Differentiate inverse tangent with its inner derivative
Use the principal inverse-tangent derivative, apply the chain rule and find tangent or normal lines using radian outputs.
Pure Mathematics 3 (Paper 3, A Level); Mathematics 9709, 2026–2027, section 3.4. Paper 1 knowledge and the shared differentiation lessons are assumed. Inverse tangent is an additional Paper 3 derivative. Derivatives of inverse sine and inverse cosine are not required in section 3.4. Inverse trigonometric outputs here are radians. Written practice is self-checked
Before you begin
- Interpret a principal inverse trigonometric value and use 1 + tan² y = sec² y.
- Apply the chain rule, point–gradient line equations and perpendicular gradients.
Work on paper. Try each question before revealing help, and explain why your method works.
Quick readiness check
What is arctan 1 on the principal radian branch?
What you will learn
- Use d(arctan x)/dx = 1/(1 + x²) and d(arctan u)/dx = u′/(1 + u²).
- Evaluate a derivative, form a tangent and normal, and distinguish inverse tangent from reciprocal tangent.
Inverse tangent names an angle, not a reciprocal
The function arctan x, also written tan⁻¹ x in inverse-function notation, returns the unique angle y in −π/2 < y < π/2 for which tan y = x. It accepts every real input x and outputs a radian angle. It is different from cot x = 1/tan x where that reciprocal representation is defined. An inverse operation undoes tangent on its principal branch; it does not turn the tangent's value upside down.
To obtain the derivative, let y = arctan x so tan y = x. Implicit differentiation gives sec² y · dy/dx = 1. Since sec² y = 1 + tan² y = 1 + x², dy/dx = 1/(1 + x²). The denominator is strictly positive for every real x; the principal inverse branch has no finite-input pole. The derivative is 1 at 0 and becomes smaller as |x| grows, consistent with the curve flattening towards its unattained horizontal limits.
Differentiate the inner input and square its whole value
For y = arctan u(x), the chain rule gives y′ = u′/(1 + u²) wherever the inner function is differentiable. For y = arctan(2x − 1), this becomes 2/[1 + (2x − 1)²]. At x = 1 the curve has y = arctan 1 = π/4 and gradient 2/(1 + 1) = 1. The tangent is y − π/4 = x − 1, and the normal is y − π/4 = −(x − 1). Square the complete inner expression; (2x − 1)² is not 4x² − 1.
For y = arctan(x²), the derivative is 2x/(1 + x⁴), not 1/(1 + x²). Its sign follows the inner derivative 2x. At x = 0 the gradient is zero, so the tangent is horizontal and the normal is the vertical line x = 0; a formula using −1/0 would not be a finite normal gradient. If inverse-tangent outputs are instead measured in degrees, their derivative carries the conversion factor 180/π. Use radian outputs for the standard calculus formula and convert a stated angular rate only after calculating it.
Pause and explain
Which derivative is correct for arctan(2x − 1)?
Worked example
For y = arctan(2x − 1), find the tangent and normal at x = 1, with y measured in radians.
Show the worked solution
- The point is P = (1, π/4), since the inner value is 1.
- Apply the chain rule: y′ = 2/[1 + (2x − 1)²]; at 1 the tangent gradient is 1.
- Use the point–gradient form with gradients 1 and −1 for the tangent and normal.
Answer Tangent y − π/4 = x − 1; normal y − π/4 = −(x − 1).
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Write your answer and reasoning first. Use a hint only if you are stuck. The model solution is for self-checking; your written work is not automatically marked.
For y = arctan(3x), find y′ and the tangent and normal at x = 0, using radian outputs.
Give me a hint
Use the inner derivative 3 and square the inner value 3x before evaluating the gradient.
Compare my reasoning
- The derivative is y′ = 3/(1 + 9x²).
- At 0 the point is (0, 0) and the tangent gradient is 3.
- The tangent is y = 3x; the perpendicular normal has gradient −1/3, so y = −x/3.
y′ = 3/(1 + 9x²); tangent y = 3x; normal y = −x/3.
Look for these in your work
- Apply the chain rule and evaluate at the actual point.
- Use perpendicular gradients and verify both lines pass through the point.
For y = arctan(x²), find y′, the tangent and normal at x = −1, and the tangent and normal at x = 0.
Give me a hint
The inner derivative changes sign, and its square produces x⁴ in the denominator.
Compare my reasoning
- Differentiate to get 2x/(1 + x⁴). At −1 the point is (−1, π/4) and the gradient is −1.
- At −1 the tangent is y − π/4 = −(x + 1), and the normal is y − π/4 = x + 1.
- At 0 the point is (0, 0) and the gradient is 0, giving horizontal tangent y = 0 and vertical normal x = 0.
y′ = 2x/(1 + x⁴); at −1: tangent y − π/4 = −(x + 1), normal y − π/4 = x + 1; at 0: tangent y = 0, normal x = 0.
Look for these in your work
- Retain the signed inner derivative and the fourth power.
- Handle the zero-gradient normal as a vertical line.
A supplied angular model is θ(t) = arctan(t/2) radians, with t measured numerically in seconds and t ≥ 0. Find its instantaneous angular rate at t = 2 in radians per second and degrees per second.
Give me a hint
Differentiate in radians first, then multiply the resulting angular rate by 180/π to convert it.
Compare my reasoning
- The inner derivative is 1/2, so θ′(t) = (1/2)/[1 + (t/2)²] = 2/(t² + 4).
- At t = 2 the rate is 2/8 = 1/4 rad/s; the angle itself is π/4 radians, a different quantity.
- The converted rate is (1/4)(180/π) = 45/π degrees per second, approximately 14.3°/s. Both rates describe the same supplied model at the same time.
θ′(t) = 2/(t² + 4); at 2 s: 1/4 rad/s = 45/π degrees per second ≈ 14.3°/s.
Look for these in your work
- Include the input scale and distinguish angle from angular rate.
- Convert output units with the explicit factor 180/π.
Common mistakes
- tan⁻¹ x as an inverse function is not cot x.
- Include the inner derivative in the numerator and the square of the whole inner input in the denominator.
- A horizontal tangent has a vertical normal; convert angular output units explicitly.
What can you explain now?
Find the derivative of arctan(3x).
Compare with the explanation
3/(1 + 9x²).
The inner derivative is 3 and the squared inner input is (3x)² = 9x².
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Make it stick
Derive the inverse-tangent formula from tan y = x, extend it to an inner function and use it to find a tangent, normal or angular rate with stated units.
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