Inverse functions and inverse trigonometric derivatives

An inverse swaps inputs and outputs, so its tangent slope is reciprocal at the corresponding point.

The idea

An inverse function reverses a function: if \(f(a)=b\), then \(f^{-1}(b)=a\). It can do this only when each output identifies one input—a one-to-one function, possibly after restricting its domain.

Swapping the coordinates reflects a graph across \(y=x\). A nonzero tangent slope becomes its reciprocal, because the horizontal and vertical changes exchange roles. Use the slope at the corresponding original input, not at the same numerical input.

Theorem

Derivative of an inverse

Suppose \(f\) is continuous and strictly monotone on an interval, differentiable at an interior point \(a\), and \(f^{\prime}(a)\ne0\). Writing \(b=f(a)\), its inverse is differentiable at \(b\), and

\[\bigl(f^{-1}\bigr)^{\prime}(b)=\frac{1}{f^{\prime}(a)}=\frac{1}{f^{\prime}(f^{-1}(b))}.\]

Strictly monotone means \(f(x_1)<f(x_2)\) whenever \(x_1<x_2\), or the reverse inequality throughout the interval; this ensures that the function is one-to-one. The nonzero derivative condition lets us take the reciprocal.

For the widget, use \(f(x)=x^2\) on \(x\ge0\) and \(f^{-1}(x)=\sqrt{x}\). A nearby point gives changes \(\Delta x\) and \(\Delta y\) along the curve. Reflection swaps these changes. The secant ratios approximate the derivatives; taking the limit gives the exact reciprocal relation.

Example 1

A derivative without solving for the inverse

Let \(f(x)=x^3+x\). Both the cubic and linear terms strictly increase as the input increases, so their sum is one-to-one and has an inverse. Its derivative is \(f^{\prime}(x)=3x^2+1\). At \(a=1\),

\[f(1)=2,\qquad f^{\prime}(1)=4.\]

The inverse point is \(b=2\), so

\[\bigl(f^{-1}\bigr)^{\prime}(2)=\frac{1}{f^{\prime}(1)}=\frac14.\]

We did not need a formula for \(f^{-1}\).

Inverse trigonometric functions

Sine and cosine repeat, so they are not one-to-one on the whole real line. Their principal branches are the chosen intervals on which we reverse them:

\(\arcsin x\) returns an angle in \([-\pi/2,\pi/2]\).
\(\arccos x\) returns an angle in \([0,\pi]\).
\(\arctan x\) returns an angle in \((-\pi/2,\pi/2)\).

The first two accept inputs in \([-1,1]\); arctangent accepts any real input. Angles are in radians.

Rules

Three common inverse trig derivatives

\[\frac{d}{dx}\arcsin x=\frac{1}{\sqrt{1-x^2}},\qquad -1<x<1.\] \[\frac{d}{dx}\arccos x=-\frac{1}{\sqrt{1-x^2}},\qquad -1<x<1.\] \[\frac{d}{dx}\arctan x=\frac{1}{1+x^2},\qquad x\in\mathbb R.\]

Arcsine rule: derivation (optional)

Let \(y=\arcsin x\) with \(-1<x<1\). Then \(\sin y=x\) and \(-\pi/2<y<\pi/2\). On this interval, \(\cos y>0\), so the identity \(\cos^2y=1-\sin^2y\) gives

\[\cos y=\sqrt{1-x^2}.\]

The inverse derivative theorem and the sine derivative therefore give

\[\frac{d}{dx}\arcsin x=\frac{1}{\cos y}=\frac{1}{\sqrt{1-x^2}}.\]

Arcsine and arccosine are defined at their endpoints, but have no finite derivative there. Their original trig functions have zero slope at the corresponding angles.

Example 2

An inverse trig function inside a composition

For \(p(x)=\arcsin(x/3)\), the inner input is \(x/3\). The chain rule gives

\[p^{\prime}(x)=\frac{1/3}{\sqrt{1-x^2/9}}=\frac{1}{\sqrt{9-x^2}},\qquad -3<x<3.\]

The function itself is defined on \([-3,3]\), but this derivative is finite only in the interior.

Example 3

A product with arctangent

For \(q(x)=x\arctan x\), use the product rule:

\[q^{\prime}(x)=\arctan x+\frac{x}{1+x^2}.\]

Both terms matter: one differentiates the factor \(x\), and the other differentiates arctangent. The result holds for every real input.

Quiz

Enable JavaScript for interactive answer checking.At a pair of corresponding points, the original graph has tangent slope \(-5\). How does the inverse tangent compare?

It is descending but less steep

It is ascending and less steep

It is descending and steeper

It is horizontal

Explanation

The inverse slope is \(-1/5\). The negative sign is preserved, while its magnitude is smaller. Reflection across \(y=x\) does not turn a decreasing branch into an increasing one.A thermometer calibration converts voltage \(V\) into temperature \(T(V)\). Near a reading, \(dT/dV=5\,{}^\circ\mathrm{C/V}\), and the inverse rule applies. What is \(dV/dT\) there?

\(5\,\mathrm{V}/{}^\circ\mathrm{C}\)

\(0.2\,\mathrm{V}/{}^\circ\mathrm{C}\)

\(-0.2\,\mathrm{V}/{}^\circ\mathrm{C}\)

Explanation

The inverse conversion has rate \(1/5=0.2\) volts per degree Celsius. The units reverse along with the slope.For \(f(x)=x^3\), does \((f^{-1})^{\prime}(0)\) exist as a finite number?

Yes: the inverse is continuous

Yes: its derivative is zero

No: its difference quotient grows without bound

Explanation

The inverse is \(g(y)=\sqrt[3]{y}\). At zero, \([g(h)-g(0)]/h=1/h^{2/3}\to+\infty\). The inverse is continuous but has a vertical tangent, not a finite derivative.What is the sign of the derivative of \(y=\arccos(2x)\) for \(-1/2<x<1/2\)?

Positive

Negative

It changes sign at zero

Explanation

The chain rule gives \(y^{\prime}=-2/\sqrt{1-4x^2}\). The denominator is positive throughout that interval, so the function is decreasing.A student claims \(\arcsin(\sin\theta)=\theta\) for every real \(\theta\). What does the left side equal when \(\theta=5\pi/6\)?

\(5\pi/6\)

\(\pi/6\)

\(-\pi/6\)

It is undefined

Explanation

Here \(\sin(5\pi/6)=1/2\), but arcsine returns the angle in \([-\pi/2,\pi/2]\). That angle is \(\pi/6\). The branch restriction matters.You need the slope of \(g(x)=\arctan x\) at \(x=2\). Which information from the original tangent function is relevant?

Its slope at angle \(2\)

Its slope at angle \(\arctan 2\)

Its height at angle \(0\)

Explanation

The corresponding original input is \(\arctan 2\), whose tangent is \(2\). Its slope is \(\sec^2(\arctan 2)=1+2^2=5\), so the inverse slope is \(1/5\).