Question

(a) One way of defining \({\sec ^{ - 1}}x\) is to say that \(y = {\sec ^{ - 1}}x \Leftrightarrow \sec y = x\) and \(0 \le y < \frac{\pi }{2}\) or \(\pi \le y < \frac{{3\pi }}{2}\). Show that with this definition,

\(\frac{d}{{dx}}\left( {{{\sec }^{ - 1}}x} \right) = \frac{1}{{x\sqrt {{x^2} - 1} }}\)

(b) Another way of defining \({\sec ^{ - 1}}x\) that is sometimes used is to say that \(y = {\sec ^{ - 1}}x \Leftrightarrow \sec y = x\) and \(0 \le y \le \pi \), \(y \ne \frac{\pi }{2}\). Show that with this definition,

\(\frac{d}{{dx}}\left( {{{\sec }^{ - 1}}x} \right) = \frac{1}{{\left| x \right|\sqrt {{x^2} - 1} }}\)

Short answer

The formula for \(\frac{d}{{dx}}\left( {{{\sec }^{ - 1}}x} \right)\) is proved.

Step-by-step solution

The chain rule

If \(g\) is differentiable at \(x\) and \(f\) is differentiable at \(g\left( x \right)\), then the composite function \(F = f \circ g\) defined by \(F\left( x \right) = f\left( {g\left( x \right)} \right)\) is differentiable at \(x\) and \(F'\) is given by the product \(F'\left( x \right) = f'\left( {g\left( x \right)} \right) \cdot g'\left( x \right)\).

In Leibniz notation, if \(y = f\left( u \right)\) and \(u = g\left( x \right)\) are both differentiable functions, then \(\frac{{dy}}{{dx}} = \frac{{dy}}{{du}} \cdot \frac{{du}}{{dx}}\).

The derivative of the function

(a)

Let \(y = {\sec ^{ - 1}}x\). This implies, \(\sec y = x\). Differentiate both the sides of the above equation with respect to \(x\) as follows:

\(\begin{aligned}\frac{d}{{dx}}\left( {\sec y} \right) & = \frac{d}{{dx}}\left( x \right)\\\sec y\tan y\frac{{dy}}{{dx}} & = 1\\\frac{{dy}}{{dx}} & = \frac{1}{{\sec y\tan y}}\\\frac{{dy}}{{dx}} & = \frac{1}{{\sec y\sqrt {{{\sec }^2}y - 1} }}\\\frac{d}{{dx}}\left( {{{\sec }^{ - 1}}x} \right) &= \frac{1}{{x\sqrt {{x^2} - 1} }}\end{aligned}\)

Here, \(\tan y > 0\), when \(0 < y < \frac{\pi }{2}\) or \(\pi < y < \frac{{3\pi }}{2}\).

Hence, the formula for \(\frac{d}{{dx}}\left( {{{\sec }^{ - 1}}x} \right)\) is proved.

(b)

Let \(y = {\sec ^{ - 1}}x\). This implies, \(\sec y = x\). Differentiate both the sides of the above equation with respect to \(x\) as follows:

\(\begin{aligned}\frac{d}{{dx}}\left( {\sec y} \right) & = \frac{d}{{dx}}\left( x \right)\\\sec y\tan y\frac{{dy}}{{dx}} & = 1\\\frac{{dy}}{{dx}} & = \frac{1}{{\sec y\tan y}}\end{aligned}\)

Here, by trigonometric identity:

\(\begin{array}{l}{\tan ^2}y = {\sec ^2}y - 1\\{\tan ^2}y = {x^2} - 1\\\tan y = \pm \sqrt {{x^2} - 1} \end{array}\)

In the interval \(\left( {0,\frac{\pi }{2}} \right)\), \(\tan y \ge 0\) and \(x \ge 1\). So, \(\sec y = x = \left| x \right|\). That implies:

\(\begin{aligned}\frac{{dy}}{{dx}} & = \frac{1}{{x\sqrt {{x^2} - 1} }}\\ & = \frac{1}{{\left| x \right|\sqrt {{x^2} - 1} }}\end{aligned}\)

In the interval \(\left( {\frac{\pi }{2},\pi } \right)\), \(\tan y = - \sqrt {{x^2} - 1} \) and \(x \le - 1\). So, \(\left| x \right| = - x\). That implies:

\(\begin{aligned}\frac{{dy}}{{dx}} & = \frac{1}{{x\left( { - \sqrt {{x^2} - 1} } \right)}}\\ & = \frac{1}{{ - x\left( {\sqrt {{x^2} - 1} } \right)}}\\ & = \frac{1}{{\left| x \right|\left( {\sqrt {{x^2} - 1} } \right)}}\end{aligned}\)

Hence, it is proved.