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Inverse trigonometric functions

Derivative of arctan(x): Formula, Proof, and Examples

Arctan is the inverse function of tangent on its principal interval. It is not the reciprocal 1/tan(x); inverse-function notation and reciprocal notation describe different operations.

Direct answer

ddx[arctan⁡x]=11+x2\frac{\mathrm{d}}{\mathrm{d}x}\left[\arctan x\right] = \frac{1}{1 + x^{2}}

Valid for every real x; angles are measured in radians.

Proof from tan(y) = x

Set y = arctan(x), so tan(y) = x. Differentiate implicitly with respect to x.

Because d/dy tan(y) = sec²(y), the chain rule gives sec²(y) · dy/dx = 1.

Use sec²(y) = 1 + tan²(y) and tan(y) = x. Therefore dy/dx = 1/(1+x²).

1
tan⁡y=x\tan y = x
2
sec²(y) · dy/dx = 1
3
dydx=11+x2\frac{\mathrm{d}y}{\mathrm{d}x} = \frac{1}{1 + x^{2}}

Worked examples

ddx[arctan⁡(2 x)]\frac{\mathrm{d}}{\mathrm{d}x}\left[\arctan\left(2\,x\right)\right]
21+4 x2\frac{2}{1 + 4\,x^{2}}

The inner derivative contributes the factor 2.

ddx[arctan⁡(x2)]\frac{\mathrm{d}}{\mathrm{d}x}\left[\arctan\left(x^{2}\right)\right]
2 x1+x4\frac{2\,x}{1 + x^{4}}

Square the entire inner expression in the denominator.

ddx[arctan⁡(1x)]\frac{\mathrm{d}}{\mathrm{d}x}\left[\arctan\left(\frac{1}{x}\right)\right]
−1x2+1\frac{-1}{x^{2} + 1}

Apply the chain rule on intervals where x ≠ 0.

Common mistakes

  • • Replacing arctan(x) by 1/tan(x). The former is an inverse function; the latter is a reciprocal.
  • • Forgetting to square the complete inner input in 1+[g(x)]².
  • • Using degree-mode derivative constants. Standard calculus derivative formulas assume radians.

Try the worked examples in the calculator

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求导:arctan⁡(2 x)\arctan\left(2\,x\right)
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