What Are The Derivatives Of Inverse Trigonometric Functions?
The derivatives of inverse trigonometric functions are the six rules that tell you the slope of each inverse trig curve, $\arcsin$, $\arccos$, $\arctan$, $\operatorname{arccot}$, $\operatorname{arcsec}$, and $\operatorname{arccsc}$. An inverse trigonometric function answers the question "which angle has this sine (or tangent, or secant)?", and its derivative measures how fast that angle changes as the input changes.
The six standard results, each on its own domain, are:
$$ \begin{aligned} \frac{d}{dx}\arcsin x &= \frac{1}{\sqrt{1-x^2}}, & -1 < x < 1\[4pt] \frac{d}{dx}\arccos x &= -\frac{1}{\sqrt{1-x^2}}, & -1 < x < 1\[4pt] \frac{d}{dx}\arctan x &= \frac{1}{1+x^2}, & \text{all real } x\[4pt] \frac{d}{dx}\operatorname{arccot} x &= -\frac{1}{1+x^2}, & \text{all real } x\[4pt] \frac{d}{dx}\operatorname{arcsec} x &= \frac{1}{|x|\sqrt{x^2-1}}, & |x| > 1\[4pt] \frac{d}{dx}\operatorname{arccsc} x &= -\frac{1}{|x|\sqrt{x^2-1}}, & |x| > 1 \end{aligned} $$
Two features stand out right away. First, none of the answers contains a trig function: the derivatives are ordinary algebra with squares and square roots. Second, the six results come in three mirrored pairs, and within each pair the only difference is a minus sign.
That mirrored structure is worth trusting, and later we show exactly why it holds. Throughout this article the derivative is written in the single notation $\frac{d}{dx}$ for consistency.
How Do You Derive The Derivative Of Arcsin?
To derive $\frac{d}{dx}\arcsin x$, start from the definition of the inverse function and use implicit differentiation. Let
$$y = \arcsin x, \qquad \text{so} \qquad \sin y = x, \qquad -\tfrac{\pi}{2} \le y \le \tfrac{\pi}{2}.$$
The restriction on $y$ is the principal branch, the standard range of $\arcsin$. Differentiate both sides of $\sin y = x$ with respect to $x$, remembering that $y$ is a function of $x$, so the left side needs the chain rule:
$$\cos y \cdot \frac{dy}{dx} = 1 \qquad\Longrightarrow\qquad \frac{dy}{dx} = \frac{1}{\cos y}.$$
Now rewrite $\cos y$ in terms of $x$. From the Pythagorean identity, $\cos y = \pm\sqrt{1 - \sin^2 y} = \pm\sqrt{1 - x^2}$. On the principal branch $-\tfrac{\pi}{2} \le y \le \tfrac{\pi}{2}$, the cosine is never negative, so $\cos y = +\sqrt{1-x^2}$. That sign choice is the whole reason the answer is clean:
$$\frac{d}{dx}\arcsin x = \frac{1}{\sqrt{1-x^2}}, \qquad -1 < x < 1.$$
The geometric reading matches the graph. Near $x = 0$ the arcsin curve rises gently with slope $1$. As $x$ approaches $\pm 1$, the denominator $\sqrt{1-x^2}$ shrinks toward $0$, so the slope shoots up: the graph has vertical tangents at its two ends. The formula and the picture agree, which is the check every derivation should pass.
The same argument, run on $\cos y = x$, gives $\frac{d}{dx}\arccos x = -\frac{1}{\sqrt{1-x^2}}$. It picks up the minus sign because $\arccos$ decreases where $\arcsin$ increases.
How Do You Derive The Derivative Of Arctan?
The arctan derivation follows the identical three moves, and the secant identity does the simplifying. Let
$$y = \arctan x, \qquad \text{so} \qquad \tan y = x, \qquad -\tfrac{\pi}{2} < y < \tfrac{\pi}{2}.$$
Differentiate $\tan y = x$ with respect to $x$, using $\frac{d}{dy}\tan y = \sec^2 y$:
$$\sec^2 y \cdot \frac{dy}{dx} = 1 \qquad\Longrightarrow\qquad \frac{dy}{dx} = \frac{1}{\sec^2 y}.$$
The identity $\sec^2 y = 1 + \tan^2 y$ converts the answer straight into $x$, with no sign ambiguity because $\sec^2 y$ is a square:
$$\frac{d}{dx}\arctan x = \frac{1}{1 + \tan^2 y} = \frac{1}{1 + x^2}, \qquad \text{all real } x.$$
Geometrically, $\frac{1}{1+x^2}$ is always positive, so $\arctan$ always rises, and as $|x| \to \infty$ the slope fades to $0$. That is the graph flattening toward its horizontal asymptotes at $y = \pm\tfrac{\pi}{2}$, the arctan curve levelling off far from the origin.
What Are The Chain-Rule Versions For Composite Functions?
Almost every exam problem hides the inverse trig inside another function, so you rarely differentiate $\arcsin x$ alone. You differentiate $\arcsin(u)$, where $u$ is some expression in $x$. The chain rule attaches the inner derivative $u'$ to each formula:
$$ \frac{d}{dx}\arcsin(u) = \frac{u'}{\sqrt{1-u^2}}, \qquad \frac{d}{dx}\arctan(u) = \frac{u'}{1+u^2}, \qquad \frac{d}{dx}\operatorname{arcsec}(u) = \frac{u'}{|u|\sqrt{u^2-1}}. $$
The other three carry the same $u'$ with their minus signs. Forgetting that $u'$ is the single most common error in the topic, so treat it as part of the formula, not an optional extra.
Example 1: Differentiate $f(x) = \arcsin(3x)$.
Here $u = 3x$, so $u' = 3$. Substitute into the arcsin chain-rule form:
$$f'(x) = \frac{u'}{\sqrt{1 - u^2}} = \frac{3}{\sqrt{1 - (3x)^2}} = \frac{3}{\sqrt{1 - 9x^2}}.$$
The expression is valid where $1 - 9x^2 > 0$, that is $-\tfrac{1}{3} < x < \tfrac{1}{3}$.
Final answer: $f'(x) = \dfrac{3}{\sqrt{1 - 9x^2}}$.
Example 2: Differentiate $f(x) = \arctan(x^2)$.
Now $u = x^2$, so $u' = 2x$. Substitute into the arctan chain-rule form:
$$f'(x) = \frac{u'}{1 + u^2} = \frac{2x}{1 + (x^2)^2} = \frac{2x}{1 + x^4}.$$
This holds for all real $x$, since $\arctan$ has no domain restriction.
Final answer: $f'(x) = \dfrac{2x}{1 + x^4}$.
Example 3: Differentiate $f(x) = x,\arccos x$.
This one needs the product rule first, then the arccos derivative for the second factor:
$$f'(x) = (1)\cdot\arccos x + x\cdot\left(-\frac{1}{\sqrt{1-x^2}}\right) = \arccos x - \frac{x}{\sqrt{1-x^2}}.$$
The product rule handles the two factors; the inverse-trig rule supplies the derivative of $\arccos x$. Both pieces are needed, and dropping either is a classic slip.
Final answer: $f'(x) = \arccos x - \dfrac{x}{\sqrt{1-x^2}}$.
What Are All Six Inverse Trig Derivative Formulas?
Keep this table beside you until the six results are automatic. The middle column is the derivative; the right column is the domain where the formula is valid.
Table: The six derivatives of inverse trigonometric functions, with the domain of each.
Function | Derivative | Valid domain |
|---|---|---|
$\arcsin x$ | $\dfrac{1}{\sqrt{1-x^2}}$ | $-1 < x < 1$ |
$\arccos x$ | $-\dfrac{1}{\sqrt{1-x^2}}$ | $-1 < x < 1$ |
$\arctan x$ | $\dfrac{1}{1+x^2}$ | all real $x$ |
$\operatorname{arccot} x$ | $-\dfrac{1}{1+x^2}$ | all real $x$ |
$\operatorname{arcsec} x$ | $\dfrac{1}{ | x |
$\operatorname{arccsc} x$ | $-\dfrac{1}{ | x |
Read the table in three pairs, not six separate lines. Sine and cosine share $\sqrt{1-x^2}$; tangent and cotangent share $1+x^2$; secant and cosecant share $|x|\sqrt{x^2-1}$. In each pair the "co-" function simply carries the minus sign.
Memorising three denominators and one sign rule replaces memorising six formulas. For the fuller trig-derivative picture, see differentiation of trigonometric functions.
Why Do The Co-Function Pairs Differ Only By A Sign?
The mirrored pairs are not a coincidence of the algebra. They come from a single identity you already know from trigonometry.
The complementary-angle identity. For every valid $x$, $\arcsin x + \arccos x = \tfrac{\pi}{2}$. The two angles always add to a right angle, because sine and cosine are co-functions.
Differentiate the identity. The right side, $\tfrac{\pi}{2}$, is a constant, and the derivative of a constant is $0$. So $\frac{d}{dx}\arcsin x + \frac{d}{dx}\arccos x = 0$, which forces $\frac{d}{dx}\arccos x = -\frac{d}{dx}\arcsin x$.
The same holds for the other pairs. Because $\arctan x + \operatorname{arccot} x = \tfrac{\pi}{2}$ and $\operatorname{arcsec} x + \operatorname{arccsc} x = \tfrac{\pi}{2}$, differentiating each identity makes the cotangent and cosecant derivatives the exact negatives of the tangent and secant ones.
So the minus sign is doing real work: it records that as one co-function goes up, its partner comes down by the same amount, keeping the sum locked at a right angle. That is why you never memorise the "co-" formulas separately. You learn three and flip a sign. The $|x|$ in the secant and cosecant rows carries the other piece of intuition: the secant curve rises on both sides of its gap, so its slope must stay positive whether $x$ is positive or negative, and the absolute value is what guarantees it.
Who Discovered The Inverse Trig Derivative Formulas?
The story starts long before Newton and Leibniz, on the other side of the world. The power series that these derivatives generate were written down in India roughly two and a half centuries before calculus had a name in Europe.
Two later figures shaped the modern form:
James Gregory (1638–1675, Scotland) rediscovered the arctangent series in 1671, which is why it is often called the Madhava–Gregory series in the West.
John Herschel (1792–1871, England) introduced the $\sin^{-1}$ notation for the inverse function in 1813, the "$-1$" superscript that students still meet today.
Where Are Derivatives Of Inverse Trigonometric Functions Used In The Real World?
Any time an angle depends on a changing distance, an inverse-trig derivative measures how fast that angle turns.
Related rates and tracking: a camera or radar following a moving object uses $\theta = \arctan(x/d)$, and the turning rate $\frac{d\theta}{dt}$ comes straight from the arctan derivative. It is largest when the object is directly ahead.
Robotics and computer vision: a robot arm or a self-driving car recovers joint angles and headings from position data by inverting trig relationships, and the sensitivity of those angles to small position errors is exactly these derivatives.
Physics and optics: the angle of a swinging pendulum, a refracted light ray, or a projectile's launch direction is often expressed as an inverse trig function, so its rate of change is one of these formulas.
Signal processing: the instantaneous phase of a wave is an arctangent of two signal components, and its derivative gives instantaneous frequency, the basis of FM radio and many audio tools.
Navigation and surveying: bearings computed as inverse tangents of coordinate differences change at rates these derivatives describe as a vehicle moves.
One small family of formulas connects a camera on a street, a robot's elbow, and a radio signal. The common thread is always the same: a rate of turning, read off from a changing length.
What Are The Most Common Mistakes With Derivatives Of Inverse Trigonometric Functions?
These three errors account for most lost marks, and each is confirmed by real student questions on Cuemath's FAQ, Physics Forums, and Math LibreTexts.
Dropping the chain-rule inner derivative.
Where it slips in:
Asked for $\frac{d}{dx}\arctan(x^2)$, a student writes $\frac{1}{1+x^4}$ and stops, using the bare formula as if the input were just $x$.
Don't do this:
Do not apply $\frac{1}{1+u^2}$ without multiplying by $u'$. The bare formula only works when the inside is exactly $x$.
The correct way:
Identify $u$, compute $u'$, and attach it: $\frac{d}{dx}\arctan(x^2) = \frac{2x}{1+x^4}$. The inner derivative $u' = 2x$ is part of the answer.
Sign confusion between a co-function pair.
Where it slips in:
A student gives $\frac{d}{dx}\arccos x = \frac{1}{\sqrt{1-x^2}}$, copying the arcsin result and forgetting the minus sign, or does the same with $\operatorname{arccot}$ and $\operatorname{arccsc}$.
Don't do this:
Do not assume the "co-" function has the same-sign derivative as its partner. The two always differ by a sign.
The correct way:
Remember the pairing rule: $\arccos$, $\operatorname{arccot}$, and $\operatorname{arccsc}$ each carry a minus sign, because each co-function decreases where its partner increases. When in doubt, check the graph's slope direction.
Forgetting the absolute value in arcsec and arccsc.
Where it slips in:
A student writes $\frac{d}{dx}\operatorname{arcsec} x = \frac{1}{x\sqrt{x^2-1}}$ without the bars, then gets a negative slope for negative $x$, which contradicts the graph.
Don't do this:
Do not drop the $|x|$. Without it the formula reports the wrong sign on the left branch, where $x < -1$.
The correct way:
Keep the absolute value: $\frac{d}{dx}\operatorname{arcsec} x = \frac{1}{|x|\sqrt{x^2-1}}$. The secant curve rises on both sides of its gap, so the derivative must stay positive for every valid $x$, and the $|x|$ enforces that.
Practice Problems On Derivatives Of Inverse Trigonometric Functions
Differentiate each function. Answers, with the key step, follow every problem.
$\frac{d}{dx}\arcsin\left(\tfrac{x}{2}\right)$.
(Answer: $u=\tfrac{x}{2}$, $u'=\tfrac{1}{2}$, so $\frac{1/2}{\sqrt{1-x^2/4}} = \dfrac{1}{\sqrt{4-x^2}}$.)$\frac{d}{dx}\arctan(5x)$.
(Answer: $u=5x$, $u'=5$, so $\dfrac{5}{1+25x^2}$.)$\frac{d}{dx}\arccos(2x)$.
(Answer: $u=2x$, $u'=2$, and arccos carries a minus sign, so $-\dfrac{2}{\sqrt{1-4x^2}}$.)$\frac{d}{dx}\left[x^2\arctan x\right]$.
(Answer: product rule, $2x\arctan x + \dfrac{x^2}{1+x^2}$.)$\frac{d}{dx}\operatorname{arccot}(x^2)$.
(Answer: $u=x^2$, $u'=2x$, and arccot carries a minus sign, so $-\dfrac{2x}{1+x^4}$.)$\frac{d}{dx}\operatorname{arcsec}(x^2)$ for $x>1$.
(Answer: $u=x^2$, $u'=2x$, so $\dfrac{2x}{x^2\sqrt{x^4-1}} = \dfrac{2}{x\sqrt{x^4-1}}$.)
Where Should You Go Next After Derivatives Of Inverse Trigonometric Functions?
Each door below builds directly on what you just learned.
Inverse trigonometric functions. Firm up the functions themselves, their domains, and their principal branches, the ground every derivative here stands on.
Derivative of tan inverse x. A focused walkthrough of the arctan case, the most common one on exams.
Derivative of cos inverse. The arccos derivation in full, including where its minus sign comes from.
The derivative, as a general idea. Step back to the definition of a derivative and see how slope and rate connect across all of calculus.
If your child is meeting these formulas for the first time, a live Bhanzu trainer teaches inverse-trig derivatives starting from the "why" of implicit differentiation in the Bhanzu math classes.
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