Derivative Of Inverse Trig Functions Proof

Ah, the dance between trigonometry and its inverse! The derivative of inverse trigonometric functions may seem daunting at first, but beneath the surface lies a beautiful blend of algebraic manipulation, trigonometric identities, and the chain rule. It's a topic that not only tests your calculus skills but also deepens your appreciation for the interconnectedness of mathematical concepts.

In this comprehensive exploration, we'll embark on a journey to derive the derivatives of all six inverse trigonometric functions: arcsin(x), arccos(x), arctan(x), arccot(x), arcsec(x), and arccsc(x). We'll start with a foundational understanding of inverse trigonometric functions and then meticulously work through each derivative proof, revealing the elegance and logic behind the formulas.

A Quick Refresher on Inverse Trigonometric Functions

Before diving into the derivatives, let's ensure we're all on the same page regarding inverse trigonometric functions.

• What they are: Inverse trigonometric functions "undo" the regular trigonometric functions (sine, cosine, tangent, etc.). For example, if sin(θ) = x, then arcsin(x) = θ. In other words, arcsin(x) answers the question: "What angle has a sine of x?"

• Notation: We use the "arc" prefix (e.g., arcsin, arccos) or the superscript "-1" (e.g., sin⁻¹, cos⁻¹) to denote inverse trigonometric functions. Both notations are common and interchangeable.

• Domain and Range: It's crucial to remember the domains and ranges of inverse trigonometric functions to avoid ambiguity and ensure correct interpretations. The ranges are often restricted to principal values to ensure a unique output for each input.

  • arcsin(x): Domain: [-1, 1], Range: [-π/2, π/2]
  • arccos(x): Domain: [-1, 1], Range: [0, π]
  • arctan(x): Domain: (-∞, ∞), Range: (-π/2, π/2)
  • arccot(x): Domain: (-∞, ∞), Range: (0, π)
  • arcsec(x): Domain: (-∞, -1] ∪ [1, ∞), Range: [0, π/2) ∪ (π/2, π]
  • arccsc(x): Domain: (-∞, -1] ∪ [1, ∞), Range: [-π/2, 0) ∪ (0, π/2]

Deriving the Derivative of arcsin(x)

Let's begin with the most fundamental inverse trigonometric function, arcsin(x).

1. Set up the equation: Let y = arcsin(x). This means sin(y) = x.

2. Differentiate both sides implicitly with respect to x: d/dx [sin(y)] = d/dx [x] cos(y) * dy/dx = 1 (Applying the chain rule on the left side)

3. Solve for dy/dx: dy/dx = 1 / cos(y)

4. Express cos(y) in terms of x: We know sin(y) = x. We can use the Pythagorean identity sin²(y) + cos²(y) = 1 to find cos(y). cos²(y) = 1 - sin²(y) cos²(y) = 1 - x² cos(y) = ±√(1 - x²)

5. Determine the correct sign for cos(y): Since the range of arcsin(x) is [-π/2, π/2], y lies in either the first or fourth quadrant. In both of these quadrants, cosine is non-negative. Therefore, we take the positive square root: cos(y) = √(1 - x²)

6. Substitute back into the dy/dx equation: dy/dx = 1 / √(1 - x²)

Therefore, the derivative of arcsin(x) is:

d/dx [arcsin(x)] = 1 / √(1 - x²)

Deriving the Derivative of arccos(x)

The derivation of arccos(x) is very similar to arcsin(x).

1. Set up the equation: Let y = arccos(x). This means cos(y) = x.

2. Differentiate both sides implicitly with respect to x: d/dx [cos(y)] = d/dx [x] -sin(y) * dy/dx = 1 (Applying the chain rule)

3. Solve for dy/dx: dy/dx = -1 / sin(y)

4. Express sin(y) in terms of x: We know cos(y) = x. Using the Pythagorean identity: sin²(y) + cos²(y) = 1 sin²(y) = 1 - cos²(y) sin²(y) = 1 - x² sin(y) = ±√(1 - x²)

5. Determine the correct sign for sin(y): The range of arccos(x) is [0, π]. This means y lies in either the first or second quadrant. In both of these quadrants, sine is non-negative. Therefore, we take the positive square root: sin(y) = √(1 - x²)

6. Substitute back into the dy/dx equation: dy/dx = -1 / √(1 - x²)

Therefore, the derivative of arccos(x) is:

d/dx [arccos(x)] = -1 / √(1 - x²)

Notice the striking similarity to the derivative of arcsin(x), differing only by a negative sign. This relationship is not coincidental, as arcsin(x) + arccos(x) = π/2, a constant. The derivative of a constant is zero, explaining the negative relationship between their derivatives.

Deriving the Derivative of arctan(x)

The derivative of arctan(x) requires a slightly different approach.

1. Set up the equation: Let y = arctan(x). This means tan(y) = x.

2. Differentiate both sides implicitly with respect to x: d/dx [tan(y)] = d/dx [x] sec²(y) * dy/dx = 1 (Applying the chain rule)

3. Solve for dy/dx: dy/dx = 1 / sec²(y)

4. Express sec²(y) in terms of x: We know tan(y) = x. We can use the trigonometric identity sec²(y) = 1 + tan²(y): sec²(y) = 1 + tan²(y) sec²(y) = 1 + x²

5. Substitute back into the dy/dx equation: dy/dx = 1 / (1 + x²)

Therefore, the derivative of arctan(x) is:

d/dx [arctan(x)] = 1 / (1 + x²)

Deriving the Derivative of arccot(x)

Following a similar pattern to arccos(x) and arcsin(x), the derivative of arccot(x) mirrors arctan(x) with a negative sign.

1. Set up the equation: Let y = arccot(x). This means cot(y) = x.

2. Differentiate both sides implicitly with respect to x: d/dx [cot(y)] = d/dx [x] -csc²(y) * dy/dx = 1 (Applying the chain rule)

3. Solve for dy/dx: dy/dx = -1 / csc²(y)

4. Express csc²(y) in terms of x: We know cot(y) = x. We can use the trigonometric identity csc²(y) = 1 + cot²(y): csc²(y) = 1 + cot²(y) csc²(y) = 1 + x²

5. Substitute back into the dy/dx equation: dy/dx = -1 / (1 + x²)

Therefore, the derivative of arccot(x) is:

d/dx [arccot(x)] = -1 / (1 + x²)

Again, we see the inverse relationship: arctan(x) + arccot(x) = π/2, leading to their derivatives being negatives of each other.

Deriving the Derivative of arcsec(x)

The derivatives of arcsec(x) and arccsc(x) are a bit trickier due to the more complex relationship between secant/cosecant and their respective inverse functions.

1. Set up the equation: Let y = arcsec(x). This means sec(y) = x.

2. Differentiate both sides implicitly with respect to x: d/dx [sec(y)] = d/dx [x] sec(y)tan(y) * dy/dx = 1 (Applying the chain rule)

3. Solve for dy/dx: dy/dx = 1 / [sec(y)tan(y)]

4. Express tan(y) in terms of x: We know sec(y) = x. We can use the trigonometric identity tan²(y) = sec²(y) - 1: tan²(y) = sec²(y) - 1 tan²(y) = x² - 1 tan(y) = ±√(x² - 1)

5. Determine the correct sign for tan(y): This is where it gets subtle. The range of arcsec(x) is [0, π/2) ∪ (π/2, π]. If 0 ≤ y < π/2, then tan(y) is positive. If π/2 < y ≤ π, then tan(y) is negative. To account for both cases, we use the absolute value of x:

   • If x > 1, then 0 ≤ y < π/2, and tan(y) = √(x² - 1)
   • If x < -1, then π/2 < y ≤ π, and tan(y) = -√(x² - 1)

   Therefore, sec(y)tan(y) = x * ±√(x² - 1). To generalize, we can write this as |x|√(x² - 1) to ensure the result is always positive, as dy/dx must be positive for all values in the domain of arcsec(x).

6. Substitute back into the dy/dx equation: dy/dx = 1 / [|x|√(x² - 1)]

Therefore, the derivative of arcsec(x) is:

d/dx [arcsec(x)] = 1 / [|x|√(x² - 1)]

Deriving the Derivative of arccsc(x)

The derivative of arccsc(x) follows a similar logic to arcsec(x), but with a negative sign.

1. Set up the equation: Let y = arccsc(x). This means csc(y) = x.

2. Differentiate both sides implicitly with respect to x: d/dx [csc(y)] = d/dx [x] -csc(y)cot(y) * dy/dx = 1 (Applying the chain rule)

3. Solve for dy/dx: dy/dx = -1 / [csc(y)cot(y)]

4. Express cot(y) in terms of x: We know csc(y) = x. We can use the trigonometric identity cot²(y) = csc²(y) - 1: cot²(y) = csc²(y) - 1 cot²(y) = x² - 1 cot(y) = ±√(x² - 1)

5. Determine the correct sign for cot(y): The range of arccsc(x) is [-π/2, 0) ∪ (0, π/2].

   • If 0 < y ≤ π/2, then cot(y) is positive.
   • If -π/2 ≤ y < 0, then cot(y) is negative.

   Again, to account for the correct sign and ensure the derivative is negative (as it should be), we use the absolute value of x: csc(y)cot(y) = x * ±√(x² - 1), which we generalize as |x|√(x² - 1).

6. Substitute back into the dy/dx equation: dy/dx = -1 / [|x|√(x² - 1)]

Therefore, the derivative of arccsc(x) is:

d/dx [arccsc(x)] = -1 / [|x|√(x² - 1)]

Summary of Derivatives of Inverse Trigonometric Functions

Here's a concise table summarizing the derivatives we've derived:

Key Takeaways and Observations

• Implicit Differentiation: The foundation of these derivations lies in the power of implicit differentiation.
• Trigonometric Identities: Mastery of trigonometric identities is essential for expressing the derivatives in terms of 'x'.
• Domain and Range Awareness: Careful consideration of the domain and range of each inverse trigonometric function is crucial for determining the correct signs in the derivatives.
• Pairings and Relationships: Notice the pairing of sine/cosine and tangent/cotangent, where their derivatives are negatives of each other. This stems from the complementary angle identities.
• The Absolute Value in arcsec(x) and arccsc(x): The absolute value ensures the derivative maintains the correct sign across the entire domain of these functions.

Applications and Further Exploration

Understanding the derivatives of inverse trigonometric functions opens doors to a wider range of calculus problems, including:

• Integration: These derivatives are essential for solving integrals involving expressions like √(1 - x²), 1/(1 + x²), and √(x² - 1).
• Optimization Problems: Inverse trigonometric functions can appear in optimization problems related to angles and distances.
• Related Rates Problems: These derivatives are useful in related rates problems involving angles changing with time.

Furthermore, you can explore the higher-order derivatives of these functions, which can lead to fascinating patterns and applications in areas like differential equations.

Conclusion

Deriving the derivatives of inverse trigonometric functions is a rewarding exercise that reinforces your understanding of calculus principles and trigonometric relationships. While the formulas themselves might seem complex at first, breaking down the derivations step-by-step reveals the underlying logic and elegance. With practice and a solid grasp of the concepts, you'll be well-equipped to tackle a wide range of calculus problems involving these fascinating functions. How do you feel about tackling these derivatives now? Ready to give it a try?