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Section Derivatives of the Inverse Trigonometric Functions

Definition 2.64.

Given a function \(f\) with domain D and range R, if there exists a function \(g\) with domain R and range D so that for all \(x \in D\) and all \(y \in R\) we have

\begin{equation*} g(f(x))=x \mbox{ and } f(g(y))=y, \end{equation*}

then we call \(g\) the inverse of \(f\text{.}\)

Example 2.65.

Discuss inverse functions and compute the inverse of \(\dsp f(x) = \frac{1}{x-3}\text{.}\)

Problem 2.66.

Find the inverse of each of the following functions.

  1. \(f(x) = \frac{2}{3}x+4\)

  2. \(g(x) = x^3 - 1\)

  3. \(h(x) = \sqrt[3]{x-4}\)

  4. \(i(x) = 4 - x^5\)

Problem 2.67.

Sketch the function \(f(x) = ax^2\text{,}\) where \(a > 0\text{.}\) Let c and d be two (distinct) real numbers. Let x be the x-coordinate of the point where the tangent line to f at \((c,f(c))\) intersects the tangent line to f at \((d,f(d))\text{.}\) Show that the distance between x and c equals the distance between x and d.

Problem 2.68.

  1. Does \(n(x) = x^2\) have an inverse? If so what is it? If not, why not?

  2. Does \(y(x) = \sqrt{x}\) have an inverse? If so what is it? If not, why not?

Notice in the next definition that we restrict the domain of sine to talk about the inverse sine. This is because if we simply reversed all the coordinates of sine, the resulting graph would not be a function. There are many ways we could have restricted the domain, so one simply must memorize the domain and range of all the inverse trigonometric functions.

Definition 2.69.

The inverse sine function, denoted by \(\invsin\) or \(\arcsin\) is the inverse of the function \(f(x)=\sin(x)\) with the domain \([-\pi/2,\pi/2].\)

The domain of the inverse sine function is the range of the sine function and the range of the inverse sine function is the restricted domain, \([-\pi/2,\pi/2]\text{,}\) of our sine function. We define all six inverse trigonometric functions in a similar manner and Table 2.70 lists these functions along with their respective domains and ranges.

Inverse Function Domain Range
\(\invsin\) \([-1,1]\) \([-\pi/2,\pi/2]\)
\(\invcos\) \([-1,1]\) \([0,\pi]\)
\(\invtan\) \(\re\) \((-\pi/2,\pi/2)\)
\(\invcsc\) \((-\infty,1] \cup [1, \infty)\) \([-\pi/2,0) \cup (0,\pi/2]\)
\(\invsec\) \((-\infty,1] \cup [1, \infty)\) \([0,\pi/2) \cup (\pi/2,\pi]\)
\(\invcot\) \(\re\) \((0,\pi)\)
Table 2.70. Inverse Trigonometric Functions

Proof. Computing the derivative of the inverse sine function is an application of the Pythagorean Theorem and the Chain Rule. Let

\begin{equation*} f(x) = \invsin(x). \end{equation*}

Since

\begin{equation*} \sin(f(x)) = \sin(\invsin(x)) = x, \end{equation*}

we have,

\begin{equation*} \sin(f(x))=x. \end{equation*}

Taking the derivative of both sides yields,

\begin{equation*} \cos(f(x)) f'(x) = 1, \end{equation*}

and solving for \(f'(x)\) yields

\begin{equation*} f'(x) = \frac{1}{\cos(f(x))}. \end{equation*}

Because we don't want \(f'\) in terms of \(f,\) we apply the Pythagorean Theorem. Draw a right triangle with one angle, \(f(x).\) Since

\begin{equation*} \sin(f(x))=x = \frac{x}{1} \end{equation*}

place \(x\) on the side opposite the angle \(f(x)\) and \(1\) on the hypotenuse. Note that

\begin{equation*} \dsp \cos(f(x)) = \frac{\sqrt{1-x^2}}{1} = \sqrt{1-x^2}. \end{equation*}

Thus

\begin{equation*} \dsp f'(x) = \frac{1}{\sqrt{1-x^2}}. \end{equation*}

q.e.d.

Problem 2.72.

Compute the derivatives of the remaining five inverse trigonometric functions and memorize them.

Problem 2.73.

Compute the derivative of each of the following functions.

  1. \(y(x) =\invsin(2x)\)

  2. \(H(t) = (t-\invcos(t))^{3}\)

  3. \(F(x)=\invcsc(x)\invtan(x)\)

  4. \(\dsp G(z) = \frac{z^{2}-7z+5}{\invcot(z)}\)

  5. \(h(x) = \invtan(\invsin(x))\)

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