Higher derivatives

The derivative $f'$ is a function. So we can differentiate it too. The result is the second derivative, $f''$. Its derivative is the third derivative, $f'''$, and so on. These are called higher derivatives.

The second derivative is the most important one. In physics, it is acceleration. In Chapter 4, it will tell us how a graph bends.

Notation

Derivative	Prime notation	Leibniz notation
First	$f'(x)$ or $y'$	$\dfrac{dy}{dx}$
Second	$f''(x)$ or $y''$	$\dfrac{d^2y}{dx^2}$
Third	$f'''(x)$ or $y'''$	$\dfrac{d^3y}{dx^3}$
Fourth	$f^{(4)}(x)$ or $y^{(4)}$	$\dfrac{d^4y}{dx^4}$
$n$th	$f^{(n)}(x)$ or $y^{(n)}$	$\dfrac{d^ny}{dx^n}$

We read $f''(x)$ as "f double prime of x." We read $\dfrac{d^2y}{dx^2}$ as "d two y d x squared." It means

$\dfrac{d^2y}{dx^2}=\dfrac{d}{dx}\left(\dfrac{dy}{dx}\right)$.

After three primes, we write the number in parentheses, as in $f^{(4)}$. The parentheses show that it is a derivative, not a power. So $f^{(4)}$ is the fourth derivative, but $f^4$ means $f$ to the fourth power.

Finding higher derivatives

To find a higher derivative, differentiate again and again. Simplify after each step. This makes the next step easier.

Example 1: Find all the derivatives of

$f(x)=x^4-3x^3+2x$.

Solution:

$\begin{align*}f'(x)&=4x^3-9x^2+2\\f''(x)&=12x^2-18x\\f'''(x)&=24x-18\\f^{(4)}(x)&=24\\f^{(5)}(x)&=0\end{align*}$

Every derivative after the fifth is also $0$.

Example 1 shows a pattern. Each derivative lowers the degree of a polynomial by $1$. So a polynomial of degree $n$ has $f^{(n+1)}(x)=0$.

Example 2: Find $y''$ for

$y=xe^{2x}$.

Solution:

First derivative: use the product rule. By the chain rule, the derivative of $e^{2x}$ is $2e^{2x}$.

$\begin{align*}&y'\\&=1\cdot e^{2x}+x\cdot 2e^{2x}\\&=e^{2x}(1+2x)\end{align*}$

Second derivative: use the product rule again, on $e^{2x}$ times $(1+2x)$.

$\begin{align*}&y''\\&=2e^{2x}(1+2x)+e^{2x}\cdot 2\\&=e^{2x}(2+4x+2)\\&=e^{2x}(4x+4)\\&=4e^{2x}(x+1)\end{align*}$

Acceleration

Let $s(t)$ be the position of a moving object at time $t$.

The velocity is the rate of change of position: $v(t)=s'(t)$.
The acceleration is the rate of change of velocity: $a(t)=v'(t)=s''(t)$.

So acceleration is the second derivative of position.

Example 3: A ball is dropped. After $t$ seconds, it has fallen

$s(t)=4.9t^2$ meters.

Find its acceleration.

Solution:

$\begin{align*}v(t)&=s'(t)=9.8t\\a(t)&=v'(t)=9.8\end{align*}$

The acceleration is $9.8$ m/s$^2$ at every moment. This is the acceleration due to gravity, often written $g$. The velocity grows by $9.8$ m/s every second.

Example 4: A particle moves along a line. Its position after $t$ seconds is

$s(t)=t^3-6t^2+9t$ meters.

(a) Find the velocity and the acceleration.

(b) When is the particle at rest?

Solution:

(a)

$\begin{align*}v(t)&=3t^2-12t+9\\a(t)&=6t-12\end{align*}$

(b) The particle is at rest when its velocity is $0$:

$\begin{align*}3t^2-12t+9&=0\\t^2-4t+3&=0\\(t-1)(t-3)&=0\end{align*}$

So it is at rest at $t=1$ s and at $t=3$ s.

At $t=3$: $a(3)=18-12=6$ m/s$^2$.

At $t=1$, the particle stops and starts moving backward. The negative acceleration pushes it back. At $t=3$, it stops again and starts moving forward. The positive acceleration pushes it forward.

Patterns in higher derivatives

Sometimes the derivatives follow a pattern. Then we can find the $n$th derivative without computing every step.

Example 5: Find the derivatives of $\sin x$. Then find

$\dfrac{d^{27}}{dx^{27}}(\sin x)$.

Solution:

Use the derivatives of sine and cosine:

$\begin{align*}f(x)&=\sin x\\f'(x)&=\cos x\\f''(x)&=-\sin x\\f'''(x)&=-\cos x\\f^{(4)}(x)&=\sin x\end{align*}$

The fourth derivative is $\sin x$ again. So the pattern repeats every $4$ derivatives.

Divide $27$ by $4$: $27=4\cdot 6+3$. After $24$ derivatives, we are back at $\sin x$. Three more derivatives give $-\cos x$. So

$\dfrac{d^{27}}{dx^{27}}(\sin x)=-\cos x$.

Example 6: Find a formula for the $n$th derivative of

$f(x)=\dfrac{1}{x}$.

Solution:

Write $f(x)=x^{-1}$ and use the power rule again and again:

$\begin{align*}f'(x)&=-1\cdot x^{-2}\\f''(x)&=(-1)(-2)x^{-3}=2x^{-3}\\f'''(x)&=2(-3)x^{-4}=-6x^{-4}\\f^{(4)}(x)&=-6(-4)x^{-5}=24x^{-5}\end{align*}$

Look at the pattern:

The signs alternate: $-,+,-,+$. This is $(-1)^n$.
The numbers are products of counting numbers:
- $1$
- $2=1\cdot 2$
- $6=1\cdot 2\cdot 3$
- $24=1\cdot 2\cdot 3\cdot 4$
The exponent of $x$ is $-(n+1)$.

The product $1\cdot 2\cdot 3\cdots n$ is called $n$ factorial. It is written $n!$.

For example, $4!=24$.

So the $n$th derivative is

$f^{(n)}(x)=\dfrac{(-1)^n\,n!}{x^{n+1}}$.

Check: For $n=3$, this gives

$\dfrac{(-1)^3\cdot 6}{x^4}=-\dfrac{6}{x^4}$.

That matches $f'''(x)$.

Second derivatives by implicit differentiation

With implicit differentiation, the first derivative usually contains $y$. To find $y''$, differentiate $y'$ again. Remember that $y$ is a function of $x$. Then replace $y'$ by its formula.

Example 7: Find $y''$ for the circle

$x^2+y^2=25$.

Solution:

In implicit differentiation (Example 1), we found

$y'=-\dfrac{x}{y}$.

Differentiate again with the quotient rule. The numerator $x$ has derivative $1$. The denominator $y$ has derivative $y'$:

$\begin{align*}&y''\\&=-\dfrac{1\cdot y-x\cdot y'}{y^2}\end{align*}$

Replace $y'$ by ${-\dfrac{x}{y}}$:

$\begin{align*}&=-\dfrac{y-x\left(-\dfrac{x}{y}\right)}{y^2}\end{align*}$

Simplify the numerator:

$\begin{align*}&=-\dfrac{y+\dfrac{x^2}{y}}{y^2}\end{align*}$

Multiply the numerator and the denominator by $y$:

$\begin{align*}&=-\dfrac{y^2+x^2}{y^3}\end{align*}$

On the circle, ${x^2+y^2=25}$:

$\begin{align*}&y''=-\dfrac{25}{y^3}\end{align*}$

Summary

The second derivative $f''$ is the derivative of $f'$. The third derivative $f'''$ is the derivative of $f''$, and so on.
After three primes, write $f^{(4)}$, $f^{(5)}$, … . In Leibniz notation, the $n$th derivative is $\dfrac{d^ny}{dx^n}$.
Acceleration is the second derivative of position: $a(t)=s''(t)$.
Look for a pattern to find the $n$th derivative.
In implicit differentiation, differentiate $y'$ again, and replace $y'$ by its formula.

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