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Chapter 3: Problem 53

Determine whether the following statements are true and give an explanation or counterexample. a. \(\frac{d}{d x}\left(\sin ^{-1} x+\cos ^{-1} x\right)=0\) b. \(\frac{d}{d x}\left(\tan ^{-1} x\right)=\sec ^{2} x\) c. The lines tangent to the graph of \(y=\sin ^{-1} x\) on the interval [-1,1] have a minimum slope of 1 d. The lines tangent to the graph of \(y=\sin x\) on the interval \([-\pi / 2, \pi / 2]\) have a maximum slope of 1 e. If \(f(x)=1 / x,\) then \(\left(f^{-1}\right)^{\prime}(x)=-1 / x^{2}\)

Short Answer

Expert verified
Question: Determine if the statements regarding derivatives of inverse trigonometric functions and their properties are true or false: a. \(\frac{d}{dx}(\sin^{-1}x+\cos^{-1}x)=0\) Answer: True b. \(\frac{d}{dx}(\tan^{-1}x)=\sec^{2}x\) Answer: False c. The lines tangent to the graph of \(y=\sin^{-1}x\) on the interval [-1,1] have a minimum slope of 1 Answer: True d. The lines tangent to the graph of \(y=\sin x\) on the interval \([-\pi/2, \pi/2]\) have a maximum slope of 1 Answer: True e. If \(f(x)=1/x\), then \((f^{-1})'(x)=-1/x^2\) Answer: True

Step by step solution

01

a. Determine if the statement is true or false: \(\frac{d}{dx}(\sin^{-1}x+\cos^{-1}x)=0\)

To verify this statement, let's find the derivative of \(\sin^{-1}x + \cos^{-1}x\). Using the addition rule for derivatives and differentiating each term separately, we get: \(\frac{d}{dx}(\sin^{-1}x+\cos^{-1}x) = \frac{d}{dx}\sin^{-1}x + \frac{d}{dx}\cos^{-1}x\) We know the derivatives of inverse sine and inverse cosine functions: \(\frac{d}{dx}\sin^{-1}{x} = \frac{1}{\sqrt{1-x^2}}\) and \(\frac{d}{dx}\cos^{-1}{x} = -\frac{1}{\sqrt{1-x^2}}\) So, \(\frac{d}{dx}(\sin^{-1}x+\cos^{-1}x) = \frac{1}{\sqrt{1-x^2}} - \frac{1}{\sqrt{1-x^2}} = 0\) Therefore, this statement is true.
02

b. Determine if the statement is true or false: \(\frac{d}{dx}(\tan^{-1}x)=\sec^{2}x\)

To verify this statement, let's find the derivative of \(\tan^{-1}x\). We know the derivative of the inverse tangent function: \(\frac{d}{dx}\tan^{-1}x = \frac{1}{1+x^2}\) However, this expression is not equal to \(\sec^2x\) (since \(\sec^2x = \frac{1}{\cos^2x}\)), so this statement is false.
03

c. Determine if the statement is true or false: The lines tangent to the graph of \(y=\sin^{-1}x\) on the interval [-1,1] have a minimum slope of 1

The minimum slope of tangent lines to the graph of \(y=\sin^{-1}x\) on the interval \([-1,1]\) can be found by analyzing the derivative of the function: \(\frac{d}{dx}\sin^{-1}x = \frac{1}{\sqrt{1-x^2}}\) Since the numerator is always 1, the minimum slope occurs when the denominator is at its maximum. The maximum value of the denominator, \(\sqrt{1-x^2}\), occurs when \(x=0\). That leads to the minimum slope: \(\frac{d}{dx}\sin^{-1}(0) = \frac{1}{\sqrt{1-0^2}} = 1\) So, this statement is true.
04

d. Determine if the statement is true or false: The lines tangent to the graph of \(y=\sin x\) on the interval \([-\pi/2, \pi/2]\) have a maximum slope of 1

The maximum slope of tangent lines to the graph of \(y=\sin x\) on the interval \([-\pi/2, \pi/2]\) can be found by analyzing the derivative of the function: \(\frac{d}{dx}\sin x = \cos x\) Since the maximum value of \(\cos x\) on the interval \([-\pi/2, \pi/2]\) is 1 (when \(x=0\)), the maximum slope of the lines tangent to the graph of \(y=\sin x\) on the given interval is indeed 1. So, this statement is true.
05

e. Determine if the statement is true or false: If \(f(x)=1/x\), then \((f^{-1})'(x)=-1/x^2\)

First, let's find the inverse function \(f^{-1}(x)\). Suppose \(y=f^{-1}(x)\), then we have: \(x=f(y) = \frac{1}{y} \Rightarrow y=\frac{1}{x}\) So, \(f^{-1}(x)=\frac{1}{x}\). Now, let's find its derivative: \(\frac{d}{dx}(f^{-1}(x)) = \frac{d}{dx}\left(\frac{1}{x}\right)\) Using the power rule for derivatives, we get: \(\frac{d}{dx}(f^{-1}(x)) = -\frac{1}{x^2}\) Hence, this statement is true.

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