What is the derivative of arcsin x?

The derivative of arcsin x is 1/√(1 − x²) for |x| < 1. It is derived by setting y = arcsin x, so sin y = x, then applying implicit differentiation. The derivative of arccos x is the negative of this, −1/√(1 − x²).

What is the derivative of arctan x?

The derivative of arctan x is 1/(1 + x²), valid for all real x. This is derived by implicit differentiation of tan y = x, using the identity sec²y = 1 + tan²y. The result appears frequently in integration as the antiderivative of 1/(1 + x²).

How do inverse trigonometric derivatives relate to each other?

The six inverse trig derivatives form three pairs (arcsin/arccos, arctan/arccot, arcsec/arccsc), and within each pair the derivatives are negatives of each other. This follows from each pair summing to π/2, so their derivatives sum to zero.

What are the derivatives of hyperbolic functions?

The hyperbolic derivatives mirror the trigonometric ones: (sinh x)' = cosh x, (cosh x)' = sinh x, and (tanh x)' = sech²x. The key difference is that (cosh x)' = sinh x has no negative sign, whereas (cos x)' = −sin x does.

How do you differentiate a piecewise function at a boundary?

At a boundary point where the formula changes, the derivative exists only if the function is continuous there and the left-hand and right-hand derivatives are equal. If either condition fails, the function is not differentiable at that point.

Special Derivatives

Q: What is the derivative of arctan x?

The derivative of arctan x is 1/(1 + x²), valid for all real x. This is derived by implicit differentiation of tan y = x, using the identity sec²y = 1 + tan²y. The result appears frequently in integration as the antiderivative of 1/(1 + x²).

Q: How do inverse trigonometric derivatives relate to each other?

The six inverse trig derivatives form three pairs (arcsin/arccos, arctan/arccot, arcsec/arccsc), and within each pair the derivatives are negatives of each other. This follows from each pair summing to π/2, so their derivatives sum to zero.

Q: What are the derivatives of hyperbolic functions?

The hyperbolic derivatives mirror the trigonometric ones: (sinh x)' = cosh x, (cosh x)' = sinh x, and (tanh x)' = sech²x. The key difference is that (cosh x)' = sinh x has no negative sign, whereas (cos x)' = −sin x does.

Q: How do you differentiate a piecewise function at a boundary?

At a boundary point where the formula changes, the derivative exists only if the function is continuous there and the left-hand and right-hand derivatives are equal. If either condition fails, the function is not differentiable at that point.

Inverse Trigonometric Functions — Arcsine and Arccosine

Inverse Trigonometric Functions — Arctangent and Arccotangent

Inverse Trigonometric Functions — Arcsecant and Arccosecant

Patterns in Inverse Trigonometric Derivatives

Hyperbolic Functions

Inverse Hyperbolic Functions

Piecewise Functions

Summary of Special Derivatives

Derivatives That Require Deeper Techniques

The common derivatives cover polynomials, trigonometric functions, exponentials, and logarithms. Beyond these lie function families whose derivatives are less immediately obvious and harder to rederive on the spot. Inverse trigonometric functions produce algebraic expressions involving square roots. Hyperbolic functions mirror their trigonometric counterparts with subtle sign differences. Piecewise functions demand separate treatment on each interval with careful attention at boundaries.

Most of these derivatives are established through implicit differentiation or the inverse function formula rather than direct computation from the limit definition. Understanding the derivation builds insight; memorizing the result builds speed. Both matter.

Inverse Trigonometric Functions — Arcsine and Arccosine

The derivatives of

\arcsin x

and

\arccos x

are:

\frac{d}{dx}[\arcsin x] = \frac{1}{\sqrt{1 - x^2}} \qquad |x| < 1

\frac{d}{dx}[\arccos x] = -\frac{1}{\sqrt{1 - x^2}} \qquad |x| < 1

The two derivatives differ only by sign. This follows from the identity

\arcsin x + \arccos x = \frac{\pi}{2}

: differentiating both sides gives

(\arcsin x)' + (\arccos x)' = 0

.

To derive the arcsine formula: let

y = \arcsin x

, so

\sin y = x

with

y \in [-\pi/2, \pi/2]

. Differentiating implicitly:

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

. Since

y

lies in the first or fourth quadrant,

\cos y \geq 0

, so

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

. Solving gives

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

.

The domain restriction

|x| < 1

reflects the range of

\sin y

[-\pi/2, \pi/2]

. At

x = \pm 1

, the denominator vanishes—the graph of

\arcsin x

has vertical tangents at its endpoints.

Inverse Trigonometric Functions — Arctangent and Arccotangent

\frac{d}{dx}[\arctan x] = \frac{1}{1 + x^2} \qquad \qquad \frac{d}{dx}[\text{arccot}\, x] = -\frac{1}{1 + x^2}

Again the derivatives are negatives of each other, following from

\arctan x + \text{arccot}\, x = \frac{\pi}{2}

.

To derive the arctangent formula: let

y = \arctan x

, so

\tan y = x

. Differentiating:

\sec^2 y \cdot \frac{dy}{dx} = 1

. Using the identity

\sec^2 y = 1 + \tan^2 y = 1 + x^2

gives

\frac{dy}{dx} = \frac{1}{1 + x^2}

.

Unlike arcsine and arccosine, arctangent is defined for all real

x

—no domain restriction. The derivative

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

is always positive, confirming that

\arctan x

is strictly increasing. As

x \to \pm\infty

, the derivative approaches zero, reflecting the horizontal asymptotes

y = \pm\frac{\pi}{2}

. The function rises everywhere but flattens out at the extremes.

This derivative appears prominently in integration:

\int \frac{1}{1 + x^2}\,dx = \arctan x + C

Inverse Trigonometric Functions — Arcsecant and Arccosecant

\frac{d}{dx}[\text{arcsec}\, x] = \frac{1}{|x|\sqrt{x^2 - 1}} \qquad |x| > 1

\frac{d}{dx}[\text{arccsc}\, x] = -\frac{1}{|x|\sqrt{x^2 - 1}} \qquad |x| > 1

The derivation for arcsecant: let

y = \text{arcsec}\, x

, so

\sec y = x

. Differentiating:

\sec y \tan y \cdot \frac{dy}{dx} = 1

. Since

\sec y = x

and

\tan y = \pm\sqrt{\sec^2 y - 1} = \pm\sqrt{x^2 - 1}

, the sign depends on the quadrant. The absolute value

|x|

in the formula resolves this:

\sec y \tan y = |x|\sqrt{x^2 - 1}

across the full domain.

These derivatives are less frequently encountered than the arcsine and arctangent versions. They appear primarily in integrals of the form

\int \frac{1}{x\sqrt{x^2 - 1}}\,dx

and in problems involving trigonometric substitution.

The domain

|x| > 1

reflects the range of secant:

\sec y

never takes values between

-1

and

1

Patterns in Inverse Trigonometric Derivatives

The six inverse trigonometric derivatives organize into three pairs, each pair summing to zero:

(\arcsin x)' + (\arccos x)' = 0

, both involving

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

(\arctan x)' + (\text{arccot}\, x)' = 0

, both involving

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

(\text{arcsec}\, x)' + (\text{arccsc}\, x)' = 0

, both involving

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

Each pair shares the same magnitude because the two functions in each pair sum to a constant (

\frac{\pi}{2}

). The cofunction always carries the negative sign—the same pattern seen in the common trigonometric derivatives where

\cos

\cot

, and

\csc

derivatives are negative.

Two distinct algebraic forms appear:

\sqrt{1 - x^2}

for the sine/cosine pair,

1 + x^2

for the tangent/cotangent pair, and

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

for the secant/cosecant pair. These three expressions reappear in integration, where recognizing them triggers the corresponding inverse trigonometric antiderivative.

Hyperbolic Functions

The hyperbolic functions are defined through exponentials:

\sinh x = \frac{e^x - e^{-x}}{2} \qquad \cosh x = \frac{e^x + e^{-x}}{2} \qquad \tanh x = \frac{\sinh x}{\cosh x}

Their derivatives parallel the trigonometric pattern:

\frac{d}{dx}[\sinh x] = \cosh x \qquad \qquad \frac{d}{dx}[\cosh x] = \sinh x

\frac{d}{dx}[\tanh x] = \text{sech}^2\, x \qquad \qquad \frac{d}{dx}[\coth x] = -\text{csch}^2\, x

\frac{d}{dx}[\text{sech}\, x] = -\text{sech}\, x \tanh x \qquad \qquad \frac{d}{dx}[\text{csch}\, x] = -\text{csch}\, x \coth x

The critical difference from trigonometric derivatives:

(\cosh x)' = \sinh x

carries no negative sign, whereas

(\cos x)' = -\sin x

does. This single sign difference propagates through all six formulas.

Each derivative is verified directly by differentiating the exponential definitions. For

\sinh x

\frac{d}{dx}\left[\frac{e^x - e^{-x}}{2}\right] = \frac{e^x + e^{-x}}{2} = \cosh x

. No special limits or implicit differentiation are needed—the exponential derivative does all the work.

Inverse Hyperbolic Functions

The inverse hyperbolic functions have derivatives that resemble the inverse trigonometric ones with sign changes under the radical:

\frac{d}{dx}[\text{arcsinh}\, x] = \frac{1}{\sqrt{x^2 + 1}} \qquad \qquad \frac{d}{dx}[\text{arccosh}\, x] = \frac{1}{\sqrt{x^2 - 1}}, \quad x > 1

\frac{d}{dx}[\text{arctanh}\, x] = \frac{1}{1 - x^2}, \quad |x| < 1

Compare with inverse trigonometric derivatives:

\sqrt{1 - x^2}

becomes

\sqrt{x^2 + 1}

\sqrt{x^2 - 1}

, and

1 + x^2

becomes

1 - x^2

. The signs inside the expressions flip.

These derivatives can be established through implicit differentiation or through the logarithmic representations. For instance,

\text{arcsinh}\, x = \ln(x + \sqrt{x^2 + 1})

. Differentiating this logarithmic form directly using the chain rule confirms

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

.

Inverse hyperbolic derivatives appear in integration:

\int \frac{1}{\sqrt{x^2 + 1}}\,dx = \text{arcsinh}\, x + C

and

\int \frac{1}{\sqrt{x^2 - 1}}\,dx = \text{arccosh}\, x + C

provide alternatives to trigonometric substitution for certain radical integrals.

Piecewise Functions

A piecewise function uses different formulas on different intervals. On each interval's interior, standard differentiation rules apply to the governing formula. The derivative of the piecewise function is itself piecewise—computed piece by piece.

The challenge lies at the boundaries. At a point

x = c

where the formula changes, the derivative exists only if three conditions hold: the function is continuous at

c

(the pieces connect), the left-hand derivative exists, and the right-hand derivative exists and equals the left-hand derivative.

Consider

f(x) = x^2

for

x \leq 1

and

f(x) = 2x - 1

for

x > 1

. Continuity at

x = 1

f(1) = 1

from the left,

2(1) - 1 = 1

from the right—they match. Left derivative at

1

\frac{d}{dx}[x^2]\big|_{x=1} = 2

. Right derivative at

1

\frac{d}{dx}[2x - 1]\big|_{x=1} = 2

. Both derivatives match, so

f'(1) = 2

.

If either continuity fails or the one-sided derivatives disagree, the function is not differentiable at that boundary. The absolute value function

|x|

is the canonical example: continuous at

x = 0

, but left derivative

-1

and right derivative

+1

do not match.

Summary of Special Derivatives

The complete set of special derivative formulas:

\frac{d}{dx}[\arcsin x] = \frac{1}{\sqrt{1-x^2}} \qquad \frac{d}{dx}[\arccos x] = -\frac{1}{\sqrt{1-x^2}}

\frac{d}{dx}[\arctan x] = \frac{1}{1+x^2} \qquad \frac{d}{dx}[\text{arccot}\, x] = -\frac{1}{1+x^2}

\frac{d}{dx}[\text{arcsec}\, x] = \frac{1}{|x|\sqrt{x^2-1}} \qquad \frac{d}{dx}[\text{arccsc}\, x] = -\frac{1}{|x|\sqrt{x^2-1}}

\frac{d}{dx}[\sinh x] = \cosh x \qquad \frac{d}{dx}[\cosh x] = \sinh x

\frac{d}{dx}[\tanh x] = \text{sech}^2\, x \qquad \frac{d}{dx}[\coth x] = -\text{csch}^2\, x

\frac{d}{dx}[\text{sech}\, x] = -\text{sech}\, x\tanh x \qquad \frac{d}{dx}[\text{csch}\, x] = -\text{csch}\, x\coth x

\frac{d}{dx}[\text{arcsinh}\, x] = \frac{1}{\sqrt{x^2+1}} \qquad \frac{d}{dx}[\text{arccosh}\, x] = \frac{1}{\sqrt{x^2-1}}

\frac{d}{dx}[\text{arctanh}\, x] = \frac{1}{1-x^2}

Together with the common derivatives, these formulas cover every standard function encountered in calculus. Recognizing the algebraic forms—

\sqrt{1-x^2}

1+x^2

\sqrt{x^2 \pm 1}

1-x^2

—is equally important for integration, where these expressions signal inverse trigonometric or inverse hyperbolic antiderivatives.