Derivatives Of Trig And Inverse Trig Functions

Derivatives of Trigonometric and Inverse Trigonometric Functions: A Comprehensive Guide

Understanding derivatives is crucial in calculus, and mastering the derivatives of trigonometric and inverse trigonometric functions is essential for tackling more complex problems in calculus, physics, and engineering. This comprehensive guide delves into the core concepts, provides detailed derivations, and offers practical examples to solidify your understanding.

Trigonometric Functions and Their Derivatives

Trigonometric functions describe the relationships between angles and sides in a right-angled triangle. The six primary trigonometric functions are sine (sin), cosine (cos), tangent (tan), cosecant (csc), secant (sec), and cotangent (cot). Their derivatives are fundamental building blocks in calculus.

1. The Derivative of Sine (sin x)

The derivative of sin x is cos x. This can be proven using the limit definition of a derivative and trigonometric identities.

Derivation (using the limit definition):

The limit definition of a derivative is:

f'(x) = lim (h→0) [(f(x + h) - f(x)) / h]

Applying this to f(x) = sin x:

f'(x) = lim (h→0) [(sin(x + h) - sin(x)) / h]

Using the trigonometric identity sin(A + B) = sin A cos B + cos A sin B:

f'(x) = lim (h→0) [(sin x cos h + cos x sin h - sin x) / h]

Rearranging and separating the limit:

f'(x) = lim (h→0) [(sin x (cos h - 1) / h) + (cos x (sin h / h))]

Since lim (h→0) (sin h / h) = 1 and lim (h→0) (cos h - 1) / h = 0, we get:

f'(x) = cos x

2. The Derivative of Cosine (cos x)

The derivative of cos x is -sin x. The derivation follows a similar process to that of sin x, utilizing the limit definition and trigonometric identities.

Derivation (using the limit definition):

f'(x) = lim (h→0) [(cos(x + h) - cos(x)) / h]

Using the trigonometric identity cos(A + B) = cos A cos B - sin A sin B:

f'(x) = lim (h→0) [(cos x cos h - sin x sin h - cos x) / h]

Rearranging and separating the limit:

f'(x) = lim (h→0) [(cos x (cos h - 1) / h) - (sin x (sin h / h))]

Considering the limits mentioned previously, we obtain:

f'(x) = -sin x

3. The Derivative of Tangent (tan x)

The derivative of tan x is sec²x. This can be derived using the quotient rule, since tan x = sin x / cos x.

Derivation (using the quotient rule):

The quotient rule states: (u/v)' = (u'v - uv') / v²

Let u = sin x and v = cos x. Then u' = cos x and v' = -sin x.

Applying the quotient rule:

(tan x)' = [(cos x)(cos x) - (sin x)(-sin x)] / (cos x)²

= (cos²x + sin²x) / cos²x

Using the Pythagorean identity cos²x + sin²x = 1:

(tan x)' = 1 / cos²x = sec²x

4. Derivatives of Other Trigonometric Functions

The derivatives of the remaining trigonometric functions (csc x, sec x, cot x) can be derived using the quotient rule and the derivatives of sin x, cos x, and tan x.

• (csc x)' = -csc x cot x
• (sec x)' = sec x tan x
• (cot x)' = -csc²x

Inverse Trigonometric Functions and Their Derivatives

Inverse trigonometric functions (also called arcus functions or cyclometric functions) are the inverse functions of trigonometric functions. They return the angle whose trigonometric ratio is a given value. The principal values of inverse trigonometric functions are usually restricted to specific ranges to ensure a one-to-one relationship.

1. The Derivative of arcsin x (sin⁻¹x)

The derivative of arcsin x is 1 / √(1 - x²). This can be derived using implicit differentiation.

Derivation (using implicit differentiation):

Let y = arcsin x. Then sin y = x.

Differentiating both sides with respect to x:

cos y (dy/dx) = 1

Solving for dy/dx:

dy/dx = 1 / cos y

Since sin²y + cos²y = 1, we have cos y = √(1 - sin²y). Substituting sin y = x:

dy/dx = 1 / √(1 - x²)

2. The Derivative of arccos x (cos⁻¹x)

The derivative of arccos x is -1 / √(1 - x²). The derivation is similar to that of arcsin x.

Derivation (using implicit differentiation):

Let y = arccos x. Then cos y = x.

Differentiating both sides with respect to x:

-sin y (dy/dx) = 1

Solving for dy/dx:

dy/dx = -1 / sin y

Since sin y = √(1 - cos²y) and cos y = x:

dy/dx = -1 / √(1 - x²)

3. The Derivative of arctan x (tan⁻¹x)

The derivative of arctan x is 1 / (1 + x²). This can also be derived using implicit differentiation.

Derivation (using implicit differentiation):

Let y = arctan x. Then tan y = x.

Differentiating both sides with respect to x:

sec²y (dy/dx) = 1

Solving for dy/dx:

dy/dx = 1 / sec²y

Since sec²y = 1 + tan²y and tan y = x:

dy/dx = 1 / (1 + x²)

4. Derivatives of Other Inverse Trigonometric Functions

The derivatives of the remaining inverse trigonometric functions (arccsc x, arcsec x, arccot x) can be derived similarly, employing implicit differentiation.

• (arccsc x)' = -1 / (|x|√(x² - 1))
• (arcsec x)' = 1 / (|x|√(x² - 1))
• (arccot x)' = -1 / (1 + x²)

Applications and Examples

Derivatives of trigonometric and inverse trigonometric functions are vital in numerous applications:

• Physics: Calculating velocities and accelerations in oscillatory motion (e.g., simple harmonic motion).
• Engineering: Analyzing circuits with sinusoidal signals, designing mechanical systems involving rotating parts.
• Calculus: Solving optimization problems, finding areas and volumes using integration.

Example 1: Finding the slope of a tangent line.

Find the slope of the tangent line to the curve y = sin x at x = π/4.

The slope is given by the derivative: dy/dx = cos x.

At x = π/4, the slope is cos(π/4) = √2/2.

Example 2: Related rates problem.

A ladder 10 meters long rests against a vertical wall. If the bottom of the ladder slides away from the wall at a rate of 2 m/s, how fast is the top of the ladder sliding down the wall when the bottom of the ladder is 6 meters from the wall?

This problem requires using implicit differentiation and trigonometric functions.

Example 3: Optimization problem.

Find the maximum area of a rectangle inscribed in a circle of radius r. This problem uses trigonometric functions to express the area and then employs calculus to find the maximum.

Conclusion

Mastering the derivatives of trigonometric and inverse trigonometric functions is crucial for success in calculus and its many applications. By understanding the derivations and practicing with examples, you can build a strong foundation for tackling more advanced mathematical concepts and real-world problems. Remember to utilize the fundamental trigonometric identities and the appropriate differentiation rules (limit definition, quotient rule, implicit differentiation) to accurately derive these essential derivatives. Consistent practice and problem-solving will further enhance your proficiency.