Problem 58 - Calculus - AP Edition

Question

Find \(d y / d x\). $$y=\sqrt{\cot ^{-1} x}$$

Step-by-Step Solution

Answer

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

Step 1: Identify the Composite Function

Recognize that we have a composite function where \( y = \sqrt{u} \) and \( u = \cot^{-1}(x) \). The outer function is the square root, and the inner function is the inverse cotangent.

Step 2: Differentiate the Outer Function

Use the chain rule to differentiate \( y = \sqrt{u} \). The derivative of \( \sqrt{u} \) with respect to \( u \) is \( \frac{1}{2\sqrt{u}} \).

Step 3: Differentiate the Inner Function

Differentiate \( u = \cot^{-1}(x) \) with respect to \( x \). The derivative \( \frac{d}{dx}[\cot^{-1}(x)] \) is \( -\frac{1}{1+x^2} \).

Step 4: Apply the Chain Rule

Apply the chain rule: \( \frac{dy}{dx} = \frac{dy}{du} \cdot \frac{du}{dx} \). Substitute the derivatives, resulting in \( \frac{dy}{dx} = \frac{1}{2\sqrt{\cot^{-1}(x)}} \cdot \left(-\frac{1}{1+x^2}\right) \).

Step 5: Simplify the Expression

Simplify \( \frac{dy}{dx} = -\frac{1}{2(1+x^2)\sqrt{\cot^{-1}(x)}} \).

Key Concepts

Chain Rule

The chain rule is a fundamental concept in calculus differentiation. It's especially useful when dealing with composite functions, which are functions that nest within each other. This rule tells us how to take the derivative of a composite function by differentiating the outer function first and then the inner function. Here's how you can think about it:

• Imagine you have a function inside another function, such as \( y = f(g(x)) \).
• The chain rule states that the derivative \( \frac{dy}{dx} \) is found by multiplying the derivative of the outer function \( f \) evaluated at the inner function \( g(x) \) by the derivative of the inner function \( g \) with respect to \( x \), or \( \frac{dy}{dx} = f'(g(x)) \cdot g'(x) \).

The power of the chain rule lies in its ability to make differentiating complicated nested functions manageable. When you apply it, be sure to keep track of each function and its derivative.

Inverse Trigonometric Functions

Inverse trigonometric functions allow us to work backward from values of a trigonometric function to find angles. In our exercise, we deal with \( \cot^{-1}(x) \), which represents the angle whose cotangent is \( x \). It’s important to remember the derivative properties of these functions:

• For \( \cot^{-1}(x) \), its derivative with respect to \( x \) is \( -\frac{1}{1+x^2} \).

Understanding the behavior of inverse trigonometric functions is crucial, as they often appear in calculus problems involving angles and their relationships.
When differentiating these functions, you must use their specific derivative formulas since they do not follow the standard polynomial rules. Thus, memorizing or having a reference for these derivatives can be extremely helpful.

Derivative of Composite Functions

To find the derivative of composite functions, like the one in our original exercise \( y = \sqrt{\cot^{-1}(x)} \), we combine our understanding of derivatives and the chain rule. Let's break it down:

• First, we recognize \( y = \sqrt{\cot^{-1}(x)} \) is a composite function, meaning it’s composed of an outer function \( y = \sqrt{u} \) and an inner function \( u = \cot^{-1}(x) \).
• The derivative of \( \sqrt{u} \) with respect to \( u \) is \( \frac{1}{2\sqrt{u}} \), while the derivative of \( \cot^{-1}(x) \) with respect to \( x \) is \( -\frac{1}{1+x^2} \).

Putting it all together, we apply the chain rule: multiply the derivative of the outer function by the derivative of the inner function.
The final result, after simplifying the expression, gives a clean form for the derivative \( \frac{dy}{dx} = -\frac{1}{2(1+x^2)\sqrt{\cot^{-1}(x)}} \).
This process showcases how a multi-step operation can be broken down into manageable parts, thanks to the framework the chain rule provides.