Derivative Of Inverse Trig Functions Proof

Proving the Derivatives of Inverse Trigonometric Functions

The inverse trigonometric functions – arcsine, arccosine, arctangent, arccotangent, arcsecant, and arccosecant – are fundamental to calculus and numerous applications in science and engineering. Understanding their derivatives is crucial for solving various problems involving angles, slopes, and rates of change. This article provides a comprehensive exploration of the proofs for the derivatives of these functions, focusing on clarity and mathematical rigor. We will use the implicit differentiation technique predominantly, leveraging the power of trigonometric identities along the way.

Understanding the Inverse Trigonometric Functions

Before diving into the proofs, let's briefly revisit the inverse trigonometric functions. Recall that the inverse trigonometric functions are the inverses of the trigonometric functions, but with restricted domains to ensure they are one-to-one functions. This restriction is essential for the existence of a unique inverse.

• arcsin(x): The inverse sine function, also denoted as sin⁻¹(x), returns the angle whose sine is x. Its domain is [-1, 1], and its range is [-π/2, π/2].

• arccos(x): The inverse cosine function, also denoted as cos⁻¹(x), returns the angle whose cosine is x. Its domain is [-1, 1], and its range is [0, π].

• arctan(x): The inverse tangent function, also denoted as tan⁻¹(x), returns the angle whose tangent is x. Its domain is (-∞, ∞), and its range is (-π/2, π/2).

• arccot(x): The inverse cotangent function, also denoted as cot⁻¹(x), returns the angle whose cotangent is x. Its domain is (-∞, ∞), and its range is (0, π).

• arcsec(x): The inverse secant function, also denoted as sec⁻¹(x), returns the angle whose secant is x. Its domain is (-∞, -1] ∪ [1, ∞), and its range is [0, π], excluding π/2.

• arccsc(x): The inverse cosecant function, also denoted as csc⁻¹(x), returns the angle whose cosecant is x. Its domain is (-∞, -1] ∪ [1, ∞), and its range is [-π/2, π/2], excluding 0.

Proving the Derivatives using Implicit Differentiation

The most efficient method to derive the formulas for the derivatives of inverse trigonometric functions is through implicit differentiation. This technique involves differentiating both sides of an equation with respect to a variable, treating the inverse function as an implicitly defined function.

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, we have:

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)

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

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

Solving for dy/dx, we have:

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 both sides with respect to x, we get:

sec²(y) * (dy/dx) = 1

Solving for dy/dx, we have:

dy/dx = 1/sec²(y) = cos²(y)

Since tan(y) = x, we can use the identity sec²(y) = 1 + tan²(y):

sec²(y) = 1 + x²

Therefore, cos²(y) = 1/(1 + x²) and the derivative of arctan(x) is:

d(arctan(x))/dx = 1/(1 + x²)

4. Derivative of arccot(x)

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

-csc²(y) * (dy/dx) = 1

Solving for dy/dx:

dy/dx = -1/csc²(y) = -sin²(y)

Using the identity csc²(y) = 1 + cot²(y) and cot(y) = x:

csc²(y) = 1 + x²

Therefore, sin²(y) = 1/(1 + x²) and the derivative of arccot(x) is:

d(arccot(x))/dx = -1/(1 + x²)

5. Derivative of arcsec(x)

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

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

Solving for dy/dx, we have:

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

Since sec(y) = x, we use the identity tan²(y) = sec²(y) - 1:

tan(y) = √(sec²(y) - 1) = √(x² - 1)

Therefore, the derivative of arcsec(x) is:

d(arcsec(x))/dx = 1/[|x|√(x² - 1)] (Note the absolute value for x to handle the domain)

6. Derivative of arccsc(x)

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

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

Solving for dy/dx, we have:

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

Since csc(y) = x, we use the identity cot²(y) = csc²(y) - 1:

cot(y) = √(csc²(y) - 1) = √(x² - 1)

Therefore, the derivative of arccsc(x) is:

d(arccsc(x))/dx = -1/[|x|√(x² - 1)] (Note the absolute value for x to handle the domain)

Applications and Significance

The derivatives of inverse trigonometric functions are indispensable in various fields:

• Calculus: They are essential for evaluating integrals involving inverse trigonometric functions and in solving differential equations.

• Physics: They appear in problems involving projectile motion, oscillations, and wave phenomena, where angles and their rates of change are crucial.

• Engineering: They are used in designing curves, analyzing rotational motion, and solving problems related to signal processing and control systems.

• Computer Graphics: They are used extensively in algorithms for calculating rotations, transformations, and other geometric operations.

Conclusion

This detailed exploration provides rigorous proofs for the derivatives of all six inverse trigonometric functions using implicit differentiation. Mastering these derivatives is vital for anyone working with calculus, and understanding their derivations strengthens foundational mathematical skills. The applications of these derivatives are vast, underscoring their importance across numerous scientific and engineering disciplines. Remember to always consider the domain restrictions of each inverse trigonometric function when applying these derivative formulas.