Higher-Order Derivatives

Since

f''

is the derivative of

f'

, it's the slope of the graph of

f'

. It measures how fast the slope of

f

is changing. In the figures below,

f(x) = x^3 - 3x

, so

f'(x) = 3x^2 - 3

and

f''(x) = 6x

y = f(x) = x³ − 3x

y = f′(x) = 3x² − 3, with its tangent lines at x = −1, 0, 1 (slopes −6, 0, 6)

y = f″(x) = 6x: its heights at x = −1, 0, 1 are those slopes, −6, 0, 6.

Computing second derivatives

Example 1

Find

\frac{d^2y}{dx^2}

\begin{aligned} &\text{(a)}\ \ y = \sin^2 x \\[4pt] &\text{(b)}\ \ y = \ln\left(1 + x^2\right) \\[4pt] &\text{(c)}\ \ y = \frac{x - 1}{x + 1} \end{aligned}

(a)

y' = 2\sin x\cos x = \sin 2x

, using the double-angle identity to keep the next step short, so

y'' = 2\cos 2x

(b)

y' = \frac{2x}{1 + x^2}

. The quotient rule gives

\begin{aligned} y'' &= \frac{2\left(1 + x^2\right) - 2x \cdot 2x}{\left(1 + x^2\right)^2} \\[4pt] &= \frac{2 - 2x^2}{\left(1 + x^2\right)^2}. \end{aligned}

(c)

y' = \frac{(x + 1) - (x - 1)}{(x + 1)^2} = 2(x + 1)^{-2}

, and then the chain rule gives

y'' = -4(x + 1)^{-3}

Patterns in repeated derivatives

Example 2

Find a formula for the

n

th derivative of (a)

y = xe^x

and (b)

y = \frac{1}{1 - x}

. (c) Find the

100

th derivative of

\cos x

(a) By the product rule,

y' = e^x + xe^x = (x + 1)e^x

. Differentiating

(x + k)e^x

gives

(x + k + 1)e^x

, so each step adds

1

y'' = (x + 2)e^x

y''' = (x + 3)e^x

, and

\frac{d^ny}{dx^n} = (x + n)e^x.

(b) Write

y = (1 - x)^{-1}

. Each derivative brings down the exponent and, from the inside, a factor of

-1

\begin{aligned} y' &= (1 - x)^{-2}, \\ y'' &= 2(1 - x)^{-3}, \\ y''' &= 6(1 - x)^{-4}, \end{aligned}

In general, differentiating

k!\,(1 - x)^{-(k + 1)}

gives

\begin{aligned} &k!\,(k + 1)(1 - x)^{-(k + 2)} \\ &\qquad = (k + 1)!\,(1 - x)^{-(k + 2)}, \end{aligned}

so each step multiplies by the next whole number, and

\frac{d^ny}{dx^n} = \frac{n!}{(1 - x)^{n + 1}}

(c) The derivatives of

\cos x

cycle with period

4

-\sin x

-\cos x

\sin x

\cos x

. Since

100

is a multiple of

4

, the

100

th derivative is

\cos x

Implicit second derivatives

Example 3

For the hyperbola

x^2 - y^2 = 7

, find

\frac{d^2y}{dx^2}

in terms of

y

, and its value at

(4, 3)

Implicitly,

2x - 2y\frac{dy}{dx} = 0

, so

\frac{dy}{dx} = \frac{x}{y}

. Differentiate again with the quotient rule, remembering that

y

is a function of

x

\begin{aligned} \frac{d^2y}{dx^2} &= \frac{y \cdot 1 - x\frac{dy}{dx}}{y^2} = \frac{y - \frac{x^2}{y}}{y^2} \\[4pt] &= \frac{y^2 - x^2}{y^3} = -\frac{7}{y^3}. \end{aligned}

This holds wherever

y \ne 0

. The last step uses the equation of the hyperbola. At

(4, 3)

\frac{d^2y}{dx^2} = -\frac{7}{27}

Equations with derivatives

From a table

Example 5

The function

f

is twice differentiable, with the values in the table.

$x$	$f(x)$	$f'(x)$	$f''(x)$
$1$	$4$	$3$	$-2$
$2$	$-1$	$\tfrac{1}{2}$	$5$

(a) Let

g(x) = \big(f(x)\big)^3

. Find

g''(2)

. (b) Let

h(x) = xf'(x)

. Find

h'(1)

(a)

g'(x) = 3\big(f(x)\big)^2 f'(x)

is a product, so the product rule and the chain rule give

\begin{aligned} g''(x) &= 6f(x)\big(f'(x)\big)^2 \\ &\qquad + 3\big(f(x)\big)^2 f''(x). \end{aligned}

Every factor is evaluated at the same input,

x = 2

, so

\begin{aligned} g''(2) &= 6(-1)\left(\frac{1}{2}\right)^2 + 3(-1)^2(5) \\[4pt] &= -\frac{3}{2} + 15 = \frac{27}{2}. \end{aligned}

(b) By the product rule,

h'(x) = f'(x) + xf''(x)

, so

h'(1) = 3 + 1 \cdot (-2) = 1