Derivative of arctan(x) = 1/(1+x²): Proof and Examples

The derivative of arctan(x) is 1/(1+x²), defined for all real numbers because its denominator is always positive and never equals zero. You prove this by taking the tangent of both sides to get tan(y) = x, differentiating implicitly to sec²(y)(dy/dx) = 1, and substituting the identity sec²(y) = 1 + tan²(y) = 1 + x².

Inverse sine and cosine produce square roots. They also require restricted domains between -1 and 1. In contrast, the derivative of arctangent is a rational function. It converts angle relationships into algebraic fractions. So, it serves as an essential tool. You can use it across differential calculus, integral calculus, and rate-of-change modeling.

What is the geometric intuition behind the derivative of arctan x?

The arctangent function, written as arctan(x) or tan⁻¹(x), is the inverse of the tangent function restricted to (-π/2, π/2). On a coordinate plane, the graph of y = arctan(x) is the reflection of y = tan(x) across y = x.

Reflecting a curve across y = x interchanges horizontal and vertical movements. As a result, swapping rise and run forces the slope of an inverse function to equal the reciprocal of the original slope. The slope of y = tan(y) is sec²(y) = 1 + tan²(y). Therefore, the inverse slope at x = tan(y) is 1 / (1 + tan²(y)). Next, replacing tan(y) with x gives the derivative 1 / (1 + x²).

This reciprocal relationship controls the geometric properties of the arctangent graph:

• At the origin (x = 0): Since tan(0) = 0, the derivative equals 1 / (1 + 0²) = 1. As a result, this represents the steepest slope on the curve.
• As x moves away from zero: The x² term in the denominator grows. Consequently, the slope 1 / (1 + x²) shrinks toward zero. The curve flattens as a result.
• As |x| approaches infinity: The derivative approaches zero. As a result, the curve flattens toward horizontal asymptotes at y = π/2 and y = -π/2.

How to prove the derivative of arctan(x) using implicit differentiation

The standard algebraic proof relies on differentiating an implicit equation using the chain rule:

1. Set up the inverse relation: Let y = arctan(x) for y in (-π/2, π/2). Next, taking tangent yields tan(y) = x.
2. Differentiate both sides with respect to x: Apply implicit differentiation using the chain rule:

   sec²(y) · (dy/dx) = 1

3. Isolate dy/dx: Divide both sides by sec²(y):

   dy/dx = 1 / sec²(y)

4. Apply the Pythagorean trigonometric identity: Substitute sec²(y) = 1 + tan²(y):

   dy/dx = 1 / (1 + tan²(y))

5. Substitute x back into the expression: Since tan(y) = x, replace tan²(y) with x²:

   d/dx [arctan(x)] = 1 / (1 + x²)

Alternative proof using a geometric right triangle

You can also complete the proof visually using right-triangle trigonometry. First, if tan(y) = x, view x as x/1. Next, construct a right triangle with acute angle y, opposite side x, and adjacent side 1.

You can use the Pythagorean theorem. It shows the hypotenuse is √(1 + x²). Then apply the secant ratio. This makes sec(y) = √(1 + x²)/1. So, sec(y) equals √(1 + x²). Next, square this value. You get sec²(y) = 1 + x². Finally, substitute sec²(y) back into dy/dx = 1 / sec²(y). This gives you 1 / (1 + x²).

How to prove the derivative using the inverse function rule

The derivative of an inverse function f⁻¹ is given by:

(f⁻¹)'(x) = 1 / f'(f⁻¹(x))

To differentiate arctangent, you should set f(x) = tan(x). So, f⁻¹(x) becomes arctan(x). The derivative of tangent is f'(x) = sec²(x). This also equals 1 + tan²(x). Next, apply the formula.

1. Evaluate f’ at f⁻¹(x): f'(arctan(x)) = 1 + tan²(arctan(x)).
2. Since tan(arctan(x)) = x, tan²(arctan(x)) = x².
3. Substitute into the reciprocal formula: d/dx [arctan(x)] = 1 / (1 + x²).

How to differentiate arctan(u) using the chain rule

However, the argument inside arctangent may be a differentiable function g(x). It might not be x. In this case, apply the chain rule.

d/dx [arctan(g(x))] = g'(x) / (1 + (g(x))²)

First, place the derivative of the inner function in the numerator. Next, place one plus the square of the inner function in the denominator.

Step-by-step worked examples

These worked examples show how to combine the arctangent rule with the chain, product, quotient, and power rules.

Example 1: Basic linear inner function
Find the derivative of f(x) = arctan(2x).
Let u = 2x. So, u’ = 2. Also, u² = (2x)² = 4x².
Next, apply the chain rule.
f'(x) = 2 / (1 + 4x²)

Example 2: Polynomial inner function
Find the derivative of f(x) = arctan(x²).
Let u = x² and u’ = 2x, so u² = (x²)² = x⁴.
f'(x) = 2x / (1 + x⁴)

Example 3: Reciprocal input
Find the derivative of f(x) = arctan(1/x).
Let u = 1/x = x⁻¹. So, u’ = -1/x². Also, 1 + u² = 1 + 1/x².
f'(x) = (-1/x²) / (1 + 1/x²)
Multiply the numerator and denominator by x²:
f'(x) = -1 / (x² + 1)
Adding d/dx [arctan(x)] and d/dx [arctan(1/x)] gives 1/(1+x²) + (-1/(1+x²)) = 0. This proves arctan(x) + arctan(1/x) is constant for all non-zero x. It equals π/2 for x > 0 and -π/2 for x < 0.

Example 4: Product rule combination
Find the derivative of f(x) = x · arctan(x).
First, apply the product rule (u · v)’ = u’v + uv’. Use u = x and v = arctan(x):
f'(x) = (1) · arctan(x) + x · (1 / (1 + x²)). So, f'(x) = arctan(x) + x / (1 + x²).

Tangent line equation example

Example 5: Tangent line equation
Find the equation of the line tangent to y = arctan(x) at x = 1.
First, find the y-value: y = arctan(1) = π/4.
Next, find the slope using the derivative: m = 1 / (1 + 1²) = 1/2.
Finally, use point-slope form y – y₁ = m(x – x₁):
y – π/4 = (1/2)(x – 1)

Example 6: Rational expression simplification
Differentiate f(x) = arctan((x – 1) / (x + 1)).
First, let u = (x – 1) / (x + 1). Differentiate u using the quotient rule:
u’ = [(1)(x + 1) – (x – 1)(1)] / (x + 1)². So, υ’ = 2 / (x + 1)².
Next, simplify 1 + u²:
1 + ((x – 1)/(x + 1))² = [(x + 1)² + (x – 1)²] / (x + 1)². This simplifies to (2x² + 2) / (x + 1)².
Then, divide u’ by 1 + u²:
f'(x) = [2 / (x + 1)²] / [(2x² + 2) / (x + 1)²]. This simplifies to 2 / (2x² + 2) = 1 / (1 + x²).
Thus, arctan((x – 1)/(x + 1)) differs from arctan(x) by a constant on x > -1. Specifically, it is arctan(x) – π/4.

Example 7: Exponential inside arctangent
Find the derivative of f(x) = arctan(e^x).
First, let u = e^x, u’ = e^x, and u² = e^(2x).
So, f'(x) = e^x / (1 + e^(2x))

Example 8: Second derivative calculation
Find the second derivative of f(x) = arctan(x).
Differentiate f'(x) = (1 + x²)⁻¹ using power and chain rules:
f”(x) = -1 · (1 + x²)⁻² · (2x) = -2x / (1 + x²)²

How does arctan compare to other inverse trigonometric derivatives?

Selecting the right formula requires recognizing structural differences across inverse trigonometric derivatives:

Arctangent is the only basic inverse trigonometric derivative producing a rational expression without radicals. Because 1 + x² ≥ 1 for all real x, its domain has no singular points or vertical asymptotes.

Why is the arctangent derivative crucial for integration and series?

Reversing the derivative of arctan(x) gives a foundational integration formula:

∫ 1 / (1 + x²) dx = arctan(x) + C

Scaling by a constant parameter ‘a’ yields:

∫ 1 / (a² + x²) dx = (1/a) arctan(x/a) + C

This form appears when integrating rational functions with irreducible quadratic denominators using partial fraction decomposition. For example, integrating 1 / (9 + x²) with a = 3 gives (1/3) arctan(x/3) + C. Prove this identity using u-substitution with u = x/a.

Constructing the arctangent power series

The derivative 1 / (1 + x²) connects algebraic functions to infinite series. So, expand 1 / (1 – r) with r = -x² (|x| < 1):

1 / (1 + x²) = 1 – x² + x⁴ – x⁶ + x⁸ – …

Integrating term by term from 0 to x yields the Maclaurin series for arctangent:

arctan(x) = x – x³/3 + x⁵/5 – x⁷/7 + …

Evaluating at x = 1 yields the Gregory-Leibniz formula for π:

π/4 = 1 – 1/3 + 1/5 – 1/7 + 1/9 – …

See Taylor series for convergence testing, interval analysis, and error bounds.

Real-world applications and rate-of-change models

Because arctangent converts linear ratios into viewing angles, its derivative measures angular velocity relative to linear distance.

Suppose an observer stands 500 meters away from a straight track, tracking a vehicle moving at speed v = dx/dt. Also, let x be the distance along the track from the point opposite the observer. In this setup, the angle θ is:

θ = arctan(x / 500)

Differentiating with respect to time t via chain rule yields angular tracking speed dθ/dt:

dθ/dt = (500 / (500² + x²)) · (dx/dt)

When the vehicle passes directly opposite (x = 0) at 30 m/s, angular speed peaks at dθ/dt = (500 / 250,000) · 30 = 0.06 rad/s. At x = 500 meters, angular speed drops to (500 / 500,000) · 30 = 0.03 rad/s, illustrating the 1 / (1 + x²) rate of decay.

Common algebraic mistakes and how to fix them

• Confusing arctangent with arcsine: Writing 1 / √(1 – x²) instead of 1 / (1 + x²). Arctangent features a plus sign and no square root.
• Expanding the denominator incorrectly: Writing (1 + x)² instead of 1 + x². The rule adds 1 to x², rather than squaring 1 + x.
• Forgetting to square inner constants in the chain rule: Differentiating arctan(3x) as 3 / (1 + 3x²) instead of 3 / (1 + (3x)²) = 3 / (1 + 9x²).
• Confusing tan⁻¹(x) with 1/tan(x): Treating tan⁻¹(x) as cotangent. The derivative of cot(x) is -csc²(x), while tan⁻¹(x) is the inverse function with derivative 1/(1+x²).
• Evaluating rate of change in degrees: Calculating rates in degrees per second instead of radians. Standard calculus formulas require radians.
• Leaving compound fractions unsimplified: Writing (-1/x²) / (1 + 1/x²) without multiplying through by x² to obtain -1 / (x² + 1).

Try it yourself and practice problems

Enter an expression below to work through derivative steps, or test complex inputs in the derivative calculator.

1. Calculate d/dx [arctan(3x)]
2. Next, calculate d/dx [arctan(e^x)]
3. Calculate d/dx [arctan(√x)]
4. Also, calculate d/dx [arctan(x/2)]
5. Calculate d/dx [(arctan(x))²]

Solutions:

1. 3 / (1 + 9x²): Inner derivative is 3; inner squared term is (3x)² = 9x².
2. e^x / (1 + e^(2x)): Inner derivative is e^x; inner squared term is (e^x)² = e^(2x).
3. 1 / (2√x (1 + x)): Inner derivative is 1/(2√x) and (√x)² = x, giving 1 / (2√x (1 + x)).
4. 2 / (4 + x²): Inner derivative is 1/2 and denominator is 1 + x²/4. Clearing fractions gives (1/2) / ((4+x²)/4) = 2 / (4 + x²).
5. 2 arctan(x) / (1 + x²): Apply the power rule: 2 arctan(x) times inner derivative 1 / (1 + x²).

Frequently asked questions

Is arctan the same as tan⁻¹?

Yes, arctan(x) and tan⁻¹(x) represent the exact same inverse trigonometric function. The -1 exponent denotes an inverse function relationship, not a reciprocal power.

What is the second derivative of arctan x?

Differentiating 1/(1+x²) using power and chain rules gives f”(x) = -2x / (1 + x²)². This second derivative is positive for x < 0 and negative for x > 0, making x = 0 an inflection point.

What is the integral of arctan x?

First, use integration by parts with u = arctan(x) and dv = dx. This yields ∫ arctan(x) dx = x arctan(x) – (1/2) ln(1 + x²) + C. Next, differentiating this result returns arctan(x).

Why is the derivative of arctan x defined for all x?

Because denominator 1 + x² ≥ 1 for all real x, it never equals zero. Unlike the derivative of arcsin x, there are no domain restrictions or singular points.

What is the derivative of arctan x at x = 0?

Evaluating 1 / (1 + 0²) gives 1. This represents the maximum slope on the curve, matching the derivative of tan x at x = 0.

Further reading

• Paul’s Online Notes: Derivatives of Inverse Trig Functions — step-by-step derivations for all inverse trigonometric functions.
• Paul’s Online Notes: Implicit Differentiation — practice problems and solutions for implicit differentiation techniques.