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

2 min readβ€’june 18, 2024

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

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

πŸ’«Β Derivatives of Advanced Trigonometric Functions

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

FunctionDerivative
Tangent Function: f(x)=tan⁑xf(x) =\tan xfβ€²(x)=sec⁑2xf'(x)=\sec^2 x
Cotangent Function: g(x)=cot⁑xg(x)=\cot xgβ€²(x)=βˆ’csc⁑2xg'(x)=-\csc^2 x
Secant Function: h(x)=sec⁑xh(x)= \sec xhβ€²(x)=sec⁑xtan⁑xh'(x)= \sec x \tan x
Cosecant Function: k(x)=csc⁑xk(x)=\csc xkβ€²(x)=βˆ’csc⁑xcot⁑xk'(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\tan x

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

f(x)=3tan⁑x+2x2f(x)=3\tan x+2x^2

To find the derivative of this equation, differentiate 3tan⁑x3\tan x and 2x22x^2 individually.

Since the derivative of tan⁑x\tan x is sec⁑2x\sec^2 x, the derivative of the first part is 3sec⁑2x3\sec^2 x. The derivative of 2x22x^2 is 4x4x. Hence, fβ€²(x)=3sec⁑2x+4xf'(x) = 3\sec^2 x + 4x.

Derivative of cot⁑x\cot x

The derivative of cot⁑x\cot x is βˆ’csc⁑2x-\csc^2 x. For example:

f(x)=5cot⁑x+xf(x)=5\cot x+x

We again have to differentiate the two terms separately! The derivative of cot⁑x\cot x is βˆ’csc⁑2x-\csc^2 x, so the derivative of the first term is βˆ’5csc⁑2x-5\csc^2 x. The derivative of xx is 11. Therefore, fβ€²(x)=βˆ’5csc⁑2x+1f'(x) = -5\csc^2 x + 1 or fβ€²(x)=1βˆ’5csc⁑2xf'(x)=1-5\csc^2 x.

Derivative of sec⁑x\sec x

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

f(x)=2sec⁑x+3x3f(x)=2\sec x+3x^3

Knowing the above trig derivative rule, the derivative of the first term is 2sec⁑xtan⁑x2\sec x \tan x. The derivative of 3x33x^3 is 9x29x^2. Thus, fβ€²(x)=2sec⁑xtan⁑x+9x2f'(x) = 2\sec x \tan x + 9x^2.

Derivative of csc⁑x\csc x

Last but not least, the derivative of csc⁑x\csc x is βˆ’csc⁑xcot⁑x-\csc x \cot x. For instance:

f(x)=4csc⁑x+7x2f(x)=4\csc x+7x^2

The derivative of the first part is βˆ’4csc⁑xcot⁑x-4\csc x \cot x. The derivative of 7x27x^2 is 14x14x. Therefore, fβ€²(x)=βˆ’4csc⁑xcot⁑x+14xf'(x) = -4\csc x \cot x + 14x.


πŸ‹οΈβ€β™‚οΈΒ 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.

  1. f(x)=2tan⁑(x)+sec⁑(x)f(x) = 2 \tan(x) + \sec(x)
  2. f(x)=cot⁑(x)csc⁑(x)f(x) = \frac{\cot(x)}{\csc(x)}
  3. g(x)=tan⁑2(6x)g(x) =\tan^2(6x)
  4. h(x)=5cot⁑(x)h(x) = 5\cot(x)

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

πŸ€” Advanced Trig Derivative Practice Solutions

  1. fβ€²(x)=2sec⁑2(x)+sec⁑(x)tan⁑(x)f'(x) = 2 \sec^2(x) + \sec(x) \tan(x)
  2. fβ€²(x)=βˆ’csc⁑2(x)f'(x)=βˆ’\csc^2(x)
  3. gβ€²(x)=2tan⁑(6x)(1cos⁑2(6x))g'(x)=2\tan(6x)(\frac{1}{\cos^2(6x)})
  4. hβ€²(x)=βˆ’5csc⁑2(x)h'(x)=βˆ’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. 🌈

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

2 min readβ€’june 18, 2024

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

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

πŸ’«Β Derivatives of Advanced Trigonometric Functions

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

FunctionDerivative
Tangent Function: f(x)=tan⁑xf(x) =\tan xfβ€²(x)=sec⁑2xf'(x)=\sec^2 x
Cotangent Function: g(x)=cot⁑xg(x)=\cot xgβ€²(x)=βˆ’csc⁑2xg'(x)=-\csc^2 x
Secant Function: h(x)=sec⁑xh(x)= \sec xhβ€²(x)=sec⁑xtan⁑xh'(x)= \sec x \tan x
Cosecant Function: k(x)=csc⁑xk(x)=\csc xkβ€²(x)=βˆ’csc⁑xcot⁑xk'(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\tan x

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

f(x)=3tan⁑x+2x2f(x)=3\tan x+2x^2

To find the derivative of this equation, differentiate 3tan⁑x3\tan x and 2x22x^2 individually.

Since the derivative of tan⁑x\tan x is sec⁑2x\sec^2 x, the derivative of the first part is 3sec⁑2x3\sec^2 x. The derivative of 2x22x^2 is 4x4x. Hence, fβ€²(x)=3sec⁑2x+4xf'(x) = 3\sec^2 x + 4x.

Derivative of cot⁑x\cot x

The derivative of cot⁑x\cot x is βˆ’csc⁑2x-\csc^2 x. For example:

f(x)=5cot⁑x+xf(x)=5\cot x+x

We again have to differentiate the two terms separately! The derivative of cot⁑x\cot x is βˆ’csc⁑2x-\csc^2 x, so the derivative of the first term is βˆ’5csc⁑2x-5\csc^2 x. The derivative of xx is 11. Therefore, fβ€²(x)=βˆ’5csc⁑2x+1f'(x) = -5\csc^2 x + 1 or fβ€²(x)=1βˆ’5csc⁑2xf'(x)=1-5\csc^2 x.

Derivative of sec⁑x\sec x

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

f(x)=2sec⁑x+3x3f(x)=2\sec x+3x^3

Knowing the above trig derivative rule, the derivative of the first term is 2sec⁑xtan⁑x2\sec x \tan x. The derivative of 3x33x^3 is 9x29x^2. Thus, fβ€²(x)=2sec⁑xtan⁑x+9x2f'(x) = 2\sec x \tan x + 9x^2.

Derivative of csc⁑x\csc x

Last but not least, the derivative of csc⁑x\csc x is βˆ’csc⁑xcot⁑x-\csc x \cot x. For instance:

f(x)=4csc⁑x+7x2f(x)=4\csc x+7x^2

The derivative of the first part is βˆ’4csc⁑xcot⁑x-4\csc x \cot x. The derivative of 7x27x^2 is 14x14x. Therefore, fβ€²(x)=βˆ’4csc⁑xcot⁑x+14xf'(x) = -4\csc x \cot x + 14x.


πŸ‹οΈβ€β™‚οΈΒ 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.

  1. f(x)=2tan⁑(x)+sec⁑(x)f(x) = 2 \tan(x) + \sec(x)
  2. f(x)=cot⁑(x)csc⁑(x)f(x) = \frac{\cot(x)}{\csc(x)}
  3. g(x)=tan⁑2(6x)g(x) =\tan^2(6x)
  4. h(x)=5cot⁑(x)h(x) = 5\cot(x)

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

πŸ€” Advanced Trig Derivative Practice Solutions

  1. fβ€²(x)=2sec⁑2(x)+sec⁑(x)tan⁑(x)f'(x) = 2 \sec^2(x) + \sec(x) \tan(x)
  2. fβ€²(x)=βˆ’csc⁑2(x)f'(x)=βˆ’\csc^2(x)
  3. gβ€²(x)=2tan⁑(6x)(1cos⁑2(6x))g'(x)=2\tan(6x)(\frac{1}{\cos^2(6x)})
  4. hβ€²(x)=βˆ’5csc⁑2(x)h'(x)=βˆ’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. 🌈