Understanding Why The First Derivative Of Arctan(x) Is 1/(1 + X²)

Introduction to arctan(x) and Its Derivative

The inverse tangent function, arctan(x), is foundational in calculus, trigonometry, and many applied sciences. Its derivative, [arctan(x)]’ = 1/(1 + x²), often appears in integration problems, differential equations, and modeling phenomena in physics and engineering. In this blog post, we will explore two clear, step-by-step proofs and explain the geometric intuition that makes this result so elegant.

This guide is useful for students, educators, and math enthusiasts.

Why Study the Derivative of arctan(x)

• Fundamental in Integration: Knowing (arctgx)’ allows integration of rational functions leading to arctangent-based antiderivatives
• Modeling Real‑World Phenomena: arctan(x) appears in formulas for wave behavior, signal processing, and probability distributions (e.g., Cauchy distribution)
• Solving Differential Equations: Many differential equations, particularly those with quadratic expressions, simplify elegantly when derivatives of inverse trigonometric functions are applied

Common Confusion and the Tabulated Derivative

By default, many students accept from tables that the derivative of arctan(x) equals 1/(1 + x²). But understanding why, requires delving into limits, implicit differentiation, and triangle geometry.

Proof 1 – Limit Definition and L’Hôpital’s Rule

1. Start with the limit definition:

   [arctan(x)]’ = lim_{h→0} [arctan(x+h) − arctan(x)] / h

2. Use the angle-difference identity:

   arctan(a) − arctan(b) = arctan[(a−b)/(1 + ab)]

3. Substitute a = x+h, b = x:

   lim_{h→0} {arctan{h/[1 + x(x+h)]}} / h

4. Apply L’Hôpital’s Rule for the 0/0 form, differentiate numerator and denominator with respect to h. As a reminder, if:

   lim_{x→a} [f(x) / g(x)] = 0/0 or ∞/∞

then it holds:

   lim_{x→a} [f(x) / g(x)] = lim_{x→a} [f'(x) / g'(x)]

5. Simplify at h = 0 to obtain 1/(1 + x²).

Proof 2 – Implicit Differentiation

1. Set y = arctan(x), so tan(y) = x, y ∈ (−π/2, π/2).

2. Differentiate implicitly:

   If tan(y) = x, then differentiating both sides with respect to x gives:

[tan(y)]’ = x’

[tan(y)]’ = 1

   Since:

(u / v)’ = (u’ v – u v’) / v²

   this means that:

[tan(x)]’ = [(sin(x) / cos(x)]’ = [(sin(x))’ cos(x) – (cos(x))’ sin(x)] / cos²(x) = [cos(x) cos(x) – (-sin(x)) sin(x)] / cos²(x) = [cos²(x) + sin²(x)] / cos²(x) = 1 / cos²(x)

   Since y is a dependent variable, then it must be:

[tan(y)]’ = [1 / cos²(y)] y’

1 = y’ / cos²(y)

y’ = cos²(y)

3. Using the identity: tan²(y) + 1 = 1 / cos²(y).

4. Conclude y’ = 1/(1 + x²).

Geometric Visualization with a Right Triangle

1. Construct a right triangle where:

  – Opposite side = x

  – Adjacent side = 1

  – Hypotenuse = √(1 + x²)

  – y = α

2. Since tan(y) = x/1, y = arctan(x).

3. The derivative y’ is cos²(y) = [1/√(1 + x²)]² = 1/(1 + x²).

Practical Applications of the arctan Derivative

1. Integration of Rational Functions
   ∫[1 / (1 + x²)] dx = arctan(x) + C
2. Signal Processing
   Phase shifts in electrical engineering often involve arctan terms; their rates of change depend on 1 / (1 + x²)
3. Probability & Statistics
   The Cauchy distribution’s cumulative distribution function is an arctan, and its density involves 1 / (1 + x²)

Conclusion

The derivative:

[arctan(x)]’ = 1 / (1 + x²)

emerges naturally from implicit differentiation, limit definitions, and geometric reasoning. This fundamental result not only enriches theoretical calculus but also underpins diverse applications, from integration techniques to modeling real‑world phenomena.

In a similar way it can be proved:

[arccotan(x)]’ = -1 / (1 + x²)

[arcsin(x)]’ = 1 / √(1 – x²)

[arccos(x)]’ = -1 / √(1 – x²)