Problem 52 - Calculus - AP Edition

Question

Find \(d y / d x\). $$y=\ln \left(\cos ^{-1} x\right)$$

Step-by-Step Solution

Answer

\( \frac{dy}{dx} = \frac{-1}{\cos^{-1}(x) \sqrt{1-x^2}} \)

Step 1: Identify the Derivative Rule Needed

The function is given as \( y = \ln(\cos^{-1}(x)) \). Since it is a composition of two functions (a logarithmic function and an inverse trigonometric function), we'll apply the chain rule to find the derivative.

Step 2: Differentiate the Outer Function

First, differentiate the outer function \( \ln(u) \) with respect to \( u \), which gives \( \frac{d}{du} \ln(u) = \frac{1}{u} \). Substitute the inner function, \( \cos^{-1}(x) \), for \( u \). This yields \( \frac{1}{\cos^{-1}(x)} \).

Step 3: Differentiate the Inner Function

Now, differentiate the inner function \( \cos^{-1}(x) \) with respect to \( x \). The derivative is \( \frac{d}{dx} \cos^{-1}(x) = -\frac{1}{\sqrt{1-x^2}} \).

Step 4: Apply the Chain Rule

Using the chain rule, \( \frac{dy}{dx} = \frac{dy}{du} \cdot \frac{du}{dx} \), combine the derivatives obtained in Steps 2 and 3: \[ \frac{dy}{dx} = \frac{1}{\cos^{-1}(x)} \cdot \left( -\frac{1}{\sqrt{1-x^2}} \right) \].

Step 5: Simplify the Expression

Simplify the expression. The derivative of \( y = \ln(\cos^{-1}(x)) \) is \( \frac{-1}{\cos^{-1}(x) \sqrt{1-x^2}} \).

Key Concepts

Inverse Trigonometric Functions

Inverse trigonometric functions are functions that reverse the action of the standard trigonometric functions such as sine, cosine, and tangent. For instance, while the cosine function takes an angle and gives you its cosine value, the inverse cosine function, denoted as \( \cos^{-1}(x) \), does the opposite. It takes a cosine value and provides the corresponding angle.These functions are important when you need to solve equations involving trigonometric functions because they allow you to determine the angles. However, they have specific domains and ranges:

• For \( \cos^{-1}(x) \), the domain is \([-1, 1]\), and the range is \([0, \pi]\).
• Each inverse trigonometric function has a derivative, which is useful for calculus.

Understanding how these derivatives work is vital for finding rates of change, as we saw in the exercise. In this case, the derivative of \( \cos^{-1}(x) \) is essential when applying the chain rule.

Logarithmic Differentiation

Logarithmic differentiation is a technique utilized to simplify the differentiation process, especially when dealing with complex functions. It can often make the process much easier by transforming multiplicative relationships into additive ones through the use of logarithms.Here's how it generally works:

• First, you take the natural logarithm of both sides of the equation, if applicable.
• The properties of logarithms, such as \( \ln(ab) = \ln(a) + \ln(b) \), help to break down the function.

In the context of the exercise, we had \( y = \ln(\cos^{-1}(x)) \). Although we directly differentiated \( \ln(\cos^{-1}(x)) \), logarithmic differentiation helps handle complex expressions inside a log function, changing the way we approach the problem and making the differentiation process clearer.The core idea is to transform difficulties into manageable pieces. Often, logarithmic differentiation provides the clarity needed for intricate problems.

Derivatives

Derivatives are a fundamental concept in calculus, representing the rate at which a quantity changes. In simpler terms, they can be thought of as the slope of a function at a particular point.Calculating derivatives can be straightforward or complicated, depending on the function. For instance, with basic power functions like \( x^2 \), you apply simple rules. However, for composite functions such as \( \ln(\cos^{-1}(x)) \), we use more advanced techniques like the chain rule:

• The chain rule is used when differentiating composed functions, where you differentiate the outer function and then the inner function.
• In our example, \( y = \ln(\cos^{-1}(x)) \), we start by differentiating \( \ln(u) \) which is \( \frac{1}{u} \), and we use \( u = \cos^{-1}(x) \).
• We then find the derivative of \( \cos^{-1}(x) \) and chain them together.

This process resulted in the combined derivative, leading to our final function derivative outcome. Through derivative calculations, we gain insights into how functions behave and change, which is fundamental in understanding motion, growth, and many other phenomena.