Derivatives and integrals of inverse trigonometric functions

What this dot point is asking

NESA wants you to know the standard derivatives of $\arcsin$, $\arccos$ and $\arctan$ cold, apply the chain rule to compositions like $\arctan(3x)$, combine them with the product and quotient rules, and run the process in reverse to integrate functions that produce these inverse-trig antiderivatives. The whole topic is the calculus of the three inverse trig functions, so the graphs they come from and the domains they live on are not optional background: they explain every feature of the derivatives, including the awkward square root and the sign on $\arccos$.

The three results look like things to memorise, and they are, but they are far easier to keep straight if you can re-derive them from $\sin y = x$ in a line of implicit differentiation. A marker can tell instantly whether you understand where $\sqrt{1 - x^2}$ comes from or are pattern-matching from a half-remembered formula sheet, and the "establish" or "hence" style questions reward the former.

The answer

Standard derivatives

Note that $\arcsin$ and $\arccos$ have the same magnitude derivative but opposite signs, consistent with $\arcsin x + \arccos x = \frac{\pi}{2}$: differentiating that identity gives $(\arcsin x)' + (\arccos x)' = 0$, so the two derivatives must be exact negatives. That is also why $\arccos$ never appears as a standard integral: it would only duplicate the $\arcsin$ result up to sign.

Where the derivative comes from: implicit differentiation

Let $y = \arcsin x$, so $\sin y = x$. Differentiate implicitly with respect to $x$:

This is correct but is in terms of $y$; we want it in terms of $x$. For $y$ in the principal range $\left[ -\frac{\pi}{2}, \frac{\pi}{2} \right]$, $\cos y \ge 0$, so we take the positive square root: $\cos y = \sqrt{1 - \sin^2 y} = \sqrt{1 - x^2}$. Hence

The same method gives $\arctan$: from $\tan y = x$, differentiating gives $\sec^2 y \cdot \frac{dy}{dx} = 1$, and $\sec^2 y = 1 + \tan^2 y = 1 + x^2$, so $(\arctan x)' = \frac{1}{1 + x^2}$. For $\arccos$, $\cos y = x$ gives $-\sin y \cdot \frac{dy}{dx} = 1$, and on the range $[0, \pi]$ we have $\sin y \ge 0$, so $\sin y = \sqrt{1 - x^2}$ and the minus sign survives: $(\arccos x)' = -\frac{1}{\sqrt{1 - x^2}}$. The choice of sign on each square root is forced by the principal range, which is exactly why the graphs matter.

The three curves and where the derivatives live, stage by stage

Each inverse function is a reflection of a restricted piece of a trig function in the line $y = x$, and that reflection explains the domain, the range and the shape of the derivative. The four panels build the picture one curve at a time.

Stage 1, restrict and reflect to build $\arcsin$. $\sin x$ is not one-to-one, so to invert it we keep only the rising piece on $\left[-\frac{\pi}{2}, \frac{\pi}{2}\right]$, where it passes the horizontal-line test. Reflecting that arc in $y = x$ swaps the axes and produces $y = \arcsin x$. The reflection sends the gentle slope of $\sin x$ near its turning points into a steep slope for $\arcsin$ near $x = \pm 1$, which is the geometric origin of the $\sqrt{1 - x^2}$ in the denominator.

Stage 2, read off the domain, range and the vertical tangents of $\arcsin$. The reflected curve has domain $[-1, 1]$ (the range of $\sin$) and range $\left[-\frac{\pi}{2}, \frac{\pi}{2}\right]$ (the restricted domain of $\sin$). At the two endpoints the tangent is vertical, so the gradient $\to \infty$: that is the derivative $\frac{1}{\sqrt{1 - x^2}}$ becoming undefined as $x \to \pm 1$. In the middle the curve is least steep at $x = 0$, where the derivative is smallest, equal to $1$.

Stage 3, the same construction for $\arccos$. Reflecting the piece of $\cos x$ on $[0, \pi]$ gives $y = \arccos x$: domain $[-1, 1]$, range $[0, \pi]$, and it decreases throughout (cosine falls on $[0, \pi]$). The decreasing graph is the geometric reason for the minus sign in $(\arccos x)' = -\frac{1}{\sqrt{1 - x^2}}$; the magnitude is identical to $\arcsin$ because the two curves are mirror images about the line $y = \frac{\pi}{4}$.

Stage 4, $\arctan$ and why its derivative is defined everywhere. $\tan x$ on $\left(-\frac{\pi}{2}, \frac{\pi}{2}\right)$ takes every real value, so its inverse $\arctan$ has domain all reals and range $\left(-\frac{\pi}{2}, \frac{\pi}{2}\right)$. The vertical asymptotes of $\tan$ become the horizontal asymptotes $y = \pm\frac{\pi}{2}$ of $\arctan$. The curve is steepest at the origin and flattens forever without reaching the asymptotes, which is exactly the behaviour of $\frac{1}{1 + x^2}$: largest at $x = 0$, tending to $0$ as $x \to \pm\infty$. There are no endpoints, so unlike $\arcsin$ there is no domain restriction to worry about when you differentiate.

Chain rule extensions

In an exam the argument is almost never a bare $x$; it is some function $f(x)$. The chain rule multiplies the standard derivative by $f'(x)$:

The single most common error in the whole topic is dropping the $f'(x)$ from the numerator, so write the inner function down explicitly, differentiate it, and place it on top before you simplify. For $\arctan(3x)$, $f = 3x$ and $f' = 3$, giving $\frac{3}{1 + 9x^2}$. For $\arcsin(x^2)$, $f = x^2$ and $f' = 2x$, giving $\frac{2x}{\sqrt{1 - x^4}}$. Note that $[f(x)]^2$ means you square the whole inner function: $(x^2)^2 = x^4$, not $x^2$.

Standard antiderivatives (the reverse direction)

Every derivative above, read right to left, is an integral. The two that are quoted on the HSC reference sheet are:

Generalised to a constant $a > 0$ (these are also on the reference sheet):

Refer to the dedicated integration-of-inverse-trig dot point for the completing-the-square and substitution techniques that reduce a messier quadratic to one of these forms.

Reading the integrand to pick the right form

The skill examined here is recognising which standard form an integrand matches. A square root of $a^2 - x^2$ in the denominator signals an $\arcsin$ result; a sum $a^2 + x^2$ in the denominator (no root) signals an $\arctan$ result. Always identify $a$ first: in $\frac{1}{9 + x^2}$ you have $a^2 = 9$, so $a = 3$ and the constant in front of $\arctan$ is $\frac{1}{a} = \frac{1}{3}$. A common slip is to read $a^2$ itself as $a$, which gives the wrong constant.

When the inside is a multiple of $x$, such as $\frac{1}{\sqrt{1 - 4x^2}}$, factor it into the form $\sqrt{1 - (2x)^2}$ and substitute $u = 2x$, or quote the chain-rule-adjusted derivative directly. Either route is acceptable, but show the substitution if the marks are for method.

Connection to the derivatives

These integrals are nothing more than the derivative results read backwards, so the fastest way to recall them under exam pressure is to recall the three derivatives and reverse each one. If $\frac{d}{dx}(\arctan x) = \frac{1}{1 + x^2}$, then $\int \frac{1}{1 + x^2}\,dx = \arctan x + C$. Keeping the derivative-integral pairs together in your memory halves what you need to learn and guards against the sign error on the $\arccos$ form, which never appears as a standard integral precisely because it duplicates the $\arcsin$ result up to sign.

How exam questions ask about inverse-trig calculus

The wording tells you which rule is being tested:

• "Differentiate $y = \arctan(\ldots)$ / $\arcsin(\ldots)$." A direct chain-rule question. Identify the inner function $f$, find $f'$, and put $f'$ on top of the standard form. Most are $2$ marks: one for the form, one for the chain-rule factor.
• "Differentiate $x \arctan x$" or "$\frac{\arcsin x}{x}$." A product or quotient rule wrapped around an inverse-trig derivative. State the rule, then substitute the inverse-trig derivative for the relevant factor.
• "Find the gradient / equation of the tangent at $x = a$." Differentiate, then substitute $x = a$. Keep surds exact unless told to round.
• "Show that $\frac{d}{dx}(\arcsin x) = \frac{1}{\sqrt{1 - x^2}}$" or "by writing $\sin y = x$ ...". A derivation: use implicit differentiation and justify the positive square root from the principal range. Do not just quote the result.
• "Hence find $\int \ldots$." The "hence" signals you should reverse the derivative you just computed. The integral is the antiderivative of the thing you differentiated.
• "State the domain of the derivative" or a question about vertical tangents. A graph question in disguise: $\arcsin$ and $\arccos$ derivatives are defined only on $(-1, 1)$; $\arctan$ everywhere.