Chapter 5 · Continuity and Differentiability

Exercise 5.7 · Q6: Second derivative of $e^x\sin 5x$

Exercise 5.7Second order derivative · Section 5.7

The question

Find the second order derivatives of the functions given in Exercises 1 to 10.

6. $e^x\sin 5x$

Exam answer

Let $y = e^x\sin 5x$ ⁠. By the product rule,

\begin{aligned} \frac{dy}{dx} &= e^x\sin 5x + 5e^x\cos 5x \\ &= e^x(\sin 5x + 5\cos 5x), \\ \frac{d^2y}{dx^2} &= e^x(\sin 5x + 5\cos 5x) \\ &\quad + e^x(5\cos 5x - 25\sin 5x) \\ &= e^x(10\cos 5x - 24\sin 5x) \end{aligned}

$\therefore\ \dfrac{d^2y}{dx^2} = 2e^x(5\cos 5x - 12\sin 5x)$

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

The one idea this problem is built on

Differentiating $e^x$ leaves it unchanged, so with the product rule $e^x$ comes out as a common factor every time:

\frac{d}{dx}\big(e^x\,g(x)\big) = e^x\big(g(x) + g'(x)\big).

So each differentiation just replaces the bracket $g$ by $g + g'$ ⁠. The chain rule supplies the factor 5 each time $\sin 5x$ or $\cos 5x$ is differentiated.

Given, to find, and the plan

Given: $y = e^x\sin 5x$ ⁠, defined and differentiable for every real $x$ ⁠.
To find: $\dfrac{d^2y}{dx^2}$ ⁠.
Plan: (1) State the domain. (2) Chain rule for $\sin 5x$ and $\cos 5x$ ⁠. (3) Product rule for $\frac{dy}{dx}$ ⁠. (4) Product rule again for $\frac{d^2y}{dx^2}$ ⁠. (5) Collect terms and match the textbook form.

Basics you need

1Derivative of $e^x$

$\frac{d}{dx}e^x = e^x$ (Theorem 5(1), Section 5.4), and $e^x \gt 0$ for every real $x$ ⁠. Basics: exponentials and logarithms

2Chain rule (Theorem 4)

If $y = f(g(x))$ ⁠, then $\frac{dy}{dx} = f'(g(x))\cdot g'(x)$ ⁠: derivative of the outer function at the inner one, times the derivative of the inner function. Lesson: chain rule

3Product rule (Leibnitz rule)

$(uv)' = u'v + uv'$ (Section 5.3, rule (2)). Basics: derivative rules

4Derivatives of $\sin$ and $\cos$

$\frac{d}{dx}\sin x = \cos x$ ⁠, $\frac{d}{dx}\cos x = -\sin x$ (Table 5.3). Basics: trigonometric identities

Solution

Name the function and its domain

Let $y = e^x\sin 5x$ ⁠. Both $e^x$ and $\sin 5x$ are defined and differentiable for all real $x$ ⁠, so the answer holds for every $x \in \mathbf{R}$ ⁠.

No logarithm, root or denominator appears, so nothing is excluded.
Chain rule for $\sin 5x$ and $\cos 5x$

For $\sin 5x$ ⁠: outer function $\sin$ ⁠, inner function $5x$ with derivative $5$ ⁠. For $\cos 5x$ ⁠: outer $\cos$ ⁠, same inner function. So
$\begin{aligned} \frac{d}{dx}\sin 5x &= \cos 5x\cdot 5 = 5\cos 5x, \\ \frac{d}{dx}\cos 5x &= -\sin 5x\cdot 5 = -5\sin 5x. \end{aligned}$
These two small results are needed in both rounds of differentiation, so we work them out once. The factor 5 is the derivative of the inner function $5x$ ⁠; forgetting it is the classic error.

Chain rule (Theorem 4, Section 5.3.1). Lesson: chain rule
First derivative: product rule

Take $u = e^x$ ⁠, $v = \sin 5x$ ⁠, so $u' = e^x$ and $v' = 5\cos 5x$ (Step 2):
$\begin{aligned} \frac{dy}{dx} &= e^x\sin 5x + e^x\cdot 5\cos 5x \\ &= e^x(\sin 5x + 5\cos 5x). \end{aligned}$
$y$ is a product of $e^x$ and $\sin 5x$ ⁠. Because $\frac{d}{dx}e^x = e^x$ ⁠, both terms of the product rule contain $e^x$ ⁠, and we take it out as a common factor to keep the next step tidy.

Product rule (Section 5.3, rule (2)); $\frac{d}{dx}e^x = e^x$ (Theorem 5).
Second derivative: product rule again

Now $\frac{dy}{dx} = e^x\,v$ with $v = \sin 5x + 5\cos 5x$ ⁠. By the sum rule and Step 2,
$\begin{aligned} v' &= 5\cos 5x + 5(-5\sin 5x) \\ &= 5\cos 5x - 25\sin 5x. \end{aligned}$
Product rule with $u = e^x$ ⁠:
$\begin{aligned} \frac{d^2y}{dx^2} &= e^x\,v + e^x\,v' \\ &= e^x(\sin 5x + 5\cos 5x) \\ &\quad + e^x(5\cos 5x - 25\sin 5x). \end{aligned}$
We differentiate the first derivative from Step 3, which is again $e^x$ times a bracket, so the product rule is the right tool. The bracket's derivative needs the sum rule and the two chain-rule results of Step 2.

Section 5.7: $\frac{d^2y}{dx^2} = \frac{d}{dx}\left(\frac{dy}{dx}\right)$ ⁠. Lesson: second order derivative
Collect like terms and factor

Inside $e^x(\ \cdots)$ ⁠: the $\cos 5x$ terms give $5 + 5 = 10$ ⁠, the $\sin 5x$ terms give $1 - 25 = -24$ ⁠. So
$\begin{aligned} \frac{d^2y}{dx^2} &= e^x(10\cos 5x - 24\sin 5x) \\ &= 2e^x(5\cos 5x - 12\sin 5x). \end{aligned}$
So $\dfrac{d^2y}{dx^2} = 2e^x(5\cos 5x - 12\sin 5x)$ for every real $x$ ⁠.

Taking out the common factor 2 from $10$ and $-24$ gives the textbook's form exactly.

$10\cos 5x - 24\sin 5x = 2(5\cos 5x - 12\sin 5x)$ ⁠.

Result

For $y = e^x\sin 5x$ (all real $x$ ⁠):

\begin{aligned} \frac{dy}{dx} &= e^x(\sin 5x + 5\cos 5x), \\ \frac{d^2y}{dx^2} &= 2e^x(5\cos 5x - 12\sin 5x). \end{aligned}

(Textbook answer: $2e^x(5\cos 5x - 12\sin 5x)$ ⁠.)

Check it with numbers

At $x = 0$ ⁠: $y'' = 2(5 - 0) = 10$ ⁠. A rough independent look: for small $x$ ⁠, $e^x \approx 1 + x$ ⁠, $\sin 5x \approx 5x$ and $5\cos 5x \approx 5$ (tangent-line approximations at $0$ ⁠, with errors at most a multiple of $x^2$ ⁠; a plausibility check only), so $y' \approx (1 + x)(5x + 5) \approx 5 + 10x$ ⁠, whose slope is 10.

At $x = 0.3$ (radians): formula $y'' \approx -31.36061$ ⁠; second difference with $h = 0.001$ ⁠, $\frac{y(0.301) - 2y(0.3) + y(0.299)}{0.000001}$ ⁠, is about $-31.36056$ ⁠. First derivative: formula $\approx 1.823904$ ⁠, central difference $\approx 1.823885$ ⁠. Both agree to about five significant figures; the small gaps shrink like $h^2$ ⁠. ✓

Common mistakes

Missing the chain-rule factor: $\frac{d}{dx}\sin 5x = 5\cos 5x$ ⁠, not $\cos 5x$ ⁠; and $\frac{d}{dx}\cos 5x = -5\sin 5x$ ⁠.
Differentiating only the bracket in the second round. $\frac{dy}{dx}$ is a product; the term $e^x\cdot v$ (from differentiating $e^x$ ⁠) is needed too.
Sign slip: $5 \times (-5\sin 5x) = -25\sin 5x$ ⁠.

What to take away

$\frac{d}{dx}\big(e^x g\big) = e^x(g + g')$ ⁠: keep $e^x$ factored out.
Work out the chain-rule derivatives of $\sin kx$ ⁠, $\cos kx$ once ( $k\cos kx$ ⁠, $-k\sin kx$ ⁠) and reuse them.

Every fact used, and where it comes from

Fact	Used in	Source
$\frac{d}{dx}e^x = e^x$	Steps 3, 4	Theorem 5(1), Section 5.4 (Lesson: exponential and logarithmic functions)
Chain rule	Step 2	Theorem 4, Section 5.3.1 (Lesson: chain rule)
$(uv)' = u'v + uv'$ ⁠; $(u + v)' = u' + v'$	Steps 3, 4	Section 5.3, rules (1), (2)
$\frac{d}{dx}\sin x = \cos x$ ⁠, $\frac{d}{dx}\cos x = -\sin x$	Step 2	Table 5.3 (Basics: derivative rules)
$\frac{d^2y}{dx^2} = \frac{d}{dx}\left(\frac{dy}{dx}\right)$	Step 4	Section 5.7 (Lesson: second order derivative)

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Exercise 5.7 · Q6: Second derivative of exsin⁡5xe^x\sin 5xexsin5x

Basics you need

Solution

Name the function and its domain

Chain rule for sin⁡5x\sin 5xsin5x and cos⁡5x\cos 5xcos5x

First derivative: product rule

Second derivative: product rule again

Collect like terms and factor

Every fact used, and where it comes from

Exercise 5.7 · Q6: Second derivative of $e^x\sin 5x$

Chain rule for $\sin 5x$ and $\cos 5x$