Exponential and trigonometric functions

Two families of functions are their own derivatives up to a factor: exponentials, and sine and cosine taken together. Everything about them follows from two limits, the one that defines $e$ and the one that says $\sin x$ behaves like $x$ near zero. Most exam questions are the product and quotient rules applied to these functions, or a tangent line at a value where they are easy to evaluate.

Start practising Browse all 22 questions

In this topic

Taught: Before the midterm
Questions: 22
Total marks: 50
Suggested time: 83 min
Difficulty: Introductory to Capstone
Show the 4 hardest

Key results

Two basic limits: $\lim_{h \to 0}\frac{e^{h} - 1}{h} = 1, \qquad \lim_{x \to 0}\frac{\sin x}{x} = 1$
The second with $x$ in radians.
The derivative of eˣ: $\frac{d}{dx}e^{x} = e^{x}$
Other bases: $\frac{d}{dx}a^{x} = a^{x}\log a$
For a constant base $a > 0$ .
Sine and cosine: $\frac{d}{dx}\sin x = \cos x, \qquad \frac{d}{dx}\cos x = -\sin x$
With $x$ in radians.
Tangent and secant: $\frac{d}{dx}\tan x = \sec^{2}x, \qquad \frac{d}{dx}\sec x = \sec x\tan x$
With $x$ in radians, wherever $\cos x \neq 0$ .
Cotangent and cosecant: $\frac{d}{dx}\cot x = -\csc^{2}x, \qquad \frac{d}{dx}\csc x = -\csc x\cot x$
With $x$ in radians, wherever $\sin x \neq 0$ .

Read the theoryUBC Math 100 §3.5 Derivatives of exponential functionsUBC Math 100 §4.2 Trigonometric functions and their derivativesCLP-1 §2.7 Derivatives of exponential functionsCLP-1 §2.8 Derivatives of trigonometric functions

Worked example

Standard3 marks · 4 min

Find the $y$ -intercept of the line tangent to $y = \sin x + 3x$ at $x = \pi$ .

Strategy

Point plus slope gives the tangent line; the intercept is its value at x = 0. Every trigonometric value at π is one you know exactly.

Working

$y(\pi) = \sin\pi + 3\pi = 3\pi, \qquad y' = \cos x + 3, \qquad y'(\pi) = -1 + 3 = 2$
Point (π, 3π), slope 2.
$y - 3\pi = 2(x - \pi) \quad\Longrightarrow\quad y = 2x + \pi$
Point-slope form, simplified.
$\text{y-intercept: } \pi$
Set x = 0.

Answer

\pi

Check

Near x = π, sin x ≈ π − x, so the curve is close to (π − x) + 3x = 2x + π, a line with intercept π.

Where this goes wrong

Taking cos π = 1 and getting slope 4, or sin π = 1 and getting the point wrong. At π the cosine is −1 and the sine is 0.

On the paper

Work in radians, write the derivative before you substitute, and give exact values such as $\pi$ , $e$ and $\sqrt{3}$ unless a decimal is asked for.

Leans on

Values of sin and cos at π, from Functions, graphs and inverses

Your turn

Standard3 marks · 4 min

Let $f(x) = \dfrac{e^{x}}{2 + \cos x}$ . Find $f'\!\left(\dfrac{\pi}{2}\right)$ .

Answer this question

Practice questions

5 of the 22, across every level. Open a solution to read the working, or answer the question yourself and it counts towards your progress.

Introductory1 mark · 2 min
Let $f(x) = 4e^{x} - 2\sin x$ . Find $f'(0)$ .
Strategy
Both functions have derivatives you should know outright. Differentiate each term with its constant, then evaluate at 0, where e⁰ = 1 and cos 0 = 1.
Working
$f'(x) = 4e^{x} - 2\cos x$
Derivative of eˣ is eˣ; derivative of sin x is cos x.
$f'(0) = 4(1) - 2(1) = 2$
e⁰ = 1 and cos 0 = 1.
Answer
$f'(0) = 2$
The answer is 2.
Check
Near 0, eˣ ≈ 1 + x and sin x ≈ x, so f(x) ≈ 4 + 2x, a line of slope 2.
Where this goes wrong
Writing the derivative of sin x as −cos x. It is cos x; the minus sign belongs to the derivative of cos x.
Leans on
Values of eˣ, sin and cos at 0, from Functions, graphs and inverses
Try a similar question
Let $f(x) = 3\tan x - 2\sec x$ . Find $f'\!\left(\dfrac{\pi}{6}\right)$ .
Answer this question
Introductory1 mark · 2 min
Which of the following is the derivative of $g(x) = 3^{x}$ ?
Strategy
An exponential with base a is e raised to (log a)·x. Differentiating that brings down the constant log a.
Working
$3^{x} = e^{x\log 3}$
Rewrite with base e.
$\frac{d}{dx}\, e^{x\log 3} = e^{x\log 3}\cdot\log 3 = 3^{x}\log 3$
The derivative of e^kx is k·e^kx, with k = log 3.
Answer
$g'(x) = 3^{x}\log 3$
The answer is 3 to the x times the natural log of 3.
Check
At x = 0, (3^0.001 − 1)/0.001 ≈ 1.099, which is log 3 ≈ 1.0986, not the 1 that 3ˣ alone would give.
Where this goes wrong
Applying the power rule to get x·3^x−1. That rule needs a constant exponent; when the variable is in the exponent the derivative picks up a factor of log 3 instead.
Leans on
Rewriting aˣ as e^{x log a}, from Functions, graphs and inverses
Try a similar question
Let $f(x) = 4e^{x} - 2\sin x$ . Find $f'(0)$ .
Answer this question
Standard3 marks · 5 min
Find every $x$ in $[0, 2\pi]$ at which the graph of $f(x) = \sin x + \cos x$ has a horizontal tangent, and enter the sum of those values.
Strategy
Horizontal tangents are zeros of the derivative. The derivative is cos x − sin x, which vanishes where sine and cosine are equal.
Working
$f'(x) = \cos x - \sin x = 0 \iff \sin x = \cos x \iff \tan x = 1$
Dividing by cos x is safe: cos x = 0 would force sin x = 0 too, impossible.
$x = \tfrac{\pi}{4},\ \tfrac{5\pi}{4}$
tan x = 1 in the first and third quadrants.
$\tfrac{\pi}{4} + \tfrac{5\pi}{4} = \tfrac{3\pi}{2}$
Sum.
Answer
$x = \frac{\pi}{4},\ \frac{5\pi}{4};\quad \text{sum } \frac{3\pi}{2}$
The answer is pi over 4 and 5 pi over 4, whose sum is 3 pi over 2.
Check
Written as √2 sin(x + π/4), sin x + cos x peaks at x = π/4 and bottoms out at x = 5π/4, its only flat points in [0, 2π].
Where this goes wrong
Stopping at x = π/4. The equation tan x = 1 has two solutions per full turn, and 5π/4 is the second.
Leans on
Solving a trigonometric equation, from Functions, graphs and inverses
Try a similar question
Find the sum of all solutions of the equation below in the interval $[0, 2\pi)$ . $2\sin^{2} x - \sin x - 1 = 0$
Answer this question
Challenging4 marks · 6 min
Find the constants $a$ and $b$ for which the function below is differentiable everywhere, and enter $a + b$ .
$f(x) = \begin{cases} ax + b, & x < \dfrac{\pi}{4} \\[8pt] 3\sin x + 4\cos x, & x \geq \dfrac{\pi}{4} \end{cases}$
Strategy
Both pieces are differentiable on their own; only the join at π/4 can fail. Matching slopes there fixes a; continuity then fixes b, remembering that the line’s value at the join is a(π/4) + b.
Working
$\text{slope at } \tfrac{\pi}{4}:\quad a = 3\cos\tfrac{\pi}{4} - 4\sin\tfrac{\pi}{4} = \tfrac{3\sqrt{2}}{2} - \tfrac{4\sqrt{2}}{2} = -\tfrac{\sqrt{2}}{2}$
The derivative of the right piece is 3cos x − 4sin x.
$\text{value at } \tfrac{\pi}{4}:\quad a\cdot\tfrac{\pi}{4} + b = 3\sin\tfrac{\pi}{4} + 4\cos\tfrac{\pi}{4} = \tfrac{7\sqrt{2}}{2}$
Continuity at the join.
$b = \tfrac{7\sqrt{2}}{2} - \left(-\tfrac{\sqrt{2}}{2}\right)\tfrac{\pi}{4} = \tfrac{7\sqrt{2}}{2} + \tfrac{\sqrt{2}\,\pi}{8}$
Solve with a = −√2/2.
$a + b = -\tfrac{\sqrt{2}}{2} + \tfrac{7\sqrt{2}}{2} + \tfrac{\sqrt{2}\,\pi}{8} = 3\sqrt{2} + \tfrac{\sqrt{2}\,\pi}{8} \approx 4.80$
Combine.
Answer
$a = -\frac{\sqrt{2}}{2},\ b = \frac{7\sqrt{2}}{2} + \frac{\sqrt{2}\,\pi}{8},\ a + b = 3\sqrt{2} + \frac{\sqrt{2}\,\pi}{8}$
The answer is a is −√2/2 and b is 7√2/2 + √2π/8, so a + b is 3√2 + √2π/8, about 4.80.
Check
At x = 1 the line ax + b has height a + b, and as a tangent to a concave curve it sits just above 3 sin 1 + 4 cos 1 ≈ 4.69.
Where this goes wrong
Setting b equal to the value at the join and getting a + b = 3√2. The line is ax + b, so its value at π/4 is a(π/4) + b, and the a(π/4) part has to be moved across.
Leans on
Differentiability at a join, from The definition of the derivative
Try a similar question
Find the values of the constants $a$ and $b$ for which the function below is differentiable everywhere, and enter $b$ .
Answer this question
Capstone5 marks · 10 min
The function below is differentiable everywhere. Find $a$ .
$f(x) = \begin{cases} ax + b, & x < \dfrac{\pi}{6} \\[8pt] \dfrac{5\cos x}{3 + \sin x + \cos x}, & x \geq \dfrac{\pi}{6} \end{cases}$
Strategy
A line joined to a smooth function is differentiable at the join exactly when the line has the function’s value and slope there. Matching slopes makes a the derivative of the quotient at π/6; matching values would then fix b, which is not asked for.
Working
$u = 5\cos x,\ u' = -5\sin x;\qquad v = 3 + \sin x + \cos x,\ v' = \cos x - \sin x$
Parts for the quotient rule.
$g'(x) = \frac{\begin{gathered} -5\sin x\,(3 + \sin x + \cos x) \\ {} - 5\cos x\,(\cos x - \sin x) \end{gathered}}{(3 + \sin x + \cos x)^{2}}$
u′v − uv′ over v².
$= \frac{-15\sin x - 5\sin^{2}x - 5\cos^{2}x}{(3 + \sin x + \cos x)^{2}} = -\frac{15\sin x + 5}{(3 + \sin x + \cos x)^{2}}$
The sin x cos x terms cancel, and 5sin² x + 5cos² x = 5.
$g'\!\left(\tfrac{\pi}{6}\right) = -\frac{\tfrac{15}{2} + 5}{\left(\tfrac{7 + \sqrt{3}}{2}\right)^{2}} = -\frac{50}{52 + 14\sqrt{3}} = -\frac{25}{26 + 7\sqrt{3}}$
At π/6, sin x = 1/2 and cos x = √3/2, so v = (7 + √3)/2; and (7 + √3)² = 52 + 14√3.
$a = g'\!\left(\tfrac{\pi}{6}\right) = -\frac{25}{26 + 7\sqrt{3}} \approx -0.656$
Matching slopes gives a; matching values would give b, which is not asked for.
Answer
$a = -\dfrac{25}{26 + 7\sqrt{3}} \approx -0.656$
The answer is negative 25 over (26 + 7√3), about −0.656.
Check
The quotient goes from 1.25 at x = 0 to about 0.573 at x = π/3, an average slope of about −0.65, near −0.656 at the midpoint π/6.
Where this goes wrong
Differentiating only the numerator, which gives −5sin x/(3 + sin x + cos x) and a ≈ −0.573. The quotient rule’s second term, −u v′, is not zero at π/6 and has to be kept.
Leans on
Differentiability at a join, from The definition of the derivative
Try a similar question
Find the constants $a$ and $b$ for which the function below is differentiable everywhere, and enter $a + b$ .
Answer this question

Browse all 22 questions

Needs
Differentiation rules
Power, sum, product and quotient rules, tangent lines and higher-order derivatives.
23 questions
Leads to
The chain rule and logarithms
Compositions, nested rules, the natural logarithm and logarithmic differentiation.
30 questions
Leads to
Exponential growth and decay
The differential equations y′ = ky and y′ = a − by: half-lives, doubling times, Newton’s law of cooling and inflow and outflow.
11 questions
Leads to
Antiderivatives
Reversing differentiation, the constant of integration and initial conditions.
11 questions

Key results

Worked example

Strategy

Working

Answer

Check

Where this goes wrong

On the paper

Leans on

Your turn

Practice questions

Strategy

Working

Answer

Check

Where this goes wrong

Leans on

Try a similar question

Strategy

Working

Answer

Check

Where this goes wrong

Leans on

Try a similar question

Strategy

Working

Answer

Check

Where this goes wrong

Leans on

Try a similar question

Strategy

Working

Answer

Check

Where this goes wrong

Leans on

Try a similar question

Strategy

Working

Answer

Check

Where this goes wrong

Leans on

Try a similar question

Related topics

Differentiation rules

The chain rule and logarithms

Exponential growth and decay

Antiderivatives