Derivative Of Inverse Trig Functions Proof

Deriving 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 derive the derivatives of the six inverse trigonometric functions: arcsin x, arccos x, arctan x, arccot x, arcsec x, and arccsc x. We'll delve into the proofs, providing a step-by-step explanation for each, utilizing implicit differentiation and the Pythagorean identities.

1. Understanding the Foundation: Implicit Differentiation

Before embarking on the derivations, it's crucial to grasp the concept of implicit differentiation. This technique allows us to find the derivative of a function that is not explicitly expressed as y = f(x). Instead, the function is defined implicitly through an equation relating x and y.

The core idea is to differentiate both sides of the implicit equation with respect to x, treating y as a function of x and applying the chain rule. This yields an equation involving dy/dx, which can then be solved for dy/dx to obtain the derivative.

2. Deriving the Derivative of arcsin x

Let y = arcsin x. This means that sin y = x. Now, we apply implicit differentiation:

Step 1: Differentiate both sides with respect to x:

cos y * (dy/dx) = 1

Step 2: Solve for dy/dx:

dy/dx = 1 / cos y

Step 3: Express cos y in terms of x:

We use the Pythagorean identity: sin²y + cos²y = 1. Since sin y = x, we have:

cos²y = 1 - sin²y = 1 - x²

cos y = ±√(1 - x²)

Step 4: Determine the sign:

The range of arcsin x is [-π/2, π/2]. In this interval, cos y is always non-negative. Therefore, we choose the positive square root:

cos y = √(1 - x²)

Step 5: Substitute and finalize:

Substituting this into our expression for dy/dx, we get:

dy/dx = 1 / √(1 - x²)

Therefore, the derivative of arcsin x is:

d(arcsin x)/dx = 1 / √(1 - x²)

3. Deriving the Derivative of arccos x

Similar to the derivation of arcsin x, let y = arccos x, which implies cos y = x. We follow the same steps:

Step 1: Implicit differentiation:

-sin y * (dy/dx) = 1

Step 2: Solve for dy/dx:

dy/dx = -1 / sin y

Step 3: Express sin y in terms of x:

Using the Pythagorean identity, sin²y = 1 - cos²y = 1 - x²

sin y = ±√(1 - x²)

Step 4: Determine the sign:

The range of arccos x is [0, π]. In this interval, sin y is always non-negative. Therefore:

sin y = √(1 - x²)

Step 5: Substitute and finalize:

Substituting into the expression for dy/dx:

dy/dx = -1 / √(1 - x²)

Therefore, the derivative of arccos x is:

d(arccos x)/dx = -1 / √(1 - x²)

4. Deriving the Derivative of arctan x

Let y = arctan x, implying tan y = x. Again, we use implicit differentiation:

Step 1: Implicit differentiation:

sec²y * (dy/dx) = 1

Step 2: Solve for dy/dx:

dy/dx = 1 / sec²y

Step 3: Express sec²y in terms of x:

Using the identity sec²y = 1 + tan²y, and since tan y = x:

sec²y = 1 + x²

Step 4: Substitute and finalize:

Substituting this into the expression for dy/dx:

dy/dx = 1 / (1 + x²)

Therefore, the derivative of arctan x is:

d(arctan x)/dx = 1 / (1 + x²)

5. Deriving the Derivative of arccot x

Let y = arccot x, meaning cot y = x. Following the same procedure:

Step 1: Implicit differentiation:

-csc²y * (dy/dx) = 1

Step 2: Solve for dy/dx:

dy/dx = -1 / csc²y

Step 3: Express csc²y in terms of x:

Using the identity csc²y = 1 + cot²y, and since cot y = x:

csc²y = 1 + x²

Step 4: Substitute and finalize:

dy/dx = -1 / (1 + x²)

Therefore, the derivative of arccot x is:

d(arccot x)/dx = -1 / (1 + x²)

6. Deriving the Derivative of arcsec x

Let y = arcsec x, so sec y = x.

Step 1: Implicit differentiation:

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

Step 2: Solve for dy/dx:

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

Step 3: Express sec y and tan y in terms of x:

We know sec y = x. Using the identity tan²y = sec²y - 1:

tan²y = x² - 1

tan y = ±√(x² - 1)

Step 4: Determine the sign:

The range of arcsec x is [0, π], excluding π/2. For x ≥ 1, tan y ≥ 0, and for x ≤ -1, tan y ≤ 0. Thus:

tan y = √(x² - 1) if x ≥ 1 tan y = -√(x² - 1) if x ≤ -1

Step 5: Substitute and finalize:

For x ≥ 1: dy/dx = 1 / (x√(x² - 1)) For x ≤ -1: dy/dx = 1 / (x(-√(x² - 1))) = -1 / (x√(x² - 1))

Combining both cases, we can write:

d(arcsec x)/dx = 1 / (|x|√(x² - 1))

7. Deriving the Derivative of arccsc x

Finally, let y = arccsc x, which implies csc y = x.

Step 1: Implicit differentiation:

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

Step 2: Solve for dy/dx:

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

Step 3: Express csc y and cot y in terms of x:

We have csc y = x. Using the identity cot²y = csc²y - 1:

cot²y = x² - 1

cot y = ±√(x² - 1)

Step 4: Determine the sign:

The range of arccsc x is [-π/2, π/2], excluding 0. Similar to arcsec x, we consider the signs:

cot y = √(x² - 1) if x ≥ 1 cot y = -√(x² - 1) if x ≤ -1

Step 5: Substitute and finalize:

Following similar logic to arcsec x:

d(arccsc x)/dx = -1 / (|x|√(x² - 1))

Conclusion: A Powerful Toolkit for Calculus

This guide has provided a detailed derivation of the derivatives of all six inverse trigonometric functions. These derivatives are essential tools in various calculus applications, particularly in integration problems involving trigonometric functions and in solving differential equations. Mastering these derivations forms a strong foundation for advanced calculus studies and its applications in various scientific and engineering disciplines. Remember to always carefully consider the domains and ranges of these functions when applying their derivatives.