Last updated on July 17th, 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.
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
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.
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.
Higher-order derivatives are obtained by differentiating a function multiple times. For trigonometric functions, these derivatives often exhibit a repetitive pattern.
For example: -
Understanding these patterns helps simplify the computation of higher-order derivatives in various applications.
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.
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:
Calculate the derivative of cos x·sec x.
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.
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.
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.
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.
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.
Derive the second derivative of the function y = sec(x).
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.
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.
Prove: d/dx (cot²(x)) = -2 cot(x) csc²(x).
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.
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.
Solve: d/dx (sec x/x).
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.
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.
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