Derivative of All Trig Functions
2026-02-28 11:06 Diff
https://brightchamps.com/en-us/math/calculus/derivative-of-all-trig-functions

230 Learners

Last updated on August 5, 2025

The derivatives of trigonometric functions are crucial tools for understanding how these functions change in response to variations in x. They are extensively used in fields like physics, engineering, and economics to model and solve real-world problems. This discussion will delve into the derivatives of all primary trigonometric functions.

What are the Derivatives of Trigonometric Functions?

The derivatives of trigonometric functions are fundamental in calculus. They help in understanding how these functions change with respect to x. The primary trigonometric functions and their derivatives are: 

Sine Function: d/dx (sin x) = cos x 

Cosine Function: d/dx (cos x) = -sin x 

Tangent Function: d/dx (tan x) = sec²x 

Cotangent Function: d/dx (cot x) = -csc²x 

Secant Function: d/dx (sec x) = sec x tan x 

Cosecant Function: d/dx (csc x) = -csc x cot x

Formulae for Derivatives of Trig Functions

The derivatives of the basic trigonometric functions are given by the following formulas: -

d/dx (sin x) = cos x 

d/dx (cos x) = -sin x 

d/dx (tan x) = sec²x 

d/dx (cot x) = -csc²x 

d/dx (sec x) = sec x tan x 

d/dx (csc x) = -csc x cot x

These formulas are applicable for all x within the domain of the respective trigonometric functions.

Proofs of the Derivatives of Trig Functions

The derivatives of trigonometric functions can be derived using various methods, including trigonometric identities and rules of differentiation.

Common methods include: 

By First Principle: Using the limit definition of the derivative. 

Using Chain Rule: Applicable when functions are compositions. 

Using Quotient and Product Rules: Useful for functions that are ratios or products.

For instance, the derivative of sin x using the first principle is:

1. Consider f(x) = sin x.

2. Its derivative is f'(x) = limₕ→₀ [sin(x + h) - sin x] / h.

3. Using the identity sin(A + B) = sin A cos B + cos A sin B, we get: f'(x) = limₕ→₀ [sin x cos h + cos x sin h - sin x] / h.

4. Rearranging gives: limₕ→₀ [sin x(cos h - 1) + cos x sin h] / h.

5. Using the limits limₕ→₀ (cos h - 1)/h = 0 and limₕ→₀ sin h/h = 1, we find: f'(x) = cos x.

Similarly, other trigonometric functions can be derived using appropriate methods.

Explore Our Programs

Higher-Order Derivatives of Trig Functions

Higher-order derivatives are obtained by differentiating a function multiple times. For trigonometric functions, these derivatives often exhibit a repetitive pattern.

For example: -

  • The first derivative of sin x is cos x. 
  • The second derivative of sin x, which is the derivative of cos x, is -sin x. 
  • The third derivative, the derivative of -sin x, is -cos x. 
  • The fourth derivative, the derivative of -cos x, is sin x, completing the cycle.

Understanding these patterns helps simplify the computation of higher-order derivatives in various applications.

Special Cases

Certain values of x lead to special considerations in the derivatives of trigonometric functions: - At x = π/2, the derivative of cos x is 0 because cos x itself becomes 0. 

At x = 0, the derivative of sin x is 1, as sin 0 = 0 but cos 0 = 1

Some trigonometric derivatives may be undefined at specific points due to vertical asymptotes, such as the derivative of tan x at x = π/2.

Common Mistakes and How to Avoid Them in Derivatives of Trig Functions

Missteps in differentiating trigonometric functions are common. These mistakes can be avoided by careful analysis and understanding of the functions and rules involved. Here are some frequent errors and solutions:

Problem 1

Calculate the derivative of cos x·sec x.

Okay, lets begin

Here, we have f(x) = cos x·sec x. Using the product rule, f'(x) = u′v + uv′. In the given equation, u = cos x and v = sec x. Differentiating each term gives: u′ = d/dx (cos x) = -sin x v′ = d/dx (sec x) = sec x tan x

Substituting into the product rule: f'(x) = (-sin x)·(sec x) + (cos x)·(sec x tan x).

Simplifying gives: f'(x) = -sin x sec x + cos x sec x tan x.

Thus, the derivative of the specified function is -sin x sec x + cos x sec x tan x.

Explanation

The derivative is found by applying the product rule. Each function is differentiated separately, and then the results are combined to obtain the final derivative.

Well explained 👍

Problem 2

A company uses a rotating sign modeled by the function y = sin(x) to display advertisements. If x = π/6 radians, determine the rate of change of the sign's height.

Okay, lets begin

We have y = sin(x) representing the sign's height. Differentiate the equation: dy/dx = cos(x). Given x = π/6, substitute this into the derivative: dy/dx = cos(π/6) = √3/2.

Thus, the rate of change of the sign's height at x = π/6 is √3/2.

Explanation

The problem requires finding the derivative of the height function at a specific angle. The derivative, cos(x), evaluated at x = π/6, provides the rate of change of the height.

Well explained 👍

Problem 3

Derive the second derivative of the function y = sec(x).

Okay, lets begin

First, find the first derivative: dy/dx = sec x tan x. Now differentiate again to find the second derivative: d²y/dx² = d/dx [sec x tan x]. Using the product rule: d²y/dx² = sec x (d/dx [tan x]) + tan x (d/dx [sec x]). = sec x sec²x + tan x sec x tan x. = sec³x + sec x tan²x. Therefore, the second derivative of the function y = sec(x) is sec³x + sec x tan²x.

Explanation

The second derivative is found by differentiating the first derivative. The product rule is applied to the first derivative, and terms are simplified to obtain the second derivative.

Well explained 👍

Problem 4

Prove: d/dx (cot²(x)) = -2 cot(x) csc²(x).

Okay, lets begin

Using the chain rule: Let y = cot²(x), which is [cot(x)]².

Differentiate using the chain rule: dy/dx = 2 cot(x) d/dx [cot(x)].

Since the derivative of cot(x) is -csc²(x): dy/dx = 2 cot(x)(-csc²(x)). = -2 cot(x) csc²(x).

Hence proved.

Explanation

The chain rule is used to differentiate the square of cotangent. After applying the rule, the derivative of cotangent is substituted, leading to the final result.

Well explained 👍

Problem 5

Solve: d/dx (sec x/x).

Okay, lets begin

Differentiate using the quotient rule: d/dx (sec x/x) = (d/dx (sec x)·x - sec x·d/dx(x))/x².

Substitute d/dx (sec x) = sec x tan x and d/dx (x) = 1: = (sec x tan x·x - sec x·1)/x². = (x sec x tan x - sec x)/x². = sec x (x tan x - 1)/x².

Therefore, d/dx (sec x/x) = sec x (x tan x - 1)/x².

Explanation

The problem is solved using the quotient rule. Each component is differentiated, and the results are combined and simplified to find the derivative of the given function.

Well explained 👍

FAQs on the Derivatives of Trig Functions

1.What are the derivatives of sine and cosine functions?

The derivatives are: - d/dx (sin x) = cos x - d/dx (cos x) = -sin x

2.How are derivatives of trigonometric functions used in real life?

They are used to model and predict changes in periodic phenomena, such as waves and oscillations, and to solve problems in physics and engineering.

3.Can the derivative of tan x be taken at x = π/2?

No, the function tan x is undefined at x = π/2, so its derivative cannot be taken at this point.

4.Which rule is used to differentiate sec x/x?

The quotient rule is used: d/dx (sec x/x) = (x sec x tan x - sec x)/x².

5.Are the derivatives of tan x and cot x the same?

No, they are different. The derivative of tan x is sec²x, while the derivative of cot x is -csc²x.

6.How do you differentiate cot²(x)?

Using the chain rule: d/dx (cot²(x)) = -2 cot(x) csc²(x).

Important Glossaries for the Derivatives of Trig Functions

  • Derivative: Represents the rate of change of a function with respect to a variable.
  • Trigonometric Functions: Functions related to angles, including sine, cosine, tangent, etc.
  • Chain Rule: A rule for differentiating compositions of functions.
  • Product Rule: A rule for differentiating products of two functions.
  • Quotient Rule: A rule for differentiating quotients of two functions.

What Is Calculus? 🔢 | Easy Tricks, Limits & 🎯 Fun Learning for Kids | ✨BrightCHAMPS Math

▶

Jaskaran Singh Saluja

About the Author

Jaskaran Singh Saluja is a math wizard with nearly three years of experience as a math teacher. His expertise is in algebra, so he can make algebra classes interesting by turning tricky equations into simple puzzles.

Fun Fact

: He loves to play the quiz with kids through algebra to make kids love it.