Proof Of Derivative Of Inverse Trig Functions

Proving the Derivatives of Inverse Trigonometric Functions: A Comprehensive Guide

The inverse trigonometric functions, also known as arcus functions or cyclometric functions, are the inverse functions of the trigonometric functions. Understanding their derivatives is crucial in various fields, including calculus, physics, and engineering. This comprehensive guide will meticulously prove the derivatives of all six inverse trigonometric functions, offering a detailed explanation of the process and providing insights into the underlying mathematical concepts.

Understanding the Inverse Trigonometric Functions

Before diving into the proofs, let's briefly review the six inverse trigonometric functions and their principal value ranges:

• arcsin(x) (inverse sine): Returns the angle whose sine is x. Principal value range: [-π/2, π/2].
• arccos(x) (inverse cosine): Returns the angle whose cosine is x. Principal value range: [0, π].
• arctan(x) (inverse tangent): Returns the angle whose tangent is x. Principal value range: (-π/2, π/2).
• arccsc(x) (inverse cosecant): Returns the angle whose cosecant is x. Principal value range: [-π/2, π/2], excluding 0.
• arcsec(x) (inverse secant): Returns the angle whose secant is x. Principal value range: [0, π], excluding π/2.
• arccot(x) (inverse cotangent): Returns the angle whose cotangent is x. Principal value range: (0, π).

These principal value ranges are essential because trigonometric functions are periodic, meaning they repeat their values infinitely. Restricting the range ensures a unique output for each input.

Proving the Derivatives using Implicit Differentiation

The most efficient way to derive the formulas for the derivatives of inverse trigonometric functions is through implicit differentiation. This method leverages the relationship between a function and its inverse.

1. Derivative of arcsin(x)

Let y = arcsin(x). Then, sin(y) = x. Differentiating both sides with respect to x, we get:

cos(y) * (dy/dx) = 1

Solving for dy/dx:

dy/dx = 1 / cos(y)

Since sin²(y) + cos²(y) = 1, we can express cos(y) in terms of sin(y) = x:

cos(y) = √(1 - sin²(y)) = √(1 - x²)

Therefore, the derivative of arcsin(x) is:

d(arcsin(x))/dx = 1 / √(1 - x²)

2. Derivative of arccos(x)

Following a similar approach, let y = arccos(x). Then, cos(y) = x. Differentiating implicitly:

-sin(y) * (dy/dx) = 1

dy/dx = -1 / sin(y)

Using the identity sin²(y) + cos²(y) = 1, and cos(y) = x:

sin(y) = √(1 - cos²(y)) = √(1 - x²)

Therefore, the derivative of arccos(x) is:

d(arccos(x))/dx = -1 / √(1 - x²)

3. Derivative of arctan(x)

Let y = arctan(x). Then, tan(y) = x. Differentiating implicitly:

sec²(y) * (dy/dx) = 1

dy/dx = 1 / sec²(y)

Since sec²(y) = 1 + tan²(y) and tan(y) = x:

sec²(y) = 1 + x²

Therefore, the derivative of arctan(x) is:

d(arctan(x))/dx = 1 / (1 + x²)

4. Derivative of arccsc(x)

Let y = arccsc(x). Then, csc(y) = x. Differentiating implicitly:

-csc(y)cot(y) * (dy/dx) = 1

dy/dx = -1 / (csc(y)cot(y))

Using the identities csc²(y) = 1 + cot²(y) and csc(y) = x:

cot(y) = ±√(csc²(y) - 1) = ±√(x² - 1)

Since the principal value range of arccsc(x) excludes 0, we consider the sign carefully. For x > 1, cot(y) > 0, and for x < -1, cot(y) < 0. This leads to:

d(arccsc(x))/dx = -1 / (|x|√(x² - 1))

5. Derivative of arcsec(x)

Let y = arcsec(x). Then, sec(y) = x. Differentiating implicitly:

sec(y)tan(y) * (dy/dx) = 1

dy/dx = 1 / (sec(y)tan(y))

Using the identity tan²(y) + 1 = sec²(y) and sec(y) = x:

tan(y) = ±√(sec²(y) - 1) = ±√(x² - 1)

Similar to arccsc(x), the sign depends on the range. For x > 1, tan(y) > 0, and for x < -1, tan(y) < 0. This gives:

d(arcsec(x))/dx = 1 / (|x|√(x² - 1))

6. Derivative of arccot(x)

Let y = arccot(x). Then, cot(y) = x. Differentiating implicitly:

-csc²(y) * (dy/dx) = 1

dy/dx = -1 / csc²(y)

Since csc²(y) = 1 + cot²(y) and cot(y) = x:

csc²(y) = 1 + x²

Therefore, the derivative of arccot(x) is:

d(arccot(x))/dx = -1 / (1 + x²)

Applications and Importance

The derivatives of inverse trigonometric functions are frequently used in various mathematical contexts:

• Integration: They are essential in solving integrals involving expressions like 1/√(1-x²), 1/(1+x²), and others.
• Differential Equations: They appear in the solutions of certain differential equations, particularly those related to oscillatory systems.
• Physics: They are applied in problems involving angles, trajectories, and oscillations.
• Engineering: They find applications in signal processing, control systems, and other areas.
• Geometry: They are used in calculations involving arc lengths and areas of sectors.

Conclusion

This detailed guide has presented rigorous proofs of the derivatives of all six inverse trigonometric functions using implicit differentiation. Understanding these derivations is crucial for mastering calculus and applying it effectively in various scientific and engineering disciplines. The formulas derived here are fundamental tools in solving numerous problems and understanding complex mathematical relationships. Remember to always consider the principal value ranges of these functions to avoid ambiguity in your calculations. By grasping these derivations and their applications, you significantly enhance your mathematical capabilities and open doors to more advanced concepts.