Question

Find the derivative of the inverse cosine function in the following two ways. a. Using Theorem 3.21 b. Using the identity \(\sin ^{-1} x+\cos ^{-1} x=\pi / 2\)

Short answer

Answer: The derivative of the inverse cosine function is \((\frac{d}{dx}\cos^{-1}(x)) = -\frac{1}{\sqrt{1 - x^2}}\).

Step-by-step solution

Derivative of an inverse function

If \(f^{-1}(x)\) is the inverse function of \(f(x)\), then the derivative of the inverse function is given by \((f^{-1})'(x) = \frac{1}{f'(f^{-1}(x))}\).

Differentiate cosine function

First, we need to find the derivative of the cosine function, which is \(-sin(x)\).

Apply Theorem 3.21

Now, we apply the theorem to find the derivative of the inverse cosine function: \((\cos^{-1})'(x) = \frac{1}{-\sin(\cos^{-1}(x))}\). However, this expression still contains the inverse function itself, so we need to replace it using the Pythagorean identity: \(\cos^2(y) + \sin^2(y) = 1\).

Replace trigonometric function using identity

To remove the inverse function, let \(y = \cos^{-1}(x)\), which means that \(\cos(y) = x\). Using the Pythagorean identity, we get: \(\sin^2(y) = 1 - \cos^2(y) = 1 - x^2\). So, \(\sin(y) = \pm\sqrt{1 - x^2}\).

Choose the correct sign for the sine function

Since the range of the inverse cosine function is \(0 \le y \le \pi\), the sine function is positive in the first and second quadrants. Hence, we take the positive sign for the sine function: \(\sin(y) = \sqrt{1 - x^2}\).

Write the final derivative

Now, we can write the final derivative of the inverse cosine function: \((\cos^{-1})'(x) = \frac{1}{-\sin(\cos^{-1}(x))} = \frac{1}{-\sqrt{1 - x^2}}\). Strategy b. Using the identity \(\sin^{-1} x+\cos^{-1} x=\pi / 2\)

Differentiate the given identity

We start by differentiating the identity with respect to \(x\): \((\frac{d}{dx}\sin^{-1}(x)) + (\frac{d}{dx}\cos^{-1}(x)) = 0\).

Differentiate the inverse sine function

We already found the derivative of the inverse cosine function in the first part (\((strategy a\))). The derivative of the inverse sine function \((\sin^{-1})'(x)\) can be found in the same way, using Theorem 3.21, and the result is: \((\sin^{-1})'(x) = \frac{1}{\sqrt{1 - x^2}}\).

Solve for \((\frac{d}{dx} \cos^{-1} (x))\)

Substitute the derivative of the inverse sine function into the differentiated identity: \(\frac{1}{\sqrt{1 - x^2}} + (\frac{d}{dx}\cos^{-1}(x)) = 0\). Now, solving for \((\frac{d}{dx}\cos^{-1}(x))\), we get: \((\frac{d}{dx}\cos^{-1}(x)) = -\frac{1}{\sqrt{1 - x^2}}\). Therefore, the derivative of the inverse cosine function by both methods is: \((\frac{d}{dx}\cos^{-1}(x)) = -\frac{1}{\sqrt{1 - x^2}}\).

Key concepts

Inverse Cosine Function

The inverse cosine function is commonly denoted as \( \cos^{-1}(x) \) or \( \arccos(x) \). Inverse trigonometric functions are used to find angles when given the cosine value. The inverse cosine of a value returns the angle whose cosine is that value. It is widely used in trigonometry, calculus, and even in practical fields such as engineering and physics.
- **Range:** The inverse cosine function returns values in the range \([0, \pi]\).- **Domain:** The function is defined for \(x\) values between \(-1\) and \(1\).
The graph of \( \cos^{-1}(x) \) is a declining curve, symbolic of the decreasing angle as the cosine value increases from \(-1\) to \(1\). With its distinct properties, knowing how to differentiate the inverse cosine function is essential for solving calculus problems effectively.

Theorem 3.21

Theorem 3.21 is a key result in calculus that helps us understand derivatives of inverse functions. Without this theorem, finding derivatives would be quite challenging, especially for trigonometric functions. For any function \( f \) possessing an inverse, the derivative formula is given by:
\[(f^{-1})'(x) = \frac{1}{f'(f^{-1}(x))}\]
This theorem underscores the reciprocal relationship between a function and its inverse. So, when you apply this to \( \cos^{-1}(x) \), it shows us the pathway:

• First, find the derivative of the original function, in this case, \(\cos(x)\).
• Next, use the formula to uncover the derivative of its inverse.

The essence of Theorem 3.21 lies in its ability to transform derivative problems into simpler components, paving the way for further manipulations using identities such as the Pythagorean identity.

Pythagorean Identity

The Pythagorean identity is perhaps one of the most recognized equations in trigonometry, fundamental to solving a variety of problems:
\[\cos^2(y) + \sin^2(y) = 1\]
This identity relates the squares of the sine and cosine of an angle. Knowing this, you can deduce other trigonometric values when one is known.

• If \( \cos(y) = x \), then with some algebra, you derive \( \sin^2(y) = 1 - x^2 \).
• From there, \( \sin(y) \) can be expressed as \( \pm \sqrt{1 - x^2} \).

For the inverse cosine, since it spans angles from 0 to π, the sine is non-negative. Thus, \( \sin(y) = \sqrt{1 - x^2} \) when \( y = \cos^{-1}(x) \). This understanding allows us to compute derivatives more straightforwardly, especially for solving trigonometric identities.

Trigonometric Identities

Trigonometric identities are equations involving trigonometric functions that are true for every value of the variable. They are tools for transforming and simplifying mathematical expressions, especially vital in calculus.
One such identity used for inverse functions is:
\[\sin^{-1}(x) + \cos^{-1}(x) = \frac{\pi}{2}\]
This equation highlights a fundamental property of the arcsine and arccosine functions - they complement each other to sum up to \( \frac{\pi}{2} \), a right angle. This relationship simplifies differentiation problems by using one known derivative to find another.

• Differentiate \( \sin^{-1}(x) \) using standard theorems.
• Substitute it into the identity to solve for the derivative of \( \cos^{-1}(x) \).

These identities aren't just formulas but pathways to solutions, assisting with integration, differentiation, and even solving equations in geometry and physics.