Second Derivatives Of Inverse Trigonometric Functions

Why second derivatives matter here

The first derivative of an inverse trigonometric function describes its local rate of change. The second derivative describes how that rate changes. For arcsine and arccosine, this change becomes especially pronounced near the edges of their domains. For arctangent, the second derivative instead approaches zero far out along the real line.

The goal is to calculate these second derivatives accurately, read their signs to determine concavity, and distinguish what happens at a finite domain endpoint from what happens at an ordinary interior point.

Domains and first derivatives

The real-valued principal functions $\arcsin x$ and $\arccos x$ have domain $[-1,1]$. Their first derivatives exist on the open interval $(-1,1)$, where the standard formulas are

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

The real-valued function $\arctan x$ has domain $\mathbb{R}$, and

\frac{d}{dx}\arctan x=\frac{1}{1+x^2}.

These first-derivative formulas are the starting point. Since a second derivative differentiates the first derivative, the formulas for $\arcsin x$ and $\arccos x$ can only be differentiated on $(-1,1)$. For $\arctan x$, they can be differentiated at every real number.

Definition: The second derivative of a function $f$, where it exists, is the derivative of its first derivative: $f''(x)=\frac{d}{dx}(f'(x))$. On an interval where $f''(x)>0$, $f$ is concave up; where $f''(x)<0$, $f$ is concave down. A point at which the concavity changes is an inflection point.

Deriving the three second derivatives

For arcsine, write the first derivative as $(1-x^2)^{-1/2}$. The chain rule gives a positive sign: differentiating the inside $1-x^2$ contributes $-2x$, which cancels the negative factor from differentiating the power $-1/2$.

\begin{aligned} \frac{d^2}{dx^2}\arcsin x &=\frac{d}{dx}(1-x^2)^{-1/2}\\ &=-\frac12(1-x^2)^{-3/2}(-2x)\\ &=\frac{x}{(1-x^2)^{3/2}}, \qquad -1<x<1. \end{aligned}

The arccosine first derivative has an extra negative sign. Differentiating it gives the negative of the arcsine second derivative:

\begin{aligned} \frac{d^2}{dx^2}\arccos x &=\frac{d}{dx}\bigl(-(1-x^2)^{-1/2}\bigr)\\ &=-\left[-\frac12(1-x^2)^{-3/2}(-2x)\right]\\ &=-\frac{x}{(1-x^2)^{3/2}}, \qquad -1<x<1. \end{aligned}

For arctangent, apply the chain rule to $(1+x^2)^{-1}$:

\begin{aligned} \frac{d^2}{dx^2}\arctan x &=\frac{d}{dx}(1+x^2)^{-1}\\ &=-(1+x^2)^{-2}(2x)\\ &=-\frac{2x}{(1+x^2)^2}, \qquad x\in\mathbb{R}. \end{aligned}

Result: On their respective intervals of differentiability, $$ (\arcsin x)''=\frac{x}{(1-x^2)^{3/2}},\qquad (\arccos x)''=-\frac{x}{(1-x^2)^{3/2}},\qquad (\arctan x)''=-\frac{2x}{(1+x^2)^2}. $$ The arcsine and arccosine formulas apply only for $-1<x<1$; the arctangent formula applies for every real $x$.

Concavity and inflection points

The denominators in the arcsine and arccosine second derivatives are positive throughout $(-1,1)$. The denominator in the arctangent formula is also positive for every real $x$. Consequently, the signs are determined by $x$, together with the minus sign in the arccosine and arctangent formulas.

Function	Concave up	Concave down	Inflection point
$\arcsin x$	$(0,1)$	$(-1,0)$	$(0,0)$
$\arccos x$	$(-1,0)$	$(0,1)$	$(0,\pi/2)$
$\arctan x$	$(-\infty,0)$	$(0,\infty)$	$(0,0)$

At $x=0$, each second derivative is zero, and in each case its sign changes across zero. This confirms a change in concavity there. The function values locate the corresponding points: $\arcsin 0=0$, $\arccos 0=\pi/2$, and $\arctan 0=0$.

Behavior near the finite domain endpoints

The second derivatives of arcsine and arccosine are not defined at $x=-1$ or $x=1$, because their formulas have a zero denominator there. Their behavior as $x$ approaches these endpoints from inside the domain can be made precise. Near $x=1$, set $t=1-x$. Then $t$ is positive and tends to zero, while

1-x^2=(1-x)(1+x)=t(2-t)\sim 2t.

Here $A(t)\sim B(t)$ means that $A(t)/B(t)$ tends to $1$. Substituting this relation into the formulas shows that the second derivatives grow in magnitude like a constant times the distance to the endpoint raised to the power $-3/2$. Their signs depend on the function and the endpoint:

\begin{aligned} (\arcsin x)''&\sim \frac{1}{[2(1-x)]^{3/2}} &&\text{as }x\to1^-,\\ (\arcsin x)''&\sim-\frac{1}{[2(1+x)]^{3/2}} &&\text{as }x\to-1^+,\\ (\arccos x)''&\sim-\frac{1}{[2(1-x)]^{3/2}} &&\text{as }x\to1^-,\\ (\arccos x)''&\sim\frac{1}{[2(1+x)]^{3/2}} &&\text{as }x\to-1^+. \end{aligned}

Thus the arcsine second derivative tends to positive infinity at the right endpoint and negative infinity at the left endpoint. For arccosine, the signs are reversed. These are one-sided limits from within $(-1,1)$; they do not assign a second-derivative value at either endpoint.

Arctangent has no finite domain endpoints. Instead, its second derivative tends to zero as $x\to\infty$ and as $x\to-\infty$, as follows from the denominator growing faster than the numerator.

Worked examples

Worked Example: Concavity and an exact second-derivative value for arcsine

Let $f(x)=\arcsin x$. Find its concavity intervals and calculate $f''(1/2)$.

The domain is $[-1,1]$, and the second derivative exists on $(-1,1)$. Using the derived formula,

f''(x)=\frac{x}{(1-x^2)^{3/2}}.

For $-1<x<0$, the numerator is negative and the denominator is positive, so $f''(x)<0$. For $0<x<1$, the numerator and denominator are positive, so $f''(x)>0$. Thus $f$ is concave down on $(-1,0)$ and concave up on $(0,1)$.

At $x=1/2$,

\begin{aligned} f''\left(\frac12\right) &=\frac{\frac12}{\left(1-\frac14\right)^{3/2}}\\ &=\frac{\frac12}{\left(\frac34\right)^{3/2}}\\ &=\frac{\frac12}{\frac{3\sqrt{3}}{8}}\\ &=\frac{4}{3\sqrt{3}}. \end{aligned}

Check: substituting $x=1/2$ into the second-derivative formula gives the same value, $4/(3\sqrt{3})$. It is positive, consistent with concavity up at that point.

Worked Example: A scaled arccosine input

Let $g(x)=\arccos(2x)$. Find its domain, second derivative, concavity intervals, and the value of $g''(-1/4)$.

The input to arccosine must lie in $[-1,1]$, so $-1\leq 2x\leq1$, which gives the domain $[-1/2,1/2]$. For interior points, the chain rule gives

\begin{aligned} g'(x) &=-\frac{1}{\sqrt{1-(2x)^2}}\,(2)\\ &=-\frac{2}{\sqrt{1-4x^2}}. \end{aligned}

Differentiate again, keeping the factor $-2$:

\begin{aligned} g''(x) &=-2\left[-\frac12(1-4x^2)^{-3/2}(-8x)\right]\\ &=-\frac{8x}{(1-4x^2)^{3/2}}, \qquad -\frac12<x<\frac12. \end{aligned}

The denominator is positive. Therefore $g''(x)>0$ for $-1/2<x<0$, and $g''(x)<0$ for $0<x<1/2$. The function is concave up on the first interval and concave down on the second.

At $x=-1/4$,

\begin{aligned} g''\left(-\frac14\right) &=\frac{2}{\left(1-4\cdot\frac{1}{16}\right)^{3/2}}\\ &=\frac{2}{\left(\frac34\right)^{3/2}}\\ &=\frac{2}{\frac{3\sqrt{3}}{8}}\\ &=\frac{16}{3\sqrt{3}}. \end{aligned}

Check: $-8(-1/4)=2$, and the denominator at that input is $(3/4)^{3/2}=3\sqrt{3}/8$, giving the same positive value.

Worked Example: A shifted arctangent

Let $h(x)=\arctan(x-1)$. Find $h''(x)$, its concavity intervals, and $h''(2)$.

The chain rule first gives

h'(x)=\frac{1}{1+(x-1)^2}.

Differentiating this expression,

\begin{aligned} h''(x) &=-(1+(x-1)^2)^{-2}\,2(x-1)\\ &=-\frac{2(x-1)}{(1+(x-1)^2)^2}. \end{aligned}

The denominator is positive for every real $x$. If $x<1$, then $x-1<0$, so $h''(x)>0$. If $x>1$, then $h''(x)<0$. Therefore $h$ is concave up on $(-\infty,1)$, concave down on $(1,\infty)$, and changes concavity at $x=1$. Since $h(1)=\arctan 0=0$, the inflection point is $(1,0)$.

At $x=2$,

h''(2)=-\frac{2(2-1)}{(1+(2-1)^2)^2}=-\frac{2}{4}=-\frac12.

Check: substituting $x=2$ into the displayed formula gives $-1/2$, which is negative as expected on the concave-down interval.

A sign error to avoid

A common error is to differentiate $(1-x^2)^{-1/2}$ and lose the negative sign from the inner derivative. The incorrect calculation

\frac{d}{dx}(1-x^2)^{-1/2} \;\not=\; -\frac{x}{(1-x^2)^{3/2}}

would incorrectly make the arcsine second derivative negative for $x>0$. The chain rule requires multiplying the outer derivative by $\frac{d}{dx}(1-x^2)=-2x$:

-\frac12(1-x^2)^{-3/2}(-2x) =\frac{x}{(1-x^2)^{3/2}}.

A second useful check is to inspect the sign near zero: arcsine’s second derivative has the sign of $x$, while arccosine’s has the opposite sign. Also, do not extend either formula to $x=\pm1$; those endpoints are in the function domains but not in the domains of the second derivatives.

Check Your Understanding

Use the formulas and sign reasoning in this tutorial. State any interval restrictions when giving a second derivative.

Find $(\arccos x)''$ and determine the intervals on which $\arccos x$ is concave up and concave down.
For $p(x)=\arctan(3x)$, find $p''(x)$ and determine its concavity intervals.
What are the one-sided limits of $(\arcsin x)''$ as $x\to-1^+$ and $x\to1^-$?
At which input does $\arctan(x+2)$ change concavity, and what is the corresponding point on its graph?
Why is the formula for $(\arccos x)''$ not defined at $x=1$, even though $\arccos 1$ is defined?

Function	Concave up	Concave down	Inflection point
\(\arcsin x\)	\((0,1)\)	\((-1,0)\)	\((0,0)\)
\(\arccos x\)	\((-1,0)\)	\((0,1)\)	\((0,\pi/2)\)
\(\arctan x\)	\((-\infty,0)\)	\((0,\infty)\)	\((0,0)\)