2.7: Higher Derivatives

Definition:

The second derivative of a real function $f$ is the derivative of the derivative of $f$, and is denoted by $f' '$. The third derivative of $f$ is the derivative of the second derivative, and is denoted by $f' ' '$, or $f^{(3)}$. In general, the $n$th derivative of $f$ is denoted by $f^{(n)}$.

If $y$ depends on $x$, $y = f(x)$, then the second differential of $y$ is defined to be $$d^{2} y = f' '(x) \ dx^{2}. \nonumber$$

In general the $n$^th differential of $y$ is defined by $$d^{n}y = f^{(n)} (x) \ dx^{n}. \nonumber$$

Here $dx^{2}$ means $(dx)^{2}$ and $dx^{n}$ means $(dx)^{n}$.

We thus have the alternative notations $$\frac{d^{2} y}{dx^{2}} = f' '(x), \quad \frac{d^{n} y}{dx^{n}} = f^{(n)} (x). \nonumber$$ for the second and $n$^th derivatives. The notation $$y' ' = f' '(x), \quad y^{(n)} = f^{(n)} (x) \nonumber$$ is also used.

The definition of the second differential can be remembered in the following way. By definition, $$dy = f'(x) \ dx. \nonumber$$ Now hold $dx$ constant and formally apply the Constant Rule for differentiation, obtaining $$d(dy) = f' '(x) \ dx \ dx, \nonumber$$ or $$d^{2} y = f' '(x) \ dx^{2}. \nonumber$$ (This is not a correct use of the Constant Rule because the rule applies to a real constant $c$, and $dx$ is not a real number. It is only a mnemonic device to remember the definition of $d^{2}y$, not a proof.)

The third and higher differentials can be motivated in the same way. If we hold $dx$ constant and formally use the Constant Rule again and again, we obtain $$\begin{align*} dy &= f'(x) \ dx, \\ d^{2}y &= f' '(x) \ dx \ dx = f' '(x) \ dx^{2}, \\ d^{3} y &= f' ' '(x) \ dx^{2} \ dx = f' ' '(x) \ dx^{3}, \\ d^{4} y &= f^{(4)} (x) \ dx^{3} \ dx = f^{(4)}(x) \ dx^{4}, \end{align*}$$ and so on.

The acceleration of a moving particle is defined to be the derivative of the velocity with respect to time, $$a= dv/dt \nonumber.$$

Thus the velocity is the first derivative of the distance and the acceleration is the second derivative of the distance. If $s$ is distance, we have $$v = \frac{ds}{dt}, \quad a = \frac{d^{2} s}{dt^{2}}. \nonumber$$

Example $\PageIndex{1}$

A ball thrown up with initial velocity $b$ moves according to the equation $$y = bt - 16t^{2} \nonumber$$ with $y$ in feet, $t$ in seconds. Then the velocity is $$v = b - 32t \ \text{ft/sec}, \nonumber$$ and the acceleration (due to gravity) is a constant, $$a = -32 \ \text{ft/sec}^{2}. \nonumber$$

Example $\PageIndex{2}$

Find the second derivative of $y = \sin (2 \theta)$.

First derivative

Put $u = 2 \theta$. Then $$y = \sin u, \quad \frac{dy}{du} = \cos u, \quad \frac{du}{d \theta} = 2. \nonumber$$

By the Chain Rule, $$\begin{align*} \frac{dy}{d \theta} &= \frac{dy}{du} \cdot \frac{du}{d \theta} = \cos (2 \theta) \cdot 2, \\ \frac{dy}{d \theta} &= 2 \cos (2 \theta) \end{align*}$$

Second derivative

Let $v = 2 \cos (2 \theta)$. We must find $dv/d \theta$. Put $u = 2 \theta$. Then $$v = 2 \cos u, \quad \frac{dv}{du} = -2 \sin u, \quad \frac{du}{d \theta} = 2. \nonumber$$

Using the Chain Rule again, $$\frac{d^{2} y}{d \theta^{2}} = \frac{dv}{d \theta} = \frac{dv}{du} \cdot \frac{du}{d \theta} = (-2 \sin (2 \theta)) \cdot 2. \nonumber$$

This simplifies to $$\frac{d^{2} y}{d \theta^{2}} = -4 \sin (2 \theta). \nonumber$$  

Example $\PageIndex{3}$

A particle moves so that at time $t$ it has gone a distance $s$ along a straight line, its velocity is $v$, and its acceleration is $a$. Show that $$a = v \frac{dv}{ds}. \nonumber$$

By definition we have $$v = \frac{ds}{dt}, \quad a = \frac{dv}{dt}, \nonumber$$

so by the Chain Rule, $$a = \frac{dv}{ds} \frac{ds}{dt} = v \frac{dv}{ds}. \nonumber$$

Example $\PageIndex{4}$

If a polynomial of degree $n$ is repeatedly differentiated, the $k$^th derivative will be a polynomial of degree $n - k$ for $k \leq n$, and the $(n + 1)$^st derivative will be zero. For example, $$\begin{align*} y &= 3x^{5} - 10x^{4} + x^{2} - 7x + 4. \\ dy/dx &= 15x^{4} - 40x^{3} + 2x - 7. \\ d^{2} y / dx^{2} &= 60x^{3} - 120x^{2} + 2. \\ d^{3} y / dx^{3} &= 180x^{2} - 240x. \\ d^{4} y / dx^{4} &= 360x - 240. \\ d^{5} y / dx^{5} &= 360, \quad d^{6} y / dx^{6} = 0. \end{align*}$$

Geometrically, the second derivative $f' '(x)$ is the slope of the curve $y' = f'(x)$ and is also the rate of change of the slope of the curve $y = f(x)$.

Problems for Section 2.7

In Problems 1-23, find the second derivative.

1.	$y = 1/x$	2.	$y = x^{5}$	3.	$y = \dfrac{-5}{x + 1}$
4.	$f(x) = 3x^{-2}$	5.	$f(x) = x^{1/2} + x^{-1/2}$	6.	$f(t) = t^{3} - 4t^{2}$
7.	$f(t) = t \sqrt{t}$	8.	$y = (3t - 1)^{10}$	9.	$y = \sin x$
10.	$y = \cos x$	11.	$y = A \sin (Bx)$	12.	$y = A \cos (Bx)$
13.	$y = e^{ax}$	14.	$y = e^{-ax}$	15.	$y = \ln x$
16.	$y = x \ln x$	17.	$y = \dfrac{1}{t^{2} + 1}$	18.	$y = \sqrt{3t + 2}$
19.	$z = \dfrac{x - 5}{x + 2}$	20.	$z = \dfrac{2x - 1}{3x - 2}$	21.	$z = x \sqrt{x + 1}$
22.	$s = \left(\dfrac{t + 1}{t + 2}\right)^{2}$	23.	$s = \sqrt{\dfrac{t}{t + 3}}$
24.	Find the third derivative of $y = x^{2} - 2/x$.
25.	A particle moves according to the equation $s = 1 - 1/t^{2}, t > 0$. Find its acceleration.
26.	An object moves in such a way that when it has moved a distances its velocity is $v = \sqrt{s}$. Find its acceleration. (Use Example $\PageIndex{3}$.)
27.	Suppose $u$ depends on $x$ and $d^{2}u/dx^{2}$ exists. If $y = 3u$, find $d^{2}y/dx^{2}$.
28.	If $d^{2}u/dx^{2}$ and $d^{2}v/dx^{2}$ exist and $y = u + v$, find $d^{2}y/dx^{2}$.
29.	If $d^{2}u/dx^{2}$ exists and $y = u^{2}$, find $d^{2}y/dx^{2}$.
30.	If $d^{2}u/dx^{2}$ and $d^{2}v/dx^{2}$ exist and $y= uv$, find $d^{2}y/dx^{2}$.
31.	Let $y = ax^{2} + bx + c$ be a polynomial of degree two. Show that $dy/dx$ is a linear function and $d^{2}y/dx^{2}$ is a constant function.
$\square$ 32.	Prove that the $n$th derivative of a polynomial of degree $n$ is constant. (Use the fact that the derivative of a polynomial of degree $k$ is a polynomial of degree $k - 1$.)