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Inverse trigonometric derivatives

Differentiate two common inverse trigonometric functions.

Calculus · Derivatives
$$(\arcsin x)'=\frac1{\sqrt{1-x^2}},\quad(\arctan x)'=\frac1{1+x^2}$$

Inverse trigonometric derivatives is one of 9 derivatives formulas in the calculus section of this library, and it is used at ap · university level.

Why inverse trigonometric derivatives works

Set y equal to arcsin x, which means sin y = x. Differentiating both sides gives cos y times y-prime equal to 1, and the Pythagorean identity turns cos y into the square root of 1 − x². The same implicit step on tan y = x replaces sec² y with 1 + x².

What each symbol means

$x$ is a real input.

Inverse trigonometric derivatives: when it holds

For the arcsine derivative, $|x|<1$; arctangent is differentiable for all real $x$.

When it stops applying

The arcsine derivative divides by 0 at x = 1 and x = −1, matching the vertical tangents at the two ends of its graph, so it says nothing there. Arctangent has no such trouble, since 1 + x² is never 0.

Inverse trigonometric derivatives: a worked example

$\frac{d}{dx}\arctan(2x)=2/(1+4x^2)$.

The mistake to avoid

What people do: Treating arcsin x as one over sin x when differentiating.

Why it goes wrong: Arcsine is the inverse function, not the reciprocal; the reciprocal of sine is cosecant, which is a completely different function with a completely different derivative.

Do this instead: Start from sin y = x and differentiate both sides implicitly, which delivers the correct root expression.

Inverse trigonometric derivatives: step by step

  1. Name the unknown, and the unit the answer has to come out in.
  2. Match the symbols to your values. $x$ is a real input.
  3. Check the conditions before substituting. For the arcsine derivative, $|x|<1$; arctangent is differentiable for all real $x$.
  4. Substitute, keep exact values to the last line, then test the sign, size, and unit against a rough estimate — the check that catches most calculus slips.

Where this formula fits

Subject
Calculus formulas — 42 entries in this library
Topic
Derivatives
Level
AP · University

Formulas are easiest to keep when they sit inside a method rather than on a list. Use the links below to see where inverse trigonometric derivatives comes from, to check a calculation against a tool, and to practise it until you can recall it without looking.

Questions about inverse trigonometric derivatives

Why is the arctangent derivative defined for every real input?

Its denominator is at least 1 for every x, so it can never vanish. The graph also flattens out toward its two horizontal asymptotes instead of turning vertical.

What is the slope of arcsine at the origin?

It is 1 over the square root of 1, which is 1, so arcsine leaves the origin travelling in the same direction as the line y = x.

Where does the square root come from?

From replacing cos y with the root of 1 minus sin squared y. The positive root is correct because arcsine only returns angles where cosine is not negative.

Do these change if the inverse function returns degrees?

Yes. A degree-valued arcsine is 180/π times the radian one, so its derivative is scaled by that same large factor of about 57.3.

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Work a inverse trigonometric derivatives problem step by step

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