Chapter 5 · Continuity and Differentiability

Example 35: $\frac{d^2y}{dx^2}$ for $y=x^3+\tan x$

ExamplesSecond order derivative · Section 5.7

The question

Find $\dfrac{d^2y}{dx^2}$ ⁠, if $y = x^3 + \tan x$ ⁠.

Exam answer

Here $y = x^3 + \tan x$ ⁠.

Differentiating w.r.t. $x$ ⁠,

\frac{dy}{dx} = 3x^2 + \sec^2 x

Differentiating again (chain rule for $\sec^2 x$ ⁠),

\frac{d^2y}{dx^2} = 6x + 2\sec x\cdot\sec x\tan x

$\therefore\ \dfrac{d^2y}{dx^2} = 6x + 2\sec^2 x\tan x$ $(x \ne (2n + 1)\dfrac{\pi}{2},\ n \in \Z)$ ⁠.

Want the reason behind every step, the basics and the checks?

The one idea this problem is built on

The second derivative is simply the derivative of the derivative:

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

So we differentiate once, then differentiate the result again. The only new ingredient in the second round is $\sec^2 x$ ⁠, which needs the chain rule.

Given, to find, and the plan

Given: $y = x^3 + \tan x$ ⁠, defined wherever $\tan x$ is, that is for $x \neq (2n+1)\frac{\pi}{2}$ ⁠, $n \in \Z$ ⁠.
To find: $\dfrac{d^2y}{dx^2}$ ⁠.
Plan: Find $\frac{dy}{dx}$ term by term. Differentiate it again: $3x^2$ by the power rule, $\sec^2 x = (\sec x)^2$ by the chain rule. Cross-check the $\sec^2 x$ part a second way.

Basics you need

1Second order derivative (Section 5.7)

If $y = f(x)$ and $f'(x)$ is itself differentiable, then $\frac{d^2y}{dx^2} = \frac{d}{dx}\left(\frac{dy}{dx}\right)$ ⁠. It is also written $f''(x)$ ⁠, $y''$ ⁠, $y_2$ or $D^2y$ ⁠. Lesson: second order derivative

2Standard derivatives

$\frac{d}{dx}x^n = n\,x^{n-1}$ and $\frac{d}{dx}\tan x = \sec^2 x$ (Table 5.3). Also $\frac{d}{dx}\sec x = \sec x\tan x$ ⁠, which follows from the quotient rule $\left(\frac{u}{v}\right)' = \frac{u'v - uv'}{v^2}$ with $u = 1$ ⁠, $u' = 0$ ⁠, $v = \cos x$ ⁠:

\begin{aligned} \frac{d}{dx}\frac{1}{\cos x} &= \frac{0 - 1\cdot(-\sin x)}{\cos^2 x} \\ &= \frac{\sin x}{\cos^2 x}. \end{aligned}

And $\frac{\sin x}{\cos^2 x} = \frac{1}{\cos x}\cdot\frac{\sin x}{\cos x} = \sec x\tan x$ ⁠. Basics: derivative rules

3Chain rule for a square

$\frac{d}{dx}\big(u^2\big) = 2u\,\frac{du}{dx}$ ⁠, where $u$ is any differentiable function of $x$ ⁠. Lesson: chain rule

4 $\sec^2 x = 1 + \tan^2 x$

Divide $\cos^2 x + \sin^2 x = 1$ by $\cos^2 x$ ⁠. Used for the cross-check in Step 5. Basics: trigonometric identities

Solution

Find the first derivative

Differentiate term by term (sum rule):
$\begin{aligned} \frac{dy}{dx} &= \frac{d}{dx}\left(x^3\right) + \frac{d}{dx}\left(\tan x\right) \\ &= 3x^2 + \sec^2 x. \end{aligned}$
The second derivative is built on the first, so we need $\frac{dy}{dx}$ first. Both pieces are standard derivatives from Table 5.3. This holds at every $x$ where $\tan x$ is defined, since $\cos x \neq 0$ there.

Power rule with $n = 3$ ⁠; $\frac{d}{dx}\tan x = \sec^2 x$ ⁠. Basics: derivative rules
Set up the second derivative
$\begin{aligned} \frac{d^2y}{dx^2} &= \frac{d}{dx}\left(3x^2 + \sec^2 x\right) \\ &= \frac{d}{dx}\left(3x^2\right) + \frac{d}{dx}\left(\sec^2 x\right). \end{aligned}$
By definition (Basic 1), the second derivative is what we get by differentiating $\frac{dy}{dx}$ once more. The sum rule lets us treat the two terms one at a time.

Section 5.7: $\frac{d^2y}{dx^2} = \frac{d}{dx}\left(\frac{dy}{dx}\right)$ ⁠. Lesson: second order derivative
Differentiate $3x^2$
$\frac{d}{dx}\left(3x^2\right) = 3\cdot 2x = 6x.$
The constant 3 stays in front and the power rule gives $2x$ ⁠.

Power rule with $n = 2$ ⁠.
Differentiate $\sec^2 x$ with the chain rule

$\sec^2 x$ means $(\sec x)^2$ ⁠: the outer function is "square", the inner function is $u = \sec x$ ⁠. By Basic 3 and Basic 2,
$\begin{aligned} \frac{d}{dx}(\sec x)^2 &= 2\sec x\cdot\frac{d}{dx}(\sec x) \\ &= 2\sec x\cdot\sec x\tan x. \end{aligned}$
So $\dfrac{d}{dx}\left(\sec^2 x\right) = 2\sec^2 x\tan x$ ⁠.

A common error is to write $\frac{d}{dx}\sec^2 x = 2\sec x$ ⁠, forgetting to multiply by the derivative of the inside. The chain rule says: differentiate the square (giving $2\sec x$ ⁠), then multiply by the derivative of $\sec x$ (giving $\sec x\tan x$ ⁠).

Chain rule, Theorem 4 (Section 5.3.1), with outer function $u^2$ and inner function $u = \sec x$ ⁠. Lesson: chain rule
Cross-check the $\sec^2 x$ derivative

Using $\sec^2 x = 1 + \tan^2 x$ (Basic 4), and the chain rule on $\tan^2 x = (\tan x)^2$ ⁠:
$\frac{d}{dx}\left(1 + \tan^2 x\right) = 0 + 2\tan x\cdot\sec^2 x.$
This is $2\sec^2 x\tan x$ again.

Reaching the same derivative by a different route confirms Step 4. It also shows that $\frac{d}{dx}\sec^2x$ can be written in several equivalent forms.

$\frac{d}{dx}\tan x = \sec^2 x$ (Table 5.3). Basics: trigonometric identities
Put the pieces together
$\frac{d^2y}{dx^2} = 6x + 2\sec^2 x\tan x,$
valid for every $x \neq (2n+1)\frac{\pi}{2}$ ⁠, the same $x$ for which $y$ is defined.

Add the results of Steps 3 and 4. The condition on $x$ is inherited from $\tan x$ ⁠: wherever $\cos x \neq 0$ ⁠, the functions $\sec x$ and $\tan x$ are differentiable, so differentiating twice is allowed.

Sum rule $(u+v)' = u' + v'$ ⁠.

Result

\frac{d^2y}{dx^2} = 6x + 2\sec^2 x\tan x,

for every $x \neq (2n+1)\frac{\pi}{2}$ ⁠, $n \in \Z$ ⁠. (Book's answer: $6x + 2\sec^2 x\tan x$ ⁠.)

Check it with numbers

Take $x = 0.5$ ⁠. The formula gives $6(0.5) + 2\sec^2(0.5)\tan(0.5)$ ⁠, which is about $3 + 1.4186890 = 4.4186890$ ⁠.

A second derivative can be estimated from three nearby values: $\frac{y(x+h) - 2y(x) + y(x-h)}{h^2}$ ⁠. With $h = 0.001$ ⁠:

$y(0.499) \approx 0.669256250958$ ⁠, $y(0.5) \approx 0.671302489844$ ⁠, $y(0.501) \approx 0.673353147420$ ⁠.

\begin{aligned} &y(0.501) - 2y(0.5) + y(0.499) \\ &\approx 0.000004418690, \end{aligned}

and dividing by $h^2 = 0.000001$ gives about $4.41869$ ⁠.

This agrees with the formula to five significant figures.

Common mistakes

Dropping the inner derivative. $\frac{d}{dx}\sec^2 x \neq 2\sec x$ ⁠. The factor $\sec x\tan x$ from the inside function is essential.
Squaring the first derivative. $\frac{d^2y}{dx^2}$ is not $\left(\frac{dy}{dx}\right)^2$ ⁠. It means "differentiate twice".
Getting $\frac{d}{dx}\sec x$ wrong. It is $\sec x\tan x$ ⁠, not $\tan^2 x$ or $\sec^2 x$ ⁠. Rederive it from $\frac{1}{\cos x}$ if unsure (Basic 2).

What to take away

Second derivative = differentiate, then differentiate the answer. No new rules are needed, only care.
Powers of trigonometric functions, such as $\sec^2 x$ and $\tan^2 x$ ⁠, always need the chain rule.
$\frac{d}{dx}\sec^2 x = 2\sec^2 x\tan x$ is worth remembering; it appears whenever $\tan x$ is differentiated twice.

Every fact used, and where it comes from

Fact	Used in	Source
$\frac{d}{dx}x^n = n\,x^{n-1}$ ⁠, $\frac{d}{dx}\tan x = \sec^2 x$	Steps 1, 3, 5	Table 5.3, Section 5.3
Sum rule	Steps 1, 2, 6	Algebra of derivatives, Section 5.3
$\frac{d^2y}{dx^2} = \frac{d}{dx}\left(\frac{dy}{dx}\right)$	Step 2	Section 5.7 (Lesson: second order derivative)
$\frac{d}{dx}\sec x = \sec x\tan x$	Step 4	Quotient rule, Section 5.3 (Basic 2)
Chain rule	Steps 4, 5	Theorem 4, Section 5.3.1 (Lesson: chain rule)
$\sec^2 x = 1 + \tan^2 x$	Step 5	Pythagorean identity divided by $\cos^2 x$

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Example 35: d2ydx2\frac{d^2y}{dx^2}dx2d2y​ for y=x3+tan⁡xy=x^3+\tan xy=x3+tanx

Basics you need

Solution

Find the first derivative

Set up the second derivative

Differentiate 3x23x^23x2

Differentiate sec⁡2x\sec^2 xsec2x with the chain rule

Cross-check the sec⁡2x\sec^2 xsec2x derivative

Put the pieces together

Every fact used, and where it comes from

Example 35: $\frac{d^2y}{dx^2}$ for $y=x^3+\tan x$

Differentiate $3x^2$

Differentiate $\sec^2 x$ with the chain rule

Cross-check the $\sec^2 x$ derivative