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 \(k - 1\).)