Inverse trigonometric rule · Derivative Calculator

What it says

Each one comes from the inverse function rule. For $y = \arcsin x$ we have $\sin y = x$; differentiating both sides gives $\cos y \cdot y' = 1$, so $y' = \frac{1}{\cos y} = \frac{1}{\sqrt{1 - x^{2}}}$, using $\cos y = \sqrt{1 - \sin^{2}y}$.

The table: $\left(\arcsin x\right)' = \frac{1}{\sqrt{1 - x^{2}}}$, $\left(\arccos x\right)' = -\frac{1}{\sqrt{1 - x^{2}}}$, $\left(\arctan x\right)' = \frac{1}{1 + x^{2}}$. The arc-co-functions are the negatives of their partners, because $\arcsin x + \arccos x = \frac{\pi}{2}$ is constant.

That these derivatives are purely algebraic is why inverse trigonometric functions turn up as the answers to integrals that look nothing like trigonometry.

When it applies

$\arcsin$, $\arccos$, $\arctan$ and their reciprocal partners.
Written either way in this calculator: $\texttt{arctan(x)}$ or $\texttt{tan\^-1(x)}$.
With the chain rule for a compound argument, as in $\arctan(3x)$.

Five worked examples

Every line is the step the calculator would show, in the order it applies them. Each graph is live: hover it to read both curves and see the tangent whose slope is the derivative, drag to pan, scroll to zoom.

Example 1

\frac{d}{dx}\left[\arcsin\left(x\right)\right]

From $\sin y = x$, differentiate both sides.
$\cos y \cdot y' = 1$
Substitute $\cos y = \sqrt{1 - x^{2}}$.
$y' = \frac{1}{\sqrt{1 - x^{2}}}$

Answer

\frac{1}{\sqrt{1 - x^{2}}}

Both curves exist only on $[-1, 1]$, which is why the graph frames itself so tightly. f′ is a positive well that blows up at the two ends, where arcsine turns vertical, and reaches its lowest value of 1 at the origin.

Defined for $-1 < x < 1$, and blowing up at both ends where the sine curve turns over.

Open arcsin(x) in the calculator

Example 2

\frac{d}{dx}\left[\arccos\left(x\right)\right]

Same derivation, or note that $\arccos x = \frac{\pi}{2} - \arcsin x$.
$= -\frac{1}{\sqrt{1 - x^{2}}}$

Answer

-\frac{1}{\sqrt{1 - x^{2}}}

The mirror image: f falls from $\pi$ to 0 while f′ is the same well flipped below the axis.

Exactly the negative of the arcsine derivative, since the two functions add to a constant.

Open arccos(x) in the calculator

Example 3

\frac{d}{dx}\left[\arctan\left(x\right)\right]

From $\tan y = x$.
$\sec^{2}y \cdot y' = 1$
And $\sec^{2}y = 1 + \tan^{2}y = 1 + x^{2}$.
$y' = \frac{1}{1 + x^{2}}$

Answer

\frac{1}{1 + x^{2}}

f levels off toward $\pm\tfrac{\pi}{2}$ at both ends, and f′ is a bump of height 1 over the origin. Nothing blows up anywhere.

Defined for every real $x$, which is why $\arctan$ is such a well-behaved function.

Open arctan(x) in the calculator

Example 4

\frac{d}{dx}\left[\arctan\left(3 x\right)\right]

Chain rule with $u = 3x$.
$= \frac{1}{1 + \left(3x\right)^{2}}\cdot 3$
Square the inside.
$= \frac{3}{1 + 9x^{2}}$

Answer

\frac{3}{1 + 9 x^{2}}

The same S-curve squeezed threefold, so f′ is three times taller and three times narrower.

The $3$ is squared in the denominator but not in the numerator — a common place to slip.

Open arctan(3x) in the calculator

Example 5

\frac{d}{dx}\left[x \arctan\left(x\right)\right]

Product rule.
$= 1 \cdot \arctan x + x\cdot\frac{1}{1 + x^{2}}$
Tidy.
$= \arctan x + \frac{x}{1 + x^{2}}$

Answer

\arctan\left(x\right) + \frac{x}{1 + x^{2}}

A smooth valley with its floor at the origin, where f′ changes sign. Far out, f grows almost linearly.

The pairing behind $\int \arctan x\,dx = x\arctan x - \tfrac{1}{2}\ln(1 + x^{2})$.

Open x arctan(x) in the calculator

Watch out for

$\tan^{-1}x$ means $\arctan x$, not $\frac{1}{\tan x}$.
The minus sign belongs to $\arccos$ and $\text{arccot}$, not to $\arcsin$ or $\arctan$.
$\arcsin$ and $\arccos$ are only defined on $[-1, 1]$, and their derivatives only inside it.

Trigonometric rule Chain rule Inverse hyperbolic rule

What it says

When it applies

Five worked examples

Watch out for

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