Problem 283 - Calculus Volume 1

Question

For the following exercises, find \(\frac{d y}{d x}\) for the given function. $$y=\left(1+\tan ^{-1} x\right)^{3}$$

Step-by-Step Solution

Answer

The derivative is \(\frac{dy}{dx} = \frac{3(1 + \tan^{-1}x)^2}{1+x^2}\)."

Step 1: Identify the Function Type

The given function is a composition of functions, specifically a power function of an inverse trigonometric function. We have \(y = (1 + \tan^{-1}x)^{3}\). Our goal is to differentiate this with respect to \(x\).

Step 2: Apply the Chain Rule

To differentiate the function \(y = (1 + \tan^{-1}x)^{3}\), we apply the chain rule. Let \(u = 1 + \tan^{-1}x\), then \(y = u^3\). According to the chain rule, \(\frac{dy}{dx} = \frac{dy}{du} \times \frac{du}{dx}\).

Step 3: Differentiate the Outer Function

First, differentiate the outer function with respect to \(u\). For \(y = u^3\), the derivative is \(\frac{dy}{du} = 3u^2\).

Step 4: Differentiate the Inner Function

Next, differentiate the inner function \(u = 1 + \tan^{-1}x\) with respect to \(x\). The derivative is \(\frac{du}{dx} = \frac{d}{dx}(1 + \tan^{-1}x) = \frac{1}{1+x^2}\), since the derivative of \(\tan^{-1}x\) is \(\frac{1}{1+x^2}\).

Step 5: Combine Results Using the Chain Rule

Substitute the results from Steps 3 and 4 into the expression from Step 2. Thus, \(\frac{dy}{dx} = 3u^2 \times \frac{1}{1+x^2}\). Since \(u = 1 + \tan^{-1}x\), we substitute back to get \(\frac{dy}{dx} = \frac{3(1 + \tan^{-1}x)^2}{1+x^2}\).

Key Concepts

Chain Rule

When dealing with complex functions made up of multiple layers, the chain rule is an essential differentiation tool. It allows us to tackle composite functions, which are functions inside other functions. In simple terms, the chain rule states that the derivative of the composite function is the derivative of the outer function evaluated at the inner function, multiplied by the derivative of the inner function.

Imagine breaking down a task into manageable steps, where you first solve a small part, and then use that result to solve the whole. This is what applying the chain rule is like in calculus.

• Outer Function: The function that encompasses another function. In our exercise, this is \(y = u^3\).
• Inner Function: The function inside another function. Here, it's \(u = 1 + \tan^{-1} x\).

By using the chain rule, we link these functions and find the overall rate of change by calculating \(\frac{dy}{du} \ imes \frac{du}{dx}\). It simplifies differentiating complex compositions without reinventing the wheel each time for every function involved.

Inverse Trigonometric Functions

Inverse trigonometric functions take us from a ratio back to an angle. They reverse what the standard trigonometric functions do. These functions, like \(\tan^{-1} x\), are of particular importance when we need to find an angle when the ratio is known.

The derivative of these functions can be slightly trickier than regular algebraic functions. For \(\tan^{-1} x\), its derivative is \(\frac{1}{1+x^2}\). This relationship harks back to the geometry underlying trigonometry, where trigonometric identities give us the tools to express rates of change in a manageable form.

• Makes angles accessible from ratios.
• Derivative of \(\tan^{-1} x \) is unique and non-intuitive: \(\frac{1}{1+x^2}\).
• Used in problems where angles need to be determined from geometric or trigonometric relationships.

Recognizing these functions and their derivatives lets us bridge gaps between angles and their logs, which appear across various application contexts, from engineering to pure math.

Derivatives

In calculus, derivatives represent change. They show how one quantity changes in response to another. With differentiation, we can examine whether a function is increasing or decreasing at any point, and it tells us the rate at which this occurs.

A derivative such as \(\frac{dy}{dx}\) shows how the variable \(y\) changes in response to changes in \(x\). Differentiation isn't limited to simple functions—it scales to complex compositions through rules like the chain rule.

• Gives us the slope of the function at any point.
• Shows how one variable changes with another.
• Useful in analyzing the behavior of functions.

The derivative of a function is emblematic of a fundamental calculus idea: understanding dynamic systems. Whether calculating speed, optimizing systems, or sketching curves, derivatives are your go-to calculus tool, unlocking understanding of how functions behave and interact.