Skip to main content

Subsection 2.12.1 Derivatives of Inverse Trig Functions

Now that we have explored the arcsine function we are ready to find its derivative. Lets call

\begin{align*} \arcsin(x) &= \theta(x), \end{align*}

so that the derivative we are seeking is \(\diff{\theta}{x}\text{.}\) The above equation is (after taking sine of both sides) equivalent to

\begin{align*} \sin(\theta) &= x \end{align*}

Now differentiate this using implicit differentiation (we just have to remember that \(\theta\) varies with \(x\) and use the chain rule carefully):

\begin{align*} \cos(\theta) \cdot \diff{\theta}{x} &= 1\\ \diff{\theta}{x} &= \frac{1}{\cos(\theta)} & \text{substitute $\theta = \arcsin x$}\\ \diff{}{x} \arcsin x &= \frac{1}{\cos(\arcsin x)} \end{align*}

This doesn't look too bad, but it's not really very satisfying because the right hand side is expressed in terms of \(\arcsin(x)\) and we do not have an explicit formula for \(\arcsin(x)\text{.}\)

However even without an explicit formula for \(\arcsin(x)\text{,}\) it is a simple matter to get an explicit formula for \(\cos\big(\arcsin(x)\big)\text{,}\) which is all we need. Just draw a right–angled triangle with one angle being \(\arcsin(x)\text{.}\) This is done in the figure below  3 The figure is drawn for the case that \(0\le\arcsin(x)\le\frac{\pi}{2}\text{.}\) Virtually the same argument works for the case \(-\frac{\pi}{2}\le\arcsin(x)\le 0\).

Since \(\sin(\theta)=x\) (see 2.12.2), we have made the side opposite the angle \(\theta\) of length \(x\) and the hypotenuse of length \(1\text{.}\) Then, by Pythagoras, the side adjacent to \(\theta\) has length \(\sqrt{1-x^2}\) and so

\begin{gather*} \cos\big(\arcsin(x)\big)=\cos(\theta)=\sqrt{1-x^2} \end{gather*}

which in turn gives us the answer we need:

\begin{gather*} \diff{}{x} \arcsin(x) =\frac{1}{\sqrt{1-x^2}} \end{gather*}

The definitions for \(\arccos\text{,}\) \(\arctan\) and \(\arccot\) are developed in the same way. Here are the graphs that are used.

The definitions for the remaining two inverse trigonometric functions may also be developed in the same way 4 In fact, there are two different widely used definitions of \(\arcsec x\text{.}\) Under our definition, below, \(\theta=\arcsec x\) takes values in \(0\le\theta\le\pi\text{.}\) Some people, perfectly legitimately, define \(\theta=\arcsec x\) to take values in the union of \(0\le \theta\lt\frac{\pi}{2}\) and \(\pi\le\theta\lt\frac{3\pi}{2}\text{.}\) Our definition is sometimes called the “trigonometry friendly” definition. The definition itself has the advantage of simplicity. The other definition is sometimes called the “calculus friendly” definition. It eliminates some absolute values and hence simplifies some computations. Similarly there are two different widely used definitions of \(\arccsc x\text{.}\) 5 One could also define \(\arccot(x)=\arctan(1/x)\) with \(\arccot(0)=\frac{\pi}{2}\text{.}\) We have chosen not to do so, because the definition we have chosen is both continuous and standard.. But it's a little easier to use

\begin{gather*} \csc x=\frac{1}{\sin x} \qquad \sec x=\frac{1}{\cos x} \end{gather*}
Definition 2.12.4

\(\arcsin x\) is defined for \(|x|\le 1\text{.}\) It is the unique number obeying

\begin{align*} \sin\big(\arcsin(x)\big)&=x &&\text{and}& -\frac{\pi}{2}\le &\arcsin(x)\le\frac{\pi}{2}\\ \end{align*}

\(\arccos x\) is defined for \(|x|\le 1\text{.}\) It is the unique number obeying

\begin{align*} \cos\big(\arccos(x)\big)&=x &&\text{and}& 0\le &\arccos(x)\le\pi\\ \end{align*}

\(\arctan x\) is defined for all \(x\in\bbbr\text{.}\) It is the unique number obeying

\begin{align*} \tan\big(\arctan(x)\big)&=x &&\text{and}& -\frac{\pi}{2} \lt &\arctan(x) \lt \frac{\pi}{2}\\ \end{align*}

\(\arccsc x=\arcsin\frac{1}{x}\) is defined for \(|x|\ge 1\text{.}\) It is the unique number obeying

\begin{align*} \csc\big(\arccsc(x)\big)&=x &&\text{and}& -\frac{\pi}{2}\le &\arccsc(x)\le\frac{\pi}{2}\\ \end{align*}

\(\ \ \ \ \ \ \ \ \)Because \(\csc(0)\) is undefined, \(\arccsc(x)\) never takes the value \(0\text{.}\)

\begin{align*} \end{align*}

\(\arcsec x=\arccos\frac{1}{x}\) is defined for \(|x|\ge 1\text{.}\) It is the unique number obeying

\begin{align*} \sec\big(\arcsec(x)\big)&=x &&\text{and}& 0\le &\arcsec(x)\le\pi\\ \end{align*}

\(\ \ \ \ \ \ \ \ \)Because \(\sec(\pi/2)\) is undefined, \(\arcsec(x)\) never takes the value \(\pi/2\text{.}\)

\begin{align*} \end{align*}

\(\arccot x\) is defined for all \(x\in\bbbr\text{.}\) It is the unique number obeying

\begin{align*} \cot\big(\arccot(x)\big)&=x &&\text{and}& 0 \lt &\arccot(x) \lt \pi \end{align*}

To find the derivative of \(\arccos\) we can follow the same steps:

  • Write \(\arccos(x) =\theta(x)\) so that \(\cos\theta = x\) and the desired derivative is \(\diff{\theta}{x}\text{.}\)
  • Differentiate implicitly, remembering that \(\theta\) is a function of \(x\text{:}\)
    \begin{align*} -\sin\theta \diff{\theta}{x} &= 1\\ \diff{\theta}{x} &= -\frac{1}{\sin\theta}\\ \diff{}{x}\arccos x &= -\frac{1}{\sin(\arccos x)}. \end{align*}
  • To simplify this expression, again draw the relevant triangle

    from which we see

    \begin{align*} \sin(\arccos x) = \sin\theta &= \sqrt{1-x^2}. \end{align*}
  • Thus
    \begin{align*} \diff{}{x}\arccos x &= -\frac{1}{\sqrt{1-x^2}}. \end{align*}

Very similar steps give the derivative of \(\arctan x\text{:}\)

  • Start with \(\theta = \arctan x\text{,}\) so \(\tan \theta = x\text{.}\)
  • Differentiate implicitly:
    \begin{align*} \sec^2 \theta \diff{\theta}{x} &= 1\\ \diff{\theta}{x} &= \frac{1}{\sec^2 \theta} = \cos^2 \theta\\ \diff{}{x}\arctan x &= \cos^2(\arctan x). \end{align*}
  • To simplify this expression, we draw the relevant triangle

    from which we see

    \begin{gather*} \cos^2(\arctan x) = \cos^2\theta = \frac{1}{1+x^2} \end{gather*}
  • Thus
    \begin{align*} \diff{}{x}\arctan x &= \frac{1}{1+x^2}. \end{align*}

An almost identical computation gives the derivative of \(\arccot x\text{:}\)

  • Start with \(\theta = \arccot x\text{,}\) so \(\cot \theta = x\text{.}\)
  • Differentiate implicitly:
    \begin{align*} -\csc^2 \theta \diff{\theta}{x} &= 1\\ \diff{}{x}\arccot x = \diff{\theta}{x} &= -\frac{1}{\csc^2 \theta} = -\sin^2 \theta = -\frac{1}{1+x^2} \end{align*}
    from the triangle

To find the derivative of \(\arccsc\) we can use its definition and the chain rule.

\begin{align*} \theta &= \arccsc x & \text{take cosecant of both sides}\\ \csc \theta &= x & \text{but $\csc \theta = \frac{1}{\sin\theta}$, so flip both sides}\\ \sin \theta &= \frac{1}{x} & \text{now take arcsine of both sides}\\ \theta &= \arcsin\left(\frac{1}{x}\right) \end{align*}

Now just differentiate, carefully using the chain rule :

\begin{align*} \diff{\theta}{x} &= \diff{}{x} \arcsin\left(\frac{1}{x}\right)\\ &= \frac{1}{\sqrt{1-x^{-2}}} \cdot \frac{-1}{x^2}\\ \end{align*}

To simplify further we will factor \(x^{-2}\) out of the square root. We need to be a little careful doing that. Take another look at examples 1.5.6 and 1.5.7 and the discussion between them before proceeding.

\begin{align*} &= \frac{1}{\sqrt{x^{-2}(x^2-1)}} \cdot \frac{-1}{x^2}\\ &= \frac{1}{|x^{-1}|\cdot \sqrt{x^2-1}} \cdot \frac{-1}{x^2} & \text{note that $x^2 \cdot |x^{-1}| = |x|$.}\\ &= - \frac{1}{|x|\sqrt{x^2-1}} \end{align*}

In the same way we can find the derivative of the remaining inverse trig function. We just use its definition, a derivative we already know and the chain rule.

\begin{gather*} \diff{}{x} \arcsec(x) = \diff{}{x} \arccos\Big(\frac{1}{x}\Big) =-\frac{1}{\sqrt{1-\frac{1}{x^2}}}\cdot\Big(-\frac{1}{x^2}\Big) =\frac{1}{|x|\sqrt{x^2-1}} \end{gather*}

By way of summary, we have