Calculus learning

Derivative of arccos(x): Negative Sign, Domain and Chain Rule

The derivative of arccos(x) has the same square-root denominator as arcsin, with a minus sign. On its principal range, cosine decreases, so its inverse decreases too.

Direct answer

d/dx arccos(x)=−1/√(1−x²), for −1<x<1.

The principal real arccos range is [0,π]. The function exists at endpoints, but the finite derivative formula does not.

Derivative of arccos(x) worked method

Set y=arccos x, so cos y=x and 0<y<π in the interior.

Differentiate: −sin y·y′=1. Since sin y is positive here, divide by it without changing its sign.

Use sin y=√(1−cos²y)=√(1−x²) to obtain −1/√(1−x²).

For arccos(2x), the chain rule gives −2/√(1−4x²), with |x|<1/2 for a finite derivative.

1
cos y=x → −sin y · y′=1
2
y′=−1/√(1−x²)
3
(arccos(g(x)))′=−g′(x)/√(1−g(x)²)

Worked examples

d/dx arccos(x)
−11−x2\frac{-1}{\sqrt{1 - x^{2}}}

The sign is negative.

d/dx arccos(2x)
−21−4 x2\frac{-2}{\sqrt{1 - 4\,x^{2}}}

Derivative domain is |x|<1/2.

d/dx [arcsin(x)+arccos(x)]
00

The sum is π/2 on [−1,1]; differentiate in the interior.

Common mistakes

  • • Copying the positive arcsin sign.
  • • Squaring g′(x) instead of g(x) inside the root.
  • • Ignoring endpoint differentiability.

Try the supported calculator preset

The preset opens a supported expression with its variable and conditions. Teaching examples elsewhere in this guide are reference explanations, not claims that the engine supports every method.

Use acos rather than an unsupported display alias. The formula’s endpoint restriction still applies.

Derivatives:acos(x)
Open the calculator without a preset: Derivatives

References

Frequently asked questions

Why is the derivative negative?+

Cosine decreases on its principal inverse interval.

Is cos⁻¹ a reciprocal?+

Here it is arccos; sec is the reciprocal of cosine.

How does the calculator spell it?+

Use acos(x), as in the runnable preset.

Conclusion

Derivative of arccos(x) starts with the stated domain and conditions. Recheck those before carrying a worked example into a different problem.

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