If $F(x)=(\sec(x))^3$, what is $F''(x)$?

$F''(x)=6\sec(x)\tan(x)+8\sec^3(x)\tan^2(x)$

$F''(x)=6\sec(x)\tan(x)-8\sec^3(x)\tan(\theta)$

$F''(x)=12\sec(x)\tan(x)+16\sec(\theta)\tan(\theta)$

$F''(x)=12\sec(x)\tan(x)+8\sec(\theta)\tan(\theta)$

Considering the function $g(x) = \cot(x)$, what does its derivative evaluate to when $x = \frac{3\pi}{4}$?

$-\csc^2\left(\frac{3\pi}{4}\right)

$\csc^2\left(\frac{3\pi}{4}\right)

$\sec^2\left(\frac{3\pi}{4}\right)

Given that $g(\theta)=\ln(\text{inverse }\sec(\theta))^5$, finding the sixth derivative would yield which term as the constant coefficient for $\sec(\theta)\cdot\csc(\theta)$ positioned at the sixth degree?

CORRECT. $(-720\sec(\theta)\csc(\theta))^6$

INCORRECT (-120\sec(\theta)\csc(\theta))^5)

INCORRECT (-5040\sec(\theta)\csc(\theta))^7)

INCORRECT (-2520\sec(\theta)\csc(\theta))^8)

If the derivative of $y = \sec(x)$ is given by $y' = \sec(x)\tan(x)$, what is the derivative of $y = \sec(3x + 2)$?

$y' = 3\sec(3x + 2)\tan(3x + 2)

$y' = \sec^2(3x + 2)\tan^2(3x + 2)

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2.10 Finding the Derivatives of Tangent, Cotangent, Secant, and/or Cosecant Functions

2 min read•february 15, 2024

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Review all units live with expert teachers & students

Welcome back to AP Calculus with Fiveable! Now that you’ve mastered finding derivatives of trigonometric functions $\sin x$ and $\cos x$ , let’s cover the rest! Remembering these rules is key to simplifying your calculus journey. 🌟

💫 Derivatives of Advanced Trigonometric Functions

First, let’s have a glance at a summary table for quick reference.

Function	Derivative
Tangent Function: $f(x) =\tan x$	$f'(x)=\sec^2 x$
Cotangent Function: $g(x)=\cot x$	$g'(x)=-\csc^2 x$
Secant Function: $h(x)= \sec x$	$h'(x)= \sec x \tan x$
Cosecant Function: $k(x)=\csc x$	$k'(x) = -\csc x \cot x$

It's important to note that these are only valid for angles in radians, not degrees.

Derivative of $\tan x$

The derivative of $\tan x$ is $\sec^2 x$ . Let’s consider an example:

f(x)=3\tan x+2x^2

To find the derivative of this equation, differentiate $3\tan x$ and $2x^2$ individually.

Since the derivative of $\tan x$ is $\sec^2 x$ , the derivative of the first part is $3\sec^2 x$ . The derivative of $2x^2$ is $4x$ . Hence, $f'(x) = 3\sec^2 x + 4x$ .

Derivative of $\cot x$

The derivative of $\cot x$ is $-\csc^2 x$ . For example:

f(x)=5\cot x+x

We again have to differentiate the two terms separately! The derivative of $\cot x$ is $-\csc^2 x$ , so the derivative of the first term is $-5\csc^2 x$ . The derivative of $x$ is $1$ . Therefore, $f'(x) = -5\csc^2 x + 1$ or $f'(x)=1-5\csc^2 x$ .

Derivative of $\sec x$

The derivative of $\sec x$ is $\sec x \tan x$ . As an example:

f(x)=2\sec x+3x^3

Knowing the above trig derivative rule, the derivative of the first term is $2\sec x \tan x$ . The derivative of $3x^3$ is $9x^2$ . Thus, $f'(x) = 2\sec x \tan x + 9x^2$ .

Derivative of $\csc x$

Last but not least, the derivative of $\csc x$ is $-\csc x \cot x$ . For instance:

f(x)=4\csc x+7x^2

The derivative of the first part is $-4\csc x \cot x$ . The derivative of $7x^2$ is $14x$ . Therefore, $f'(x) = -4\csc x \cot x + 14x$ .

🏋️‍♂️ Practice Problems

Here are a couple of questions for you to get the concepts down!

❓ Advanced Trig Derivative Practice Questions

Find the derivatives for the following problems.

$f(x) = 2 \tan(x) + \sec(x)$
$f(x) = \frac{\cot(x)}{\csc(x)}$
$g(x) =\tan^2(6x)$
$h(x) = 5\cot(x)$

💡 Before we reveal the answers, remember to use the chain rule, sum rule, and quotient rules.

🤔 Advanced Trig Derivative Practice Solutions

$f'(x) = 2 \sec^2(x) + \sec(x) \tan(x)$
$f'(x)=−\csc^2(x)$
$g'(x)=2\tan(6x)(\frac{1}{\cos^2(6x)})$
$h'(x)=−5\csc^2(x)$

🌟 Closing

Practice these rules, and you’ll soon find them as intuitive as the basic derivatives! Keep up the great work. 🌈

Key Terms to Review (10)

(-csc(x)cot(x))

: The term (-csc(x)cot(x)) represents the product of the cosecant and cotangent functions of an angle x, with a negative sign in front. It is commonly used in trigonometric identities and calculations.

(-csc^2(x))

: The term (-csc^2(x)) represents the derivative of the cosecant function squared. It measures how fast the cosecant function is changing at a specific point on its graph, multiplied by -1.

(cot(x))'

: The term (cot(x))' represents the derivative of the cotangent function. It measures how fast the cotangent function is changing at a specific point on its graph.

(sec^2(x))

: The term (sec^2(x)) represents the derivative of the secant function. It measures how fast the secant function is changing at a specific point on its graph.

Chain Rule

: The chain rule is a formula used to find the derivative of a composition of two or more functions. It states that the derivative of a composite function is equal to the derivative of the outermost function times the derivative of the innermost function.

Cotangent

: Cotangent is one of six trigonometric ratios used in right triangles. It represents the ratio between adjacent side length and opposite side length in relation to an acute angle.

Derivatives

: Derivatives are the rates at which quantities change. They measure how a function behaves as its input (x-value) changes.

sec(x)'

: The derivative of sec(x), which represents the rate of change of the secant function with respect to x.

Secant

: The secant of an angle in a right triangle is the ratio of the length of the hypotenuse to the length of the adjacent side.

Tangent

: A tangent line is a straight line that touches a curve at only one point and has the same slope as the curve at that point.

2.10 Finding the Derivatives of Tangent, Cotangent, Secant, and/or Cosecant Functions

2 min read•february 15, 2024

Attend a live cram event

Review all units live with expert teachers & students

Welcome back to AP Calculus with Fiveable! Now that you’ve mastered finding derivatives of trigonometric functions $\sin x$ and $\cos x$ , let’s cover the rest! Remembering these rules is key to simplifying your calculus journey. 🌟

💫 Derivatives of Advanced Trigonometric Functions

First, let’s have a glance at a summary table for quick reference.

Function	Derivative
Tangent Function: $f(x) =\tan x$	$f'(x)=\sec^2 x$
Cotangent Function: $g(x)=\cot x$	$g'(x)=-\csc^2 x$
Secant Function: $h(x)= \sec x$	$h'(x)= \sec x \tan x$
Cosecant Function: $k(x)=\csc x$	$k'(x) = -\csc x \cot x$

It's important to note that these are only valid for angles in radians, not degrees.

Derivative of $\tan x$

The derivative of $\tan x$ is $\sec^2 x$ . Let’s consider an example:

f(x)=3\tan x+2x^2

To find the derivative of this equation, differentiate $3\tan x$ and $2x^2$ individually.

Since the derivative of $\tan x$ is $\sec^2 x$ , the derivative of the first part is $3\sec^2 x$ . The derivative of $2x^2$ is $4x$ . Hence, $f'(x) = 3\sec^2 x + 4x$ .

Derivative of $\cot x$

The derivative of $\cot x$ is $-\csc^2 x$ . For example:

f(x)=5\cot x+x

We again have to differentiate the two terms separately! The derivative of $\cot x$ is $-\csc^2 x$ , so the derivative of the first term is $-5\csc^2 x$ . The derivative of $x$ is $1$ . Therefore, $f'(x) = -5\csc^2 x + 1$ or $f'(x)=1-5\csc^2 x$ .

Derivative of $\sec x$

The derivative of $\sec x$ is $\sec x \tan x$ . As an example:

f(x)=2\sec x+3x^3

Knowing the above trig derivative rule, the derivative of the first term is $2\sec x \tan x$ . The derivative of $3x^3$ is $9x^2$ . Thus, $f'(x) = 2\sec x \tan x + 9x^2$ .

Derivative of $\csc x$

Last but not least, the derivative of $\csc x$ is $-\csc x \cot x$ . For instance:

f(x)=4\csc x+7x^2

The derivative of the first part is $-4\csc x \cot x$ . The derivative of $7x^2$ is $14x$ . Therefore, $f'(x) = -4\csc x \cot x + 14x$ .

🏋️‍♂️ Practice Problems

Here are a couple of questions for you to get the concepts down!

❓ Advanced Trig Derivative Practice Questions

Find the derivatives for the following problems.

$f(x) = 2 \tan(x) + \sec(x)$
$f(x) = \frac{\cot(x)}{\csc(x)}$
$g(x) =\tan^2(6x)$
$h(x) = 5\cot(x)$

💡 Before we reveal the answers, remember to use the chain rule, sum rule, and quotient rules.

🤔 Advanced Trig Derivative Practice Solutions

$f'(x) = 2 \sec^2(x) + \sec(x) \tan(x)$
$f'(x)=−\csc^2(x)$
$g'(x)=2\tan(6x)(\frac{1}{\cos^2(6x)})$
$h'(x)=−5\csc^2(x)$

🌟 Closing

Practice these rules, and you’ll soon find them as intuitive as the basic derivatives! Keep up the great work. 🌈

Key Terms to Review (10)

(-csc(x)cot(x))

: The term (-csc(x)cot(x)) represents the product of the cosecant and cotangent functions of an angle x, with a negative sign in front. It is commonly used in trigonometric identities and calculations.

(-csc^2(x))

: The term (-csc^2(x)) represents the derivative of the cosecant function squared. It measures how fast the cosecant function is changing at a specific point on its graph, multiplied by -1.

(cot(x))'

: The term (cot(x))' represents the derivative of the cotangent function. It measures how fast the cotangent function is changing at a specific point on its graph.

(sec^2(x))

: The term (sec^2(x)) represents the derivative of the secant function. It measures how fast the secant function is changing at a specific point on its graph.

Chain Rule

: The chain rule is a formula used to find the derivative of a composition of two or more functions. It states that the derivative of a composite function is equal to the derivative of the outermost function times the derivative of the innermost function.

Cotangent

: Cotangent is one of six trigonometric ratios used in right triangles. It represents the ratio between adjacent side length and opposite side length in relation to an acute angle.

Derivatives

: Derivatives are the rates at which quantities change. They measure how a function behaves as its input (x-value) changes.

sec(x)'

: The derivative of sec(x), which represents the rate of change of the secant function with respect to x.

Secant

: The secant of an angle in a right triangle is the ratio of the length of the hypotenuse to the length of the adjacent side.

Tangent

: A tangent line is a straight line that touches a curve at only one point and has the same slope as the curve at that point.

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Resources

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AP® and SAT® are trademarks registered by the College Board, which is not affiliated with, and does not endorse this website.

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2.10 Finding the Derivatives of Tangent, Cotangent, Secant, and/or Cosecant Functions

Attend a live cram event

💫 Derivatives of Advanced Trigonometric Functions

Derivative of tan⁡x\tan xtanx

Derivative of cot⁡x\cot xcotx

Derivative of sec⁡x\sec xsecx

Derivative of csc⁡x\csc xcscx

🏋️‍♂️ Practice Problems

❓ Advanced Trig Derivative Practice Questions

🤔 Advanced Trig Derivative Practice Solutions

🌟 Closing

Key Terms to Review (10)

2.10 Finding the Derivatives of Tangent, Cotangent, Secant, and/or Cosecant Functions

Attend a live cram event

💫 Derivatives of Advanced Trigonometric Functions

Derivative of tan⁡x\tan xtanx

Derivative of cot⁡x\cot xcotx

Derivative of sec⁡x\sec xsecx

Derivative of csc⁡x\csc xcscx

🏋️‍♂️ Practice Problems

❓ Advanced Trig Derivative Practice Questions

🤔 Advanced Trig Derivative Practice Solutions

🌟 Closing

Key Terms to Review (10)

About Us

Resources

Stay Connected

About Us

Resources

Derivative of $\tan x$

Derivative of $\cot x$

Derivative of $\sec x$

Derivative of $\csc x$

Derivative of $\tan x$

Derivative of $\cot x$

Derivative of $\sec x$

Derivative of $\csc x$