Calculating Higher-Order Derivatives

The derivative $f'$ is a function in its own right, so it has a derivative too. That derivative, the second derivative $f''$, measures how fast the slope of $f$ is changing. If $f$ gives position, $f'$ is velocity and $f''$ is acceleration; on a graph, $f''$ tells you which way the curve bends.

Differentiating again

The second derivative is the derivative of the first: $f''(x) = \frac{d}{dx}\big[f'(x)\big]$, at every $x$ where $f'$ is differentiable. Differentiating $f''$ gives the third derivative $f'''$, and so on. Each one is computed with the same rules you already have, applied to the previous derivative.

For $f(x) = x^4 - 2x^3 + 5$:

Each differentiation lowers the degree of a polynomial by one. So a polynomial of degree $n$ has a constant $n$th derivative, and every derivative after that is 0. The constant comes from the leading term: the 4th derivative of $x^4$ is $4\cdot 3\cdot 2\cdot 1 = 24$.

The units of $f''$ are the units of $f$ divided by the units of $x$ twice. If $s(t)$ is a position in meters and $t$ is in seconds, $s'(t)$ is in meters per second, and $s''(t)$ is in meters per second per second ($\text{m/s}^2$). Topic 4.2 uses $s''$ as acceleration, and topic 5.6 reads concavity from the sign of $f''$.

Notation

Each notation for the first derivative extends to higher ones.

The Leibniz form comes from applying $\frac{d}{dx}$ to $\frac{dy}{dx}$: the "$d$" is applied twice on top, and $dx$ appears twice on the bottom, written $dx^2$. That $dx^2$ means $(dx)^2$, not the differential of $x^2$, and $\frac{d^2y}{dx^2}$ is not $\left(\frac{dy}{dx}\right)^2$.

After three primes, the order goes in parentheses: $f^{(4)}(x)$, $f^{(5)}(x)$. The parentheses matter, because $f^4(x)$ would be read as $[f(x)]^4$. To say "evaluated at $x = 2$", write $f''(2)$ or $\left.\dfrac{d^2y}{dx^2}\right|_{x=2}$.

When a second derivative doesn't exist

$f''(a)$ exists only if $f'$ exists near $a$ and is differentiable at $a$. A function can be differentiable everywhere while its derivative has a corner, and then $f''$ fails at the corner.

Take $f(x) = x|x|$. For $x \ge 0$ it equals $x^2$, and for $x < 0$ it equals $-x^2$. Both pieces have slope 0 at $x = 0$, so $f'(0) = 0$, and $f'(x) = 2x$ for $x > 0$ and $-2x$ for $x < 0$. Together, $f'(x) = 2|x|$, which is continuous but has a corner at 0. Its slope is 2 just right of 0 and $-2$ just left, so $f''(0)$ does not exist, even though $f$ looks perfectly smooth at the origin.

When an exam question says $f$ is "twice differentiable", it is guaranteeing that $f''$ exists everywhere, which also makes $f$ and $f'$ continuous (topic 2.4).

Patterns in repeated derivatives

Some functions repeat or scale in a predictable way each time you differentiate, so you can jump to any order without computing every step.

• Exponentials. Each derivative of $e^{kx}$ multiplies by $k$, so $\frac{d^n}{dx^n}e^{kx} = k^ne^{kx}$. For example, the 10th derivative of $e^{-x}$ is $(-1)^{10}e^{-x} = e^{-x}$.
• Sine and cosine. The derivatives of $\sin x$ run $\cos x$, $-\sin x$, $-\cos x$, $\sin x$, and then repeat. Only the remainder of $n$ divided by 4 matters.
• Reciprocals. For $\frac{1}{x} = x^{-1}$, the power rule pulls down $-1$, then $-2$, then $-3$: the derivatives are $-\frac{1}{x^2}$, $\frac{2}{x^3}$, $-\frac{6}{x^4}$, $\frac{24}{x^5}$. The signs alternate and the numerators are $1, 2, 6, 24$.

Second derivatives with the other rules

Computing $f''$ means differentiating $f'$, whatever $f'$ looks like, so every rule from Units 2 and 3 can appear in the second round.

Simplify the first derivative before differentiating it again. Combining fractions, cancelling and factoring after the first derivative shrinks the second one. Example 1 shows the payoff.

Use the product rule on products inside the first derivative. Differentiating a product $uv$ twice gives

because $u'v'$ shows up once from each term of $(uv)'$. A chain-rule derivative like $2x\,g'(x^2)$ is itself a product, so its derivative needs the product rule too. Example 3 works through both.

With implicit differentiation (topic 3.2), $\frac{dy}{dx}$ usually contains $y$. When you differentiate it again, $y$ is still a function of $x$, so each $y$ produces a $\frac{dy}{dx}$ by the chain rule. Then you replace that $\frac{dy}{dx}$ with the expression you already found, so the final answer is in terms of $x$ and $y$ only. Example 4 is this kind of question.

Worked examples

Where students lose points

On the exam

A free-response question that gives $\frac{dy}{dx}$ as an expression in $x$ and $y$ (a differential equation or an implicit curve) can ask for $\frac{d^2y}{dx^2}$ next. The method is Example 4(b): differentiate implicitly, then substitute the given $\frac{dy}{dx}$. Second derivatives of functions known only through tables of values are computed like Example 3. On no-calculator parts, an unsimplified numerical value such as $\frac{12(1)(9) - 9(1)}{4(27)}$ is acceptable.