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Chapter 3: Problem 4

How are the derivatives of \(\sin ^{-1} x\) and \(\cos ^{-1} x\) related?

Short Answer

Expert verified
Answer: The derivatives of \(\sin^{-1}x\) and \(\cos^{-1}x\) are related by being negative of each other, i.e., \(\frac{d(\sin^{-1}x)}{dx} = -\frac{d(\cos^{-1}x)}{dx}\).

Step by step solution

01

Differentiate \(\sin^{-1}x\) with respect to \(x\).

Recall that if \(y = \sin^{-1}x\), then \(x = \sin y\). To find the derivative \(\frac{dy}{dx}\), we differentiate both sides of \(x = \sin y\) implicitly: $$ \frac{d}{dx}\left(x\right) = \frac{d}{dx}\left(\sin y\right) $$ We have \(\frac{dx}{dx} = 1\), and using the chain rule, we get \(\frac{d}{dx}\left(\sin y\right)= \cos y\frac{dy}{dx}\). So we obtain: $$ 1 = \cos y\frac{dy}{dx} $$ To express \(\frac{dy}{dx}\) in terms of \(x\), note that \(\cos y = \sqrt{1 - \sin^2 y}\). Using the identity \(x = \sin y\), we get \(\cos y = \sqrt{1 - x^2}\). Thus, the derivative of \(\sin^{-1}x\) is: $$ \frac{dy}{dx} = \frac{1}{\sqrt{1 - x^2}} $$
02

Differentiate \(\cos^{-1}x\) with respect to \(x\).

As we did for \(\sin^{-1}x\), let \(y = \cos^{-1}x\), so \(x = \cos y\). Differentiating both sides of \(x = \cos y\) implicitly, we have: $$ \frac{d}{dx}\left(x\right) = \frac{d}{dx}\left(\cos y\right) $$ We get \(\frac{dx}{dx} = 1\) and using the chain rule, \(\frac{d}{dx}\left(\cos y\right)= -\sin y\frac{dy}{dx}\). So we have: $$ 1 = -\sin y\frac{dy}{dx} $$ Now, we need to express \(\frac{dy}{dx}\) in terms of \(x\). Using the identity \(\sin y = \sqrt{1 - \cos^2 y}\) and noting that \(x = \cos y\), we get \(\sin y = \sqrt{1 - x^2}\). Thus, the derivative of \(\cos^{-1}x\) is: $$ \frac{dy}{dx} = \frac{-1}{\sqrt{1 - x^2}} $$
03

Find the relationship between the two derivatives.

Now that we have the derivatives of \(\sin^{-1}x\) and \(\cos^{-1}x\), we can establish their relationship. The derivatives are as follows: - \(\frac{d(\sin^{-1}x)}{dx} = \frac{1}{\sqrt{1 - x^2}}\) - \(\frac{d(\cos^{-1}x)}{dx}= \frac{-1}{\sqrt{1 - x^2}}\) Observing the expressions for the derivatives, we can see that they are negatives of each other. In other words: $$ \frac{d(\sin^{-1}x)}{dx} = -\frac{d(\cos^{-1}x)}{dx} $$ So, the derivatives of \(\sin^{-1}x\) and \(\cos^{-1}x\) are related by being negative of each other.

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