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Chapter 3: Problem 1

$$\text { State the derivative formulas for } \sin ^{-1} x, \tan ^{-1} x, \text { and } \sec ^{-1} x$$

Short Answer

Expert verified
Question: State the derivative formulas for \(\sin^{-1}x\), \(\tan^{-1}x\), and \(\sec^{-1}x\). Answer: The derivative formulas for the given inverse trigonometric functions are: 1. \(\sin^{-1}x\): \(\frac{d}{dx}(\sin^{-1}x) = \frac{1}{\sqrt{1-x^2}}\) 2. \(\tan^{-1}x\): \(\frac{d}{dx}(\tan^{-1}x) = \frac{1}{1+x^2}\) 3. \(\sec^{-1}x\): \(\frac{d}{dx}(\sec^{-1}x) = \frac{1}{x \sqrt{x^2 - 1}}\)

Step by step solution

01

Derivative of \(\sin^{-1}x\)

To find the derivative of \(\sin^{-1}x\), we can use implicit differentiation. Let \(y = \sin^{-1}x\), then \(x = \sin y\). Differentiating both sides with respect to \(x\) gives: $$ \frac{d}{dx}(x) = \frac{d}{dx}(\sin y) $$ Using the chain rule, we get: $$ 1 = \cos y \frac{dy}{dx} $$ Now, we use a trigonometric identity to express \(\cos y\) in terms of \(x\). From the Pythagorean identity, we have \(\sin^2 y + \cos^2 y = 1\), so \(\cos^2 y = 1 - \sin^2 y\). Since \(x=\sin y\), we get \(\cos^2 y = 1 - x^2\). Taking the square root and noting that \(\cos y\) must be positive (because the range of \(\sin^{-1} x\) is \([-\pi/2, \pi/2]\)), we obtain \(\cos y = \sqrt{1-x^2}\). Thus, the derivative of \(\sin^{-1}x\) is: $$ \frac{dy}{dx} = \frac{1}{\cos y} = \frac{1}{\sqrt{1-x^2}} $$
02

Derivative of \(\tan^{-1}x\)

To find the derivative of \(\tan^{-1}x\), we use a similar approach as for \(\sin^{-1}x\). Let \(y = \tan^{-1}x\), then \(x = \tan y\). Differentiating both sides with respect to \(x\) gives: $$ \frac{d}{dx}(x) = \frac{d}{dx}(\tan y) $$ Using the chain rule, we get: $$ 1 = (\sec^2 y) \frac{dy}{dx} $$ Now, we use a trigonometric identity to express \(\sec y\) in terms of \(x\). Since \(x=\tan y\), we have \(\tan^2 y = x^2\). Also, from the Pythagorean identity, we have \(1 + \tan^2 y = \sec^2 y\). Thus, we obtain \(\sec^2 y = 1 + x^2\). Now, the derivative of \(\tan^{-1}x\) is: $$ \frac{dy}{dx} = \frac{1}{\sec^2 y} = \frac{1}{1+x^2} $$
03

Derivative of \(\sec^{-1}x\)

To find the derivative of \(\sec^{-1}x\), we will use the same approach as before. Let \(y = \sec^{-1}x\), then \(x = \sec y\). Differentiating both sides with respect to \(x\) gives: $$ \frac{d}{dx}(x) = \frac{d}{dx}(\sec y) $$ Using the chain rule, we get: $$ 1 = (\sec y \tan y) \frac{dy}{dx} $$ Now, we use a trigonometric identity to express \(\tan y\) in terms of \(x\). Since \(x = \sec y\), we have \(\sec^2 y = x^2\). Also, from the Pythagorean identity, we have \(\sec^2 y = 1 + \tan^2 y\). Thus, we obtain \(\tan^2 y = x^2 -1\), so \(\tan y = \sqrt{x^2 - 1}\). Now, the derivative of \(\sec^{-1}x\) is: $$ \frac{dy}{dx} = \frac{1}{\sec y \tan y} = \frac{1}{x \sqrt{x^2 - 1}} $$

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