Single Variable Calculus Questions and Answers

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Single Variable Calculus, Chapter 7, 7.6, Section 7.6, Problem 20

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

If we let $y = \sec^{-1} x$, then

$\sec y = x$

By Implicit Differentiation,

$\displaystyle \frac{d}{dx} \sec y = \frac{d}{dx} (x)$

$\displaystyle \sec y \tan y \left( \frac{dy}{dx} \right) = 1$

$\displaystyle \frac{dy}{dx} = \frac{1}{\sec y \tan y}$

By applying Pythagorean Identity,

$ \begin{equation} \begin{aligned} 1 + \tan^y &= \sec^2 y\\ \\ \tan y &= \sqrt{\sec^2 y - 1} \end{aligned} \end{equation} $

Thus,

$\displaystyle \frac{dy}{dx} = \frac{1}{\sec y \sqrt{\sec^2 y - 1}}$

But $\sec y = x$, therefore,

$\displaystyle \frac{dy}{dx} = \frac{1}{x \sqrt{x^2 - 1}}$