In short

The derivative of \sin^{-1}x is \dfrac{1}{\sqrt{1-x^2}}, the derivative of \cos^{-1}x is \dfrac{-1}{\sqrt{1-x^2}}, and the derivative of \tan^{-1}x is \dfrac{1}{1+x^2}. Each one is derived by differentiating both sides of an identity like \sin(\sin^{-1}x) = x and solving for the unknown derivative.

Here is a puzzle. The function \sin^{-1}x — the arc sine — takes an input between -1 and 1 and returns an angle between -\pi/2 and \pi/2. Plot it, and you see a curve that starts at (-1, -\pi/2) and ends at (1, \pi/2), passing through the origin.

Look at the curve more carefully. At x = 0, it looks like a straight line of slope 1 — gentle, well-behaved. But as x approaches \pm 1, something dramatic happens: the curve gets steeper and steeper, becoming nearly vertical at the endpoints. The slope at x = 0.9 is much larger than the slope at x = 0, and the slope at x = 0.99 is larger still.

So what is the slope at each point? The answer — the derivative of \sin^{-1}x — turns out to be \frac{1}{\sqrt{1 - x^2}}. At x = 0, this is \frac{1}{\sqrt{1}} = 1. At x = 0.9, it is \frac{1}{\sqrt{1 - 0.81}} = \frac{1}{\sqrt{0.19}} \approx 2.29. At x = 0.99, it is \frac{1}{\sqrt{1 - 0.9801}} = \frac{1}{\sqrt{0.0199}} \approx 7.09. At x = 0.999, it is about 22.4. The slope grows without bound as x approaches \pm 1 — matching the picture of a curve that becomes nearly vertical at its endpoints.

The formula \frac{1}{\sqrt{1-x^2}} is not something you would guess. It has to be derived. And the derivation uses one beautiful trick: differentiate both sides of the identity \sin(\sin^{-1}x) = x.

The core technique: implicit differentiation of an identity

Every inverse trig derivative uses the same three-step approach, so let's see the technique clearly before applying it six times.

Step A: Rewrite. Start with y = \sin^{-1}x. By the definition of the inverse function, this means \sin y = x, with the restriction y \in [-\pi/2, \pi/2].

Step B: Differentiate both sides. Differentiate the equation \sin y = x with respect to x. On the left, y is a function of x, so the chain rule applies:

\frac{d}{dx}(\sin y) = \cos y \cdot \frac{dy}{dx}

On the right:

\frac{d}{dx}(x) = 1

Set them equal and solve for \frac{dy}{dx}:

\cos y \cdot \frac{dy}{dx} = 1 \quad \Longrightarrow \quad \frac{dy}{dx} = \frac{1}{\cos y}

Step C: Convert back to x. The answer should be a function of x, not y. Since \sin y = x and y \in [-\pi/2, \pi/2] (where cosine is non-negative):

\cos y = +\sqrt{1 - \sin^2 y} = \sqrt{1 - x^2}

The positive square root is chosen because \cos y \geq 0 when y \in [-\pi/2, \pi/2]. This is where the domain restriction matters — a different branch of the inverse sine would give a different sign.

So:

\frac{d}{dx}\sin^{-1}x = \frac{1}{\sqrt{1 - x^2}}

That is the complete derivation. Three moves: rewrite the inverse function as an equation, differentiate both sides using the chain rule, and convert back to x using a Pythagorean identity. Every inverse trig derivative follows this exact template — only the trig function and the identity change.

Deriving all six inverse trig derivatives

Derivative of \sin^{-1}x

Already derived above:

\frac{d}{dx}\sin^{-1}x = \frac{1}{\sqrt{1 - x^2}}, \qquad -1 < x < 1

The domain restriction |x| < 1 is essential. At x = \pm 1, the denominator \sqrt{1-x^2} equals zero and the derivative does not exist. Geometrically, the curve y = \sin^{-1}x has a vertical tangent at these endpoints — the slope is infinite.

Derivative of \cos^{-1}x

Let y = \cos^{-1}x, so \cos y = x with y \in [0, \pi].

Differentiate both sides with respect to x:

\frac{d}{dx}(\cos y) = -\sin y \cdot \frac{dy}{dx} = 1

Solve: \frac{dy}{dx} = \frac{-1}{\sin y}.

Convert to x: since \cos y = x and y \in [0, \pi] (where sine is non-negative):

\sin y = +\sqrt{1 - \cos^2 y} = \sqrt{1 - x^2}
\frac{d}{dx}\cos^{-1}x = \frac{-1}{\sqrt{1 - x^2}}, \qquad -1 < x < 1

This is exactly the negative of the derivative of \sin^{-1}x. This is no coincidence. There is a standard identity: \sin^{-1}x + \cos^{-1}x = \frac{\pi}{2} for all x \in [-1, 1]. Differentiate both sides: \frac{d}{dx}\sin^{-1}x + \frac{d}{dx}\cos^{-1}x = 0. So the two derivatives must be negatives of each other. The identity gives you the second derivative for free once you know the first.

Derivative of \tan^{-1}x

Let y = \tan^{-1}x, so \tan y = x with y \in (-\pi/2, \pi/2).

Differentiate both sides with respect to x:

\frac{d}{dx}(\tan y) = \sec^2 y \cdot \frac{dy}{dx} = 1

Solve: \frac{dy}{dx} = \frac{1}{\sec^2 y}.

Convert to x using the identity \sec^2 y = 1 + \tan^2 y:

\frac{dy}{dx} = \frac{1}{1 + \tan^2 y} = \frac{1}{1 + x^2}
\frac{d}{dx}\tan^{-1}x = \frac{1}{1 + x^2}

This formula is defined for all real x — no domain restrictions. The denominator 1 + x^2 is always at least 1, so the derivative is always between 0 and 1. At x = 0, the derivative is 1 — maximum slope. As x \to \pm\infty, the derivative approaches 0 — the curve flattens out. This matches the shape of \tan^{-1}x, which has horizontal asymptotes at y = \pm\pi/2: the curve gets flatter and flatter as it approaches these lines, but never reaches them.

The derivative \frac{1}{1+x^2} is itself a beautiful bell-shaped curve (the Cauchy distribution in probability). It peaks at 1 when x = 0 and decays symmetrically on both sides. The area under this curve from -\infty to \infty is \pi — a fact that follows directly from \tan^{-1}(\infty) - \tan^{-1}(-\infty) = \pi/2 - (-\pi/2) = \pi.

Derivative of \cot^{-1}x

Let y = \cot^{-1}x, so \cot y = x with y \in (0, \pi).

Differentiate: -\csc^2 y \cdot \frac{dy}{dx} = 1, so \frac{dy}{dx} = \frac{-1}{\csc^2 y}.

Use \csc^2 y = 1 + \cot^2 y:

\frac{d}{dx}\cot^{-1}x = \frac{-1}{1 + \cot^2 y} = \frac{-1}{1 + x^2}

This is the negative of the \tan^{-1}x derivative, which follows from the identity \tan^{-1}x + \cot^{-1}x = \frac{\pi}{2}.

The solid curve is $y = \tan^{-1}x$ and the faint dashed curve is $y = -\tan^{-1}x$ (which equals $\cot^{-1}x - \pi/2$). The horizontal asymptotes at $y = \pm\pi/2$ show why the derivative of $\tan^{-1}x$ approaches zero for large $|x|$: the function flattens out as it approaches its limits.

Derivative of \sec^{-1}x

Let y = \sec^{-1}x, so \sec y = x with y \in [0, \pi], y \neq \pi/2.

Differentiate: \sec y \tan y \cdot \frac{dy}{dx} = 1, so \frac{dy}{dx} = \frac{1}{\sec y \tan y}.

Now \sec y = x. For \tan y: since \sec^2 y = 1 + \tan^2 y, you get \tan^2 y = x^2 - 1, so |\tan y| = \sqrt{x^2 - 1}.

The product \sec y \cdot \tan y can be positive or negative depending on the quadrant. For y \in [0, \pi/2), both \sec y and \tan y are positive. For y \in (\pi/2, \pi], \sec y < 0 and \tan y < 0, so their product is positive. In both cases, \sec y \cdot \tan y has the same sign as |x| \cdot \sqrt{x^2 - 1}/|x| \cdot |x|. Working through the signs carefully:

\frac{d}{dx}\sec^{-1}x = \frac{1}{|x|\sqrt{x^2 - 1}}, \qquad |x| > 1

The domain requires |x| > 1 because \sec y is always at least 1 in absolute value, and at |x| = 1 the denominator is zero (vertical tangent).

Derivative of \csc^{-1}x

Let y = \csc^{-1}x, so \csc y = x with y \in [-\pi/2, \pi/2], y \neq 0.

Differentiate: -\csc y \cot y \cdot \frac{dy}{dx} = 1, so \frac{dy}{dx} = \frac{-1}{\csc y \cot y}.

By the same analysis as for \sec^{-1}x:

\frac{d}{dx}\csc^{-1}x = \frac{-1}{|x|\sqrt{x^2 - 1}}, \qquad |x| > 1

The minus sign matches the "co-" pattern: \sec^{-1}x + \csc^{-1}x = \frac{\pi}{2}, so their derivatives are negatives of each other.

The complete table

Function Derivative Domain
\sin^{-1}x \dfrac{1}{\sqrt{1-x^2}} -1 < x < 1
\cos^{-1}x \dfrac{-1}{\sqrt{1-x^2}} -1 < x < 1
\tan^{-1}x \dfrac{1}{1+x^2} all x
\cot^{-1}x \dfrac{-1}{1+x^2} all x
\sec^{-1}x \dfrac{1}{\lvert x\rvert\sqrt{x^2-1}} \lvert x\rvert > 1
\csc^{-1}x \dfrac{-1}{\lvert x\rvert\sqrt{x^2-1}} \lvert x\rvert > 1

Three patterns to notice:

  1. Sign pattern. The "co-" functions (\cos^{-1}, \cot^{-1}, \csc^{-1}) all have a minus sign. The others do not. This mirrors the sign pattern for the regular trig derivatives.

  2. Paired derivatives. The six derivatives come in three pairs that sum to zero: \sin^{-1} and \cos^{-1}; \tan^{-1} and \cot^{-1}; \sec^{-1} and \csc^{-1}. Each pair sums to a constant (\pi/2), so their derivatives must be negatives of each other.

  3. Three denominators. There are only three distinct denominators: \sqrt{1-x^2}, 1+x^2, and |x|\sqrt{x^2-1}. Each appears twice (once with a plus sign, once with a minus). This means you only need to remember three formulas, not six.

Why domain restrictions matter

The inverse trig functions exist because you restrict the domain of the original trig functions. Without restrictions, \sin x is not one-to-one — infinitely many angles have the same sine value — so no inverse exists. The principal branch is a deliberate choice of one interval where the function is one-to-one.

For \sin^{-1}x: the chosen range is y \in [-\pi/2, \pi/2]. In this interval, \cos y \geq 0, which is why the derivation used the positive square root: \cos y = +\sqrt{1-x^2}. If someone chose a different branch — say y \in [\pi/2, 3\pi/2] — then \cos y would be non-positive, the sign of the square root would flip, and the derivative formula would be \frac{-1}{\sqrt{1-x^2}} instead of \frac{+1}{\sqrt{1-x^2}}.

For \tan^{-1}x: the restriction y \in (-\pi/2, \pi/2) ensures \sec^2 y > 0, which means the chain rule step \sec^2 y \cdot \frac{dy}{dx} = 1 has a well-defined, positive denominator. The derivation works out cleanly.

For \sec^{-1}x: the absolute value |x| in the denominator comes from the fact that y can be in either the first quadrant (0 \leq y < \pi/2, where x = \sec y > 1) or the second quadrant (\pi/2 < y \leq \pi, where x = \sec y < -1). The product \sec y \tan y turns out positive in both cases, but its magnitude is |x|\sqrt{x^2-1}. The absolute value handles both signs of x in a single formula.

The solid black curve is $y = \sin^{-1}x$, defined on $[-1, 1]$. The dashed red curve is its derivative $\frac{1}{\sqrt{1-x^2}}$. At $x = 0$, the derivative is $1$: the arc sine curve has the same slope as the line $y = x$. At $x = \sqrt{3}/2 \approx 0.866$, the derivative is $\frac{1}{\sqrt{1-3/4}} = \frac{1}{\sqrt{1/4}} = 2$ — already twice as steep. As $x \to \pm 1$, the derivative goes to infinity.
The solid black curve is $y = \tan^{-1}x$ and the dashed red curve is its derivative $\frac{1}{1+x^2}$. The function approaches the horizontal asymptotes $y = \pm\pi/2$ as $x \to \pm\infty$, and correspondingly the derivative approaches $0$. The maximum slope is $1$, at the origin. The derivative curve has a bell shape — it peaks at $x = 0$ and decays on both sides.

Worked examples

Example 1: Derivative involving sin⁻¹ and the chain rule

Find \dfrac{d}{dx}\sin^{-1}(2x).

Step 1. Identify the outer and inner functions.

\text{Outer: } f(u) = \sin^{-1}u, \qquad \text{Inner: } u = g(x) = 2x

Why: the arc sine is applied to 2x, not to x directly, so this is a composite function and the chain rule applies.

Step 2. Differentiate the outer function.

f'(u) = \frac{1}{\sqrt{1 - u^2}}

Why: this is the derivative of \sin^{-1}u derived earlier in this article.

Step 3. Evaluate the outer derivative at the inner function.

f'(g(x)) = f'(2x) = \frac{1}{\sqrt{1 - (2x)^2}} = \frac{1}{\sqrt{1 - 4x^2}}

Why: replace u with 2x, so u^2 = 4x^2.

Step 4. Differentiate the inner function.

g'(x) = \frac{d}{dx}(2x) = 2

Why: the derivative of 2x with respect to x is just 2.

Step 5. Apply the chain rule: multiply the two pieces.

\frac{d}{dx}\sin^{-1}(2x) = \frac{1}{\sqrt{1 - 4x^2}} \cdot 2 = \frac{2}{\sqrt{1 - 4x^2}}

Result: \dfrac{d}{dx}\sin^{-1}(2x) = \dfrac{2}{\sqrt{1 - 4x^2}}, valid for |x| < \frac{1}{2}.

The solid black curve is $y = \sin^{-1}(2x)$ and the faint dashed curve is $y = \sin^{-1}(x)$ for comparison. The chain rule doubles the slope everywhere: at $x = 0$, $\sin^{-1}(x)$ has slope $1$ but $\sin^{-1}(2x)$ has slope $2$. The compressed curve reaches $\pm\pi/2$ at $x = \pm 1/2$ instead of $x = \pm 1$, making it steeper throughout its domain.

The domain is |x| < 1/2 because \sin^{-1}(2x) is only defined when |2x| \leq 1, i.e., |x| \leq 1/2. The derivative is undefined at the endpoints x = \pm 1/2 where the curve has vertical tangents. Doubling the argument compressed the entire arc sine curve into half the width, making it steeper everywhere — and the chain rule factor of 2 captures exactly this effect.

Example 2: Using tan⁻¹ in a geometric setting

A building of height 50 metres stands at a horizontal distance d metres from where you are standing. The angle of elevation to the top of the building is \theta = \tan^{-1}(50/d). Find \frac{d\theta}{dd} — the rate at which the angle changes per metre you walk further away — when d = 50 metres.

Step 1. Write the function and identify the composition.

\theta(d) = \tan^{-1}\!\left(\frac{50}{d}\right)

The outer function is \tan^{-1}u, and the inner function is u = 50/d = 50d^{-1}.

Why: from basic trigonometry, \tan\theta = \text{opposite}/\text{adjacent} = 50/d, so \theta = \tan^{-1}(50/d).

Step 2. Differentiate using the chain rule.

\frac{d\theta}{dd} = \frac{1}{1 + (50/d)^2} \cdot \frac{d}{dd}\!\left(\frac{50}{d}\right)

Why: the derivative of \tan^{-1}u is \frac{1}{1+u^2}, and the chain rule multiplies by the derivative of the inner function.

Step 3. Compute the derivative of the inner function.

\frac{d}{dd}\!\left(\frac{50}{d}\right) = \frac{d}{dd}(50d^{-1}) = 50 \cdot (-1) \cdot d^{-2} = -\frac{50}{d^2}

Why: the power rule gives \frac{d}{dd}(d^{-1}) = -d^{-2}, multiplied by the constant 50.

Step 4. Combine and simplify.

\frac{d\theta}{dd} = \frac{1}{1 + 2500/d^2} \cdot \left(-\frac{50}{d^2}\right)

The term 1 + 2500/d^2 = \frac{d^2 + 2500}{d^2}, so:

\frac{d\theta}{dd} = \frac{d^2}{d^2 + 2500} \cdot \left(-\frac{50}{d^2}\right) = \frac{-50}{d^2 + 2500}

Why: multiplying through by d^2/d^2 cleans up the compound fraction into a single clean fraction.

Step 5. Evaluate at d = 50.

\frac{d\theta}{dd}\bigg|_{d=50} = \frac{-50}{2500 + 2500} = \frac{-50}{5000} = -\frac{1}{100} = -0.01 \text{ rad/m}

Result: At a distance of 50 metres, the elevation angle decreases by 0.01 radians (about 0.57°) for each additional metre you walk away from the building.

The elevation angle $\theta = \tan^{-1}(50/d)$ as a function of distance. At $d = 50$, the angle is exactly $\pi/4$ ($45°$) — the building height equals your distance. At $d = 25$, you are closer and looking up more steeply ($63.4°$). At $d = 100$, the building looks much smaller ($26.6°$). The derivative $-50/(d^2 + 2500)$ tells you the rate of decrease: at $d = 50$ it is $-0.01$ rad/m, but at $d = 100$ it is only $-0.004$ rad/m — walking further has diminishing effect.

The negative sign confirms what geometry tells you: walking further away makes the building look shorter (the angle decreases). The formula \frac{-50}{d^2 + 2500} also reveals that the rate of decrease is fastest when d is small (close to the building, where each step makes a big difference in angle) and slowest when d is large (far away, where each step barely matters). At d = 0, the rate would be -50/2500 = -0.02 rad/m; at d = 100, it is -50/12500 = -0.004 rad/m.

Common confusions

Going deeper

If you came here to learn the six inverse trig derivatives and see how each is derived, you have them — you can stop here. The rest is for readers who want to see the connection to integration, a useful simplification technique for JEE, and a surprising link to \pi.

These derivatives are integration formulas in disguise

Every derivative formula can be read backwards as an integration formula. The inverse trig derivatives give you three of the most important integrals in calculus — integrals that have no algebraic antiderivative:

\int \frac{1}{\sqrt{1-x^2}}\,dx = \sin^{-1}x + C
\int \frac{1}{1+x^2}\,dx = \tan^{-1}x + C
\int \frac{1}{|x|\sqrt{x^2-1}}\,dx = \sec^{-1}|x| + C

The integral \int \frac{1}{\sqrt{1-x^2}}\,dx looks algebraic — the integrand is a ratio of polynomials under a square root — but there is no algebraic function whose derivative is \frac{1}{\sqrt{1-x^2}}. The arc sine is the simplest function that works, and it is fundamentally transcendental (not expressible in terms of polynomials and roots). This is one reason inverse trig derivatives matter beyond differentiation: they are the gateway to an entire class of integrals.

Simplification using substitution: a JEE technique

A common technique in competitive exams is to simplify complicated expressions by recognising them as inverse trig functions in disguise.

Take the expression \sin^{-1}\!\left(\frac{2x}{1+x^2}\right). You want its derivative with respect to x.

You could apply the chain rule directly: differentiate \sin^{-1}u where u = \frac{2x}{1+x^2}, compute u' by the quotient rule, and divide by \sqrt{1-u^2}. This works, but the algebra is a mess.

The cleaner approach: substitute x = \tan\theta. Then:

\frac{2x}{1+x^2} = \frac{2\tan\theta}{1+\tan^2\theta} = \frac{2\tan\theta}{\sec^2\theta} = 2\sin\theta\cos\theta = \sin(2\theta)

So \sin^{-1}\!\left(\frac{2x}{1+x^2}\right) = \sin^{-1}(\sin 2\theta) = 2\theta (provided 2\theta \in [-\pi/2, \pi/2], i.e., |\theta| \leq \pi/4, i.e., |x| \leq 1).

Since \theta = \tan^{-1}x:

\sin^{-1}\!\left(\frac{2x}{1+x^2}\right) = 2\tan^{-1}x, \qquad |x| \leq 1

The derivative is simply:

\frac{d}{dx}\left(2\tan^{-1}x\right) = \frac{2}{1+x^2}

This is dramatically simpler than the direct chain-rule computation. The substitution x = \tan\theta transformed the complex expression inside the arc sine into a standard double-angle formula, which collapsed to a clean answer.

The general strategy: when you see \sin^{-1}, \cos^{-1}, or \tan^{-1} applied to an expression involving \frac{2x}{1+x^2}, \frac{1-x^2}{1+x^2}, or \frac{2x}{1-x^2}, try x = \tan\theta. These three expressions are exactly \sin 2\theta, \cos 2\theta, and \tan 2\theta under that substitution.

The derivative $\frac{1}{1+x^2}$ (solid black) compared with its polynomial approximations: $1 - x^2$ (dashed, two terms) and $1 - x^2 + x^4$ (dotted, three terms). Near $x = 0$, the approximations are excellent. Integrating these polynomials term by term produces the Madhava-Leibniz series for $\tan^{-1}x$.

The Madhava-Leibniz series for \pi

The formula \frac{d}{dx}\tan^{-1}x = \frac{1}{1+x^2} has a remarkable consequence that connects differentiation to the number \pi.

The function \frac{1}{1+x^2} can be expanded as a geometric series when |x| < 1:

\frac{1}{1+x^2} = 1 - x^2 + x^4 - x^6 + x^8 - \cdots

(This is \frac{1}{1-r} with r = -x^2.)

Since \frac{1}{1+x^2} is the derivative of \tan^{-1}x, integrating both sides from 0 to x gives:

\tan^{-1}x = x - \frac{x^3}{3} + \frac{x^5}{5} - \frac{x^7}{7} + \frac{x^9}{9} - \cdots

Now set x = 1. Since \tan^{-1}1 = \pi/4:

\frac{\pi}{4} = 1 - \frac{1}{3} + \frac{1}{5} - \frac{1}{7} + \frac{1}{9} - \cdots

This is the Madhava-Leibniz series — an infinite series whose sum is exactly \pi/4. It was discovered by Madhava of Sangamagrama in the 14th century, more than two hundred years before Leibniz and Gregory arrived at it independently. A formula for the derivative of an inverse trig function leads, through a few clean steps, to one of the most beautiful expressions for \pi in all of mathematics.

The series converges slowly (you need about 300 terms to get two decimal places of \pi), but Madhava also discovered acceleration techniques that made it practical. The deeper point is that a derivative formula you learned today connects directly to the circle constant \pi — a reminder that calculus and geometry are never far apart.

Where this leads next

With the inverse trig derivatives in hand, you can now handle a much wider class of problems: