Problem 50 - Calculus - AP Edition

Question

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

Step-by-Step Solution

Answer

\( \frac{dy}{dx} = -\frac{1}{(\tan^{-1}(x))^2 \cdot (1+x^2)} \).

Step 1: Understand the Problem

We need to find the derivative of the function \( y = \frac{1}{\tan^{-1}(x)} \) with respect to \( x \). This requires using the chain rule and the derivative of the inverse tangent function.

Step 2: Apply the Chain Rule

The function is a composition of \( u = \tan^{-1}(x) \) and \( y = \frac{1}{u} \). The chain rule states that \( \frac{dy}{dx} = \frac{dy}{du} \cdot \frac{du}{dx} \).

Step 3: Differentiate \( y = \frac{1}{u} \) Respect to \( u \)

We have \( y = u^{-1} \). The derivative \( \frac{dy}{du} = -u^{-2} = -\frac{1}{u^2} \).

Step 4: Differentiate \( u = \tan^{-1}(x) \) with Respect to \( x \)

The derivative of \( u = \tan^{-1}(x) \) is \( \frac{du}{dx} = \frac{1}{1+x^2} \).

Step 5: Combine the Results Using the Chain Rule

Using the chain rule, \( \frac{dy}{dx} = \frac{dy}{du} \cdot \frac{du}{dx} = -\frac{1}{u^2} \cdot \frac{1}{1+x^2} \).

Step 6: Substitute Back \( u = \tan^{-1}(x) \)

Now we replace \( u \) with \( \tan^{-1}(x) \): \( \frac{dy}{dx} = -\frac{1}{(\tan^{-1}(x))^2 \cdot (1+x^2)} \).

Key Concepts

Chain Rule

The Chain Rule is a key concept in calculus used to find the derivative of composite functions. This means functions that are nested within each other, like in our example where we have the function \( y = \frac{1}{\tan^{-1}(x)} \). Applying the Chain Rule involves:

• Identifying the inner and outer functions: The inner function is \( u = \tan^{-1}(x) \), and the outer function \( y = \frac{1}{u} \).
• Taking the derivative of the outer function with respect to the inner function. Here, you imagine \( u \) as a single variable and find \( \frac{dy}{du} \).
• Taking the derivative of the inner function with respect to the actual variable \( x \), finding \( \frac{du}{dx} \).

Finally, you multiply these derivatives together to get the overall derivative: \( \frac{dy}{dx} = \frac{dy}{du} \cdot \frac{du}{dx} \). This allows us to differentiate complex compositions effectively by dealing with smaller, more manageable parts.

Inverse Trigonometric Functions

Inverse Trigonometric Functions are important in calculus because they help us solve equations involving trigonometric expressions. These functions are essentially the reverse of regular trigonometric functions. For example, if \( y = \tan^{-1}(x) \), it means \( x = \tan(y) \).When working with derivatives, it is crucial to recall the standard derivative formulas of inverse trigonometric functions:

• \( \frac{d}{dx}[\tan^{-1}(x)] = \frac{1}{1+x^2} \)
• Similarly, \( \frac{d}{dx}[\sin^{-1}(x)] = \frac{1}{\sqrt{1-x^2}} \)
• \( \frac{d}{dx}[\cos^{-1}(x)] = -\frac{1}{\sqrt{1-x^2}} \)

These formulas provide the tools needed to find the derivative of inverse trigonometric expressions within broader equations. Understanding how these derivatives work allows you to handle problems involving inverse trigonometric functions effectively.

Derivative Calculation

Calculating derivatives is a fundamental skill in calculus used to assess how a function changes. In this problem, calculating the derivative of the given function involves several clear steps:First, recognize the components of the function and decide on the rules needed, like the Chain Rule and standard derivative rules for inverse trigonometric functions. Starting with the function \( y = \frac{1}{\tan^{-1}(x)} \), we set \( u = \tan^{-1}(x) \). Calculating \( \frac{dy}{du} \) with the formula \( -u^{-2} \) gives \( -\frac{1}{u^2} \). Then for \( \frac{du}{dx} \), use the derivative of \( \tan^{-1}(x) \), which is \( \frac{1}{1+x^2} \).Finally, substitute these results into the Chain Rule equation \( \frac{dy}{dx} = \frac{dy}{du} \cdot \frac{du}{dx} \). Ultimately, substitute back \( u = \tan^{-1}(x) \) to finalize the calculation, resulting in \( \frac{dy}{dx} = -\frac{1}{(\tan^{-1}(x))^2 \cdot (1+x^2)} \). This methodical approach streamlines derivative calculation and enhances understanding.