answer entirely.