3. Higher Derivatives

Video coming soon3 worked examples

Theory

The derivative $\frac{dy}{dx}$ is itself a function of $x$ , so it can be differentiated again. The result is the second derivative, written in any of these ways:

\frac{d^2y}{dx^2} = \frac{d}{dx}\!\left(\frac{dy}{dx}\right), \qquad f''(x), \qquad y''

Repeating the process gives the third, fourth and in general the $n$ th derivative, $\dfrac{d^ny}{dx^n}$ or $f^{(n)}(x)$ .

The second derivative measures how the gradient itself is changing, which is what makes it a test for the nature of a stationary point. At a stationary point $x = a$ (where $f'(a) = 0$ ):

$f''(a) > 0$ — the gradient is increasing, so a minimum turning point
$f''(a) < 0$ — the gradient is decreasing, so a maximum turning point
$f''(a) = 0$ — the test fails; fall back on a nature table

Questions also ask you to compute the first few derivatives of a function and then conjecture a formula for the $n$ th, so look for a pattern in three things at once: the sign, the numerical coefficient, and the power.

The Golden Rule: differentiate one step at a time, simplifying fully before starting the next step — an untidy first derivative makes the second far harder than it needs to be.

⚠️ Common Examiner Traps

Getting the test backwards: positive second derivative means minimum. Picture a valley — the gradient rises from negative, through zero, to positive.
Assuming $f''(a)=0$ means a point of inflection: it does not. The test is simply inconclusive, and you must use a nature table instead.
Notation: $\frac{d^2y}{dx^2}$ is the second derivative, which is not the same as $\left(\frac{dy}{dx}\right)^2$ .
Forgetting the $y$ -coordinates: a stationary point is a point — substitute back into the original equation, not the derivative.
Conjectures need the sign pattern: alternating signs are captured by a factor of $(-1)^n$ or $(-1)^{n-1}$ — check which by testing $n=1$ .

Worked examples

Example 1

Find the first and second derivatives of $f(x) = x^4 - 3x^3 + 2x$ .

Step 1: Differentiate term by term:

f'(x) = 4x^3 - 9x^2 + 2

Step 2: Differentiate the result again. The constant $2$ disappears:

f''(x) = 12x^2 - 18x

Example 2

Find the stationary points of $y = x^3 - 3x^2 - 9x + 5$ and determine their nature using the second derivative.

Step 1: Differentiate and set equal to zero:

\frac{dy}{dx} = 3x^2 - 6x - 9 = 3(x^2 - 2x - 3) = 3(x-3)(x+1)

\frac{dy}{dx} = 0 \implies x = 3 \ \text{ or } \ x = -1

Step 2: Find the $y$ -coordinates from the original equation:

\begin{aligned} x=3: &\quad y = 27 - 27 - 27 + 5 = -22 \\ x=-1: &\quad y = -1 - 3 + 9 + 5 = 10 \end{aligned}

Step 3: Find the second derivative:

\frac{d^2y}{dx^2} = 6x - 6

Step 4: Evaluate it at each stationary point:

\begin{aligned} x=3: &\quad 6(3) - 6 = 12 > 0 \implies \text{minimum} \\ x=-1: &\quad 6(-1) - 6 = -12 < 0 \implies \text{maximum} \end{aligned}

So $(3, -22)$ is a minimum turning point and $(-1, 10)$ is a maximum turning point.

Example 3

Given $f(x) = \dfrac{1}{x}$ , find the first four derivatives and hence make a conjecture for $f^{(n)}(x)$ .

Step 1: Write the function as $x^{-1}$ and differentiate repeatedly, keeping every result as a negative power so the pattern stays visible:

\begin{aligned} f'(x) &= -x^{-2} \\ f''(x) &= 2x^{-3} \\ f'''(x) &= -6x^{-4} \\ f^{(4)}(x) &= 24x^{-5} \end{aligned}

Step 2: Examine the three patterns separately. The signs alternate starting from negative; the coefficients are $1, 2, 6, 24$ , which are $1!, 2!, 3!, 4!$ ; and the power is $-(n+1)$ .

Step 3: Since the first derivative is negative, the sign factor must be $(-1)^n$ . Combining all three patterns:

f^{(n)}(x) = (-1)^n\,n!\,x^{-(n+1)} = \frac{(-1)^n\,n!}{x^{n+1}}

Step 4: Check the conjecture against a known case. For $n=3$ :

\frac{(-1)^{3}\,3!}{x^{4}} = \frac{-6}{x^4} = -6x^{-4} \ \checkmark