Why is the derivative of x written 2 x and not with sines and cosines?

The quotient rule first gives 12 x, and 2 x is the same thing written shorter. Either form is correct.

Unit 2 · Topic 2.10 · about 30 minutes · 7 parts · 12 scored checks, 3 of them multiple choice

AP Calculus AB 2.10: Finding the Derivatives of Tangent, Cotangent, Secant, and/or Cosecant Functions

Q: Do these rules work if the angle is in degrees?

No. They come from the derivatives of x and x, which hold only when x is in radians. Every angle in AP Calculus is in radians.

Built by Tanner Crow from College Board's course framework · Every answer key independently checked

Unit 2: 10 to 15% of the multiple-choice section

What you'll be able to do

Calculate the derivatives of tangent, cotangent, secant and cosecant, and of products and quotients built from them, by rewriting with identities.

Key vocabulary

Term	What it means	The exam trap
Tangent function, $\tan x$	The quotient $\tan x = \frac{\sin x}{\cos x}$ , defined wherever $\cos x \neq 0$ .	The quotient rule's numerator has a fixed order, bottom times the derivative of the top first; swapping the order flips the sign of the whole answer.
Cotangent function, $\cot x$	The quotient $\cot x = \frac{\cos x}{\sin x}$ , the reciprocal of $\tan x$ , defined wherever $\sin x \neq 0$ .	Before you differentiate, check that you wrote cosine on top; a function written upside down sends every later step wrong.
Secant function, $\sec x$	The reciprocal $\sec x = \frac{1}{\cos x}$ , defined wherever $\cos x \neq 0$ .	Secant goes with cosine and cosecant goes with sine, the opposite of what the names suggest.
Cosecant function, $\csc x$	The reciprocal $\csc x = \frac{1}{\sin x}$ , defined wherever $\sin x \neq 0$ .	Cosecant has no value where $\sin x = 0$ , such as $x = 0$ , so neither does its derivative, even though secant is defined there.
Pythagorean identity	The identity $\sin^{2} x + \cos^{2} x = 1$ , true for every $x$ .	Look for $\sin^{2} x + \cos^{2} x$ inside a numerator after the quotient rule; leaving it unsimplified hides a short answer.

Graph of y = sec x between the dashed vertical asymptotes x = -pi/2 and x = pi/2. Short tangent lines touch the curve at x = -1, x = 0 and x = 1, with slopes -2.88, 0 and 2.88.

Part 1 Think first How fast is the helicopter climbing, per radian?

Guess before you learn. A wrong guess still makes the answer stick.

Predict first

A rescue helicopter lifts straight up from a helipad. A tracking station on level ground 100 meters from the pad measures the angle of elevation $θ$ , in radians, and the helicopter's height is $h (θ) = 100 \tan θ$ meters. At $θ = \frac{π}{4}$ it is 100 meters up. At that angle, how fast is the height changing with respect to the angle?

In short. The derivatives of $\tan x$ , $\cot x$ , $\sec x$ and $\csc x$ come from writing each one in sines and cosines and using the quotient rule, with $\sin^{2} x + \cos^{2} x = 1$ to simplify tangent and cotangent.

$\frac{d}{d x} \tan x = \sec^{2} x$ and $\frac{d}{d x} \cot x = - \csc^{2} x$ .
$\frac{d}{d x} \sec x = \sec x \tan x$ and $\frac{d}{d x} \csc x = - \csc x \cot x$ .
Rewriting a function with an identity before you differentiate often turns a long quotient into terms whose derivatives you already know.

Part 2 Learn it Four new derivatives from rules you have

2a. Tangent and cotangent are quotients

You already know the derivatives of $\sin x$ and $\cos x$ , and you know the quotient rule. The other four trigonometric functions are quotients of sine and cosine, so rewrite first, then differentiate.

Start with $\tan x = \frac{\sin x}{\cos x}$ . The quotient rule (bottom times the derivative of the top, minus top times the derivative of the bottom, all over the bottom squared) gives

\frac{d}{d x} \tan x = \frac{\cos x \cdot \cos x - \sin x \cdot (- \sin x)}{\cos^{2} x} = \frac{\cos^{2} x + \sin^{2} x}{\cos^{2} x} .

Now an identity does the work. Since $\sin^{2} x + \cos^{2} x = 1$ , the numerator is 1, and

\frac{d}{d x} \tan x = \frac{1}{\cos^{2} x} = \sec^{2} x .

This holds wherever $\cos x \neq 0$ , which is everywhere $\tan x$ is defined. At $x = \frac{π}{6}$ , $\cos^{2} x = \frac{3}{4}$ , so the slope of the graph of $\tan x$ there is $\frac{4}{3}$ .

Cotangent works the same way, starting from $\cot x = \frac{\cos x}{\sin x}$ . Watch the signs: the derivative of the top, $\cos x$ , is $- \sin x$ , and that minus sign survives to the end.

Quick check · Fill the statement

No calculator. Tap a blank, then tap an answer.

To differentiate $\cot x$ , first write it as the quotient . The quotient rule gives the numerator $\sin x \cdot (- \sin x) - \cos x \cdot \cos x$ , which the identity $\sin^{2} x + \cos^{2} x = 1$ turns into . So the derivative of $\cot x$ is .

2b. Secant and cosecant are reciprocals

Secant and cosecant are reciprocals: $\sec x = \frac{1}{\cos x}$ and $\csc x = \frac{1}{\sin x}$ . A reciprocal is a quotient whose top is the constant 1, and the derivative of a constant is 0.

For secant,

\frac{d}{d x} \sec x = \frac{\cos x \cdot 0 - 1 \cdot (- \sin x)}{\cos^{2} x} = \frac{\sin x}{\cos^{2} x} .

That is correct, and it is usually split into two familiar pieces:

\frac{\sin x}{\cos^{2} x} = \frac{1}{\cos x} \cdot \frac{\sin x}{\cos x} = \sec x \tan x .

Cosecant follows the same steps with sine on the bottom. The numerator is $\sin x \cdot 0 - 1 \cdot \cos x = - \cos x$ , so

\frac{d}{d x} \csc x = \frac{- \cos x}{\sin^{2} x} = - \frac{1}{\sin x} \cdot \frac{\cos x}{\sin x} = - \csc x \cot x .

Set the four results side by side and a pattern shows. The two co-functions, cotangent and cosecant, have derivatives with a minus sign, and those derivatives are built from the co-functions: $- \csc^{2} x$ for $\cot x$ against $\sec^{2} x$ for $\tan x$ , and $- \csc x \cot x$ for $\csc x$ against $\sec x \tan x$ for $\sec x$ . The pattern helps you remember them. The quotient rule is how you check one you are unsure of.

Quick check · Match

No calculator. Tap a function, then its derivative.

2c. Rewrite first, then differentiate

The identities run both ways. A function written in sines and cosines can often be rewritten in tangents, secants and their co-functions, whose derivatives you now know.

Take $f (x) = \frac{1 + \sin x}{\cos x}$ . The quotient rule works, but splitting the fraction is shorter:

f (x) = \frac{1}{\cos x} + \frac{\sin x}{\cos x} = \sec x + \tan x,

so $f^{'} (x) = \sec x \tan x + \sec^{2} x$ .

Two answers can look different and still be the same function. Divide $\sin^{2} x + \cos^{2} x = 1$ by $\cos^{2} x$ and you get $\tan^{2} x + 1 = \sec^{2} x$ . So $\sec^{2} x$ and $1 + \tan^{2} x$ are one derivative written two ways. When your form differs from a classmate's, check whether an identity turns one into the other before you decide either is wrong.

Each of these derivatives exists only where its function is defined. Tangent and secant are not defined where $\cos x = 0$ , such as $x = \frac{π}{2}$ , and neither are their derivatives. Cotangent and cosecant are not defined where $\sin x = 0$ , such as $x = 0$ .

Quick check · Multiple choice

No calculator. Let $g (x) = \frac{2 + \cos x}{\sin x}$ . Which expression is $g^{'} (x)$ ?

Common mistakes

The mistake	Why it is wrong	Do this instead
Dropping the minus sign on the derivative of $\cot x$ or $\csc x$ .	Each comes from a quotient whose numerator picks up $- \sin x$ or $- \cos x$ , and that sign does not cancel.	Check every cotangent and cosecant term for its minus sign before you simplify or substitute.
Dividing the derivative of the top by the derivative of the bottom, as in $\frac{d}{d x} \tan x = \frac{\cos x}{- \sin x}$ .	The derivative of a quotient is not the quotient of the derivatives. That shortcut gives $- \cot x$ , which is negative in places where $\tan x$ is rising.	Write the quotient rule's full structure, bottom times the derivative of the top minus top times the derivative of the bottom, over the bottom squared.
Mixing up $\sec^{2} x$ and $\sec x \tan x$ .	Both begin with secant, but $\sec^{2} x$ is the derivative of $\tan x$ and $\sec x \tan x$ is the derivative of $\sec x$ .	When you are unsure, rebuild the one you need from sine and cosine. It takes two lines.

Part 3 Watch it done Rewrite, then use the product rule

Watch each step, then do the last one yourself.

Let $f (x) = \frac{\sin x}{\cos^{2} x}$ . Find $f^{'} (\frac{π}{3})$ .

Step 1 · Rewrite with secant and tangent
$f (x) = \frac{1}{\cos x} \cdot \frac{\sin x}{\cos x} = \sec x \tan x$ .
Why this step?
Why rewrite $f$ before differentiating, when the quotient rule could be used on it as given?
The quotient rule would need the derivative of $\cos^{2} x$ , which is a product of its own. The rewritten form is a product of two functions whose derivatives you already know, $\sec x$ and $\tan x$ , so there are fewer steps and fewer signs to lose.
Step 2 · Write the product rule's structure
$f^{'} (x) = \sec x \cdot \frac{d}{d x} (\tan x) + \tan x \cdot \frac{d}{d x} (\sec x)$ .
Step 3 · Use the two derivatives
$f^{'} (x) = \sec x \cdot \sec^{2} x + \tan x \cdot \sec x \tan x = \sec^{3} x + \sec x \tan^{2} x$ .
Step 4 · Find the values at the angle
$\cos \frac{π}{3} = \frac{1}{2}$ , so $\sec \frac{π}{3} = 2$ , and $\tan \frac{π}{3} = \sqrt{3}$ , so $\tan^{2} \frac{π}{3} = 3$ .
Your step · Calculate

No calculator. Finish it. For $f (x) = \frac{\sin x}{\cos^{2} x}$ , $f^{'} (x) = \sec^{3} x + \sec x \tan^{2} x$ , with $\sec \frac{π}{3} = 2$ and $\tan^{2} \frac{π}{3} = 3$ . What is $f^{'} (\frac{π}{3})$ ?

Your answer

Part 4 Try it Two new functions

Try it · Calculate

No calculator. Let $f (x) = 3 \tan x - \cot x$ . Find $f^{'} (\frac{π}{4})$ .

Your answer

Try it · Decide and justify

No calculator. A classmate says that $f (x) = \sec^{2} x$ and $g (x) = \tan^{2} x$ have the same derivative at every $x$ where both are defined. Is the classmate right?

Which reason supports your answer?

Part 5 Explain it Write it the way the exam asks

Free response on the exam is handwritten, so write your answer on paper first. Nothing you type here leaves this page. Then open the model answer and score yourself on each point.

Free response is 50% of the exam score. This one is shaped like Question 6, from free-response Part B (calculator not permitted).

Free-response practice · Question 6 style

No calculator. A spotlight sits on level ground 20 meters from a vertical wall and is tilted up at an angle of $θ$ radians, for $0 \leq θ < \frac{π}{2}$ . The bright spot on the wall is $H (θ) = 20 \tan θ$ meters above the ground, and the beam from the light to the spot is $L (θ) = 20 \sec θ$ meters long.

(a) Write $H (θ)$ in terms of $\sin θ$ and $\cos θ$ , and use that form to show that $H^{'} (θ) = 20 \sec^{2} θ$ .

(b) Find $L^{'} (\frac{π}{4})$ . Using correct units, interpret the meaning of this value in the context of the problem.

Your answer stays on this page. It is never sent or saved.

Model answer

(a) $H (θ) = 20 \cdot \frac{\sin θ}{\cos θ}$ . By the quotient rule, $H^{'} (θ) = 20 \cdot \frac{\cos θ \cdot \cos θ - \sin θ \cdot (- \sin θ)}{\cos^{2} θ} = 20 \cdot \frac{\cos^{2} θ + \sin^{2} θ}{\cos^{2} θ} = \frac{20}{\cos^{2} θ} = 20 \sec^{2} θ$ , because $\sin^{2} θ + \cos^{2} θ = 1$ .

(b) $L (θ) = \frac{20}{\cos θ}$ , so $L^{'} (θ) = 20 \cdot \frac{\cos θ \cdot 0 - 1 \cdot (- \sin θ)}{\cos^{2} θ} = 20 \sec θ \tan θ$ . Then $L^{'} (\frac{π}{4}) = 20 \cdot \sqrt{2} \cdot 1 = 20 \sqrt{2}$ meters per radian. When the light is tilted at $\frac{π}{4}$ radians, the length of the beam is increasing at $20 \sqrt{2}$ meters per radian of tilt.

(c) $H^{'} (θ) - L^{'} (θ) = \frac{20}{\cos^{2} θ} - \frac{20 \sin θ}{\cos^{2} θ} = \frac{20 (1 - \sin θ)}{\cos^{2} θ}$ . For $0 \leq θ < \frac{π}{2}$ , $\sin θ < 1$ , so the numerator is positive, and $\cos^{2} θ > 0$ . The difference is positive, so $H^{'} (θ) > L^{'} (θ)$ .

Score your own answer. This is a self-check, not a grade. Tick each point your answer makes.

Writes $H (θ) = 20 \cdot \frac{\sin θ}{\cos θ}$ and the quotient rule's structure, and uses $\sin^{2} θ + \cos^{2} θ = 1$ to reach $20 \sec^{2} θ$ .
Finds $L^{'} (θ) = 20 \sec θ \tan θ$ and gets $L^{'} (\frac{π}{4}) = 20 \sqrt{2}$ .
Says the beam's length is increasing at $20 \sqrt{2}$ meters per radian when $θ = \frac{π}{4}$ , with units.
Writes $H^{'} (θ) - L^{'} (θ)$ as $\frac{20 (1 - \sin θ)}{\cos^{2} θ}$ , or an equal form, and explains it is positive because $\sin θ < 1$ and $\cos^{2} θ > 0$ on the interval.

Part 6 Check yourself One function, a table of values

The two case questions use the same table. Answer them from it.

A function $u$ is differentiable for all $x$ . The table gives some values of $u$ and its derivative $u^{'}$ .

Values of u and its derivative
$x$	$u (x)$	$u^{'} (x)$
$\frac{π}{4}$	3	$- 1$
$\frac{π}{3}$	2	4

Case question 1 · Calculate

No calculator. Let $h (x) = u (x) \tan x$ . Find $h^{'} (\frac{π}{4})$ .

Your answer

Case question 2 · Multiple choice

No calculator. Let $k (x) = \frac{\sec x}{u (x)}$ . What is $k^{'} (\frac{π}{3})$ ?

The rest stand on their own and do not use the case.

Check yourself · Spot the error

No calculator. A student finds the derivative of $\csc x$ and its value at $x = \frac{π}{6}$ . Tap the first line with a mistake.

Check yourself · Sort it

No calculator. Rewrite each function with an identity, wherever it is defined. Tap a card, then tap its derivative.

$\sec^{2} x$

$- \csc^{2} x$

$\sec x \tan x$

Part 7 Keep it Bring back earlier ideas

This lesson stands on the quotient rule and on the sine and cosine rules. Pulling them back up is what makes them last.

Bring it back · Calculate

No calculator. A function $p$ is the quotient $p (x) = \frac{g (x)}{h (x)}$ , where $g$ and $h$ are differentiable, $g (2) = 6$ , $g^{'} (2) = 7$ , $h (2) = 2$ and $h^{'} (2) = - 1$ . Find $p^{'} (2)$ .

Your answer

Bring it back · Multiple choice

No calculator. Let $f (x) = 3 \sin x - 2 \cos x$ . Which expression is $f^{'} (x)$ ?

In three lines

Tangent, cotangent, secant and cosecant are quotients of sine and cosine, so the quotient rule differentiates each one.
For tangent and cotangent, $\sin^{2} x + \cos^{2} x = 1$ collapses the numerator to $1$ or $- 1$ ; for secant and cosecant, splitting the fraction gives $\sec x \tan x$ and $- \csc x \cot x$ .
Rewrite with an identity before you differentiate when it turns a quotient into terms whose derivatives you know.

Unit 2 check: 10 questions from every lesson in Unit 2

Exam strategy for this topic

Section I Part A, Multiple Choice (calculator not permitted): 29 questions in 62 minutes.
Section I Part B, Multiple Choice (graphing calculator required): 13 questions in 38 minutes.
Section II Part A, Free Response (graphing calculator required): 2 questions in 30 minutes.
Section II Part B, Free Response (calculator not permitted): 4 questions in 60 minutes.

When a function is given in sines and cosines, look for an identity that rewrites it with tangent, cotangent, secant or cosecant before you reach for the quotient rule. When you evaluate a derivative at an angle, write the rule's structure first, then substitute exact values such as $\sec \frac{π}{3} = 2$ and $\tan \frac{π}{4} = 1$ .

Questions students ask

Do I have to memorize all four derivatives?

Knowing them saves time on the exam, and the pattern in this lesson makes that easier. If you blank on one, rebuild it from sine and cosine with the quotient rule; it takes two or three lines.

Why is the derivative of

\tan x

written

\sec^{2} x

and not with sines and cosines?

The quotient rule first gives $\frac{1}{\cos^{2} x}$ , and $\sec^{2} x$ is the same thing written shorter. Either form is correct.

Do these rules work if the angle is in degrees?

No. They come from the derivatives of $\sin x$ and $\cos x$ , which hold only when $x$ is in radians. Every angle in AP Calculus is in radians.

What about

\tan (2 x)

\sec (x^{2})

Those have a function inside the tangent or secant, so they need the chain rule, which comes in the next topic. In this lesson the input is always plain $x$ .

Course alignment, for teachers

AP Calculus AB topic 2.10, Unit 2: Differentiation: Definition and Fundamental Properties.

Skill 1.D: Identify an appropriate mathematical rule or procedure based on the relationship between concepts (e.g., rate of change and accumulation) or processes (e.g., differentiation and its inverse process, anti-differentiation) to solve problems.

This lesson maps the one essential knowledge statement of the topic to the part that teaches each one: 1 is checked by a scored question. The map is in the lesson file.