Problem 32 - Calculus Early Transcendentals: Pearson New International Edition

Question

Find \(d y / d x\). \(y=\sqrt{\tan ^{2} x+\sin ^{2} x}\)

Step-by-Step Solution

Answer

\( \frac{\tan x \sec^2 x + \sin 2x}{\sqrt{\tan^2 x + \sin^2 x}} \)

Step 1: Recognize the Function Type

The given function \( y = \sqrt{\tan^2 x + \sin^2 x} \) is a composite function where the outer function is a square root and the inner function is \( \tan^2 x + \sin^2 x \).

Step 2: Apply Chain Rule

To differentiate \( y \) with respect to \( x \), we apply the chain rule. Start by letting \( u = \tan^2 x + \sin^2 x \), so that \( y = \sqrt{u} = u^{1/2} \). The chain rule states \( \frac{dy}{dx} = \frac{dy}{du} \cdot \frac{du}{dx} \).

Step 3: Differentiate Outer Function

Differentiate the outer function \( y = u^{1/2} \) with respect to \( u \). The derivative is \( \frac{dy}{du} = \frac{1}{2} u^{-1/2} = \frac{1}{2\sqrt{u}} \).

Step 4: Differentiate Inner Function

Differentiate the inner function \( u = \tan^2 x + \sin^2 x \) with respect to \( x \). Notice that \( \tan^2 x = (\tan x)^2 \), so use the chain rule and derivative of \( \tan x \): \( \frac{d}{dx}[\tan x] = \sec^2 x \). The derivative is \( \frac{du}{dx} = 2\tan x \sec^2 x + 2\sin x \cos x \).

Step 5: Simplify the Result

Substitute back the expressions found: \( \frac{dy}{dx} = \frac{1}{2\sqrt{u}} \cdot (2\tan x \sec^2 x + 2\sin x \cos x) \). Replacing \( u \) with \( \tan^2 x + \sin^2 x \), the final expression becomes \[ \frac{dy}{dx} = \frac{\tan x \sec^2 x + \sin 2x}{\sqrt{\tan^2 x + \sin^2 x}} \].

Key Concepts

Chain Rule

The chain rule is an essential concept in calculus that helps us differentiate composite functions effectively. When a function is composed of two or more functions, the chain rule allows us to take the derivative of the composite function by using the derivatives of its components.

In simple terms, if you have a function that's made up of an inner function and an outer function, like in this exercise, the chain rule tells us to:

• First, differentiate the outer function with respect to the inner function.
• Then, multiply that result by the derivative of the inner function.

Consider the step where we expressed the function as \(y = u^{1/2}\) with \(u = \tan^2 x + \sin^2 x\). By applying the chain rule, we first differentiate the outer part (square root) with respect to \(u\), and then multiply by the derivative of the inner part \(u\) with respect to \(x\).

This approach is crucial for working through complex functions, especially when dealing with combinations of trigonometric functions and algebraic operations.

Composite Function

Composite functions are functions composed of two or more simpler functions. In shorthand, if \(f(x)\) and \(g(x)\) are functions, then the composition \(f(g(x))\) is a composite function.

In our exercise, \(y = \sqrt{\tan^2 x + \sin^2 x}\) is a composite function where the outer function is the square root and the inner function is \(\tan^2 x + \sin^2 x\). Handling these functions often involves identifying the layers of functions first, ensuring we apply the correct differentiation techniques, such as the chain rule, on each layer.

Recognizing and understanding composite functions allows us to simplistically break down and solve complex problems.

• The outer layer, here represented by the square root, dictates the general behavior of the function.
• The inner layer, combining various trigonometric expressions, introduces specific characteristics that need careful handling during differentiation.

This layering structure is essential for applying the correct rules for differentiation and integration.

Trigonometric Functions

Trigonometric functions like \(\sin\), \(\cos\), and \(\tan\) provide important tools in calculus and play a significant role in solving this problem. They appear often in physics and engineering, where wave and oscillating patterns must be described mathematically.

In this problem, we differentiate \(\tan^2 x\) and \(\sin^2 x\) by applying rules specific to trigonometric derivatives. It helps to remember:

• The derivative of \(\tan x\) is \(\sec^2 x\).
• The chain rule is used when differentiating powers of these functions, as seen with \(\tan^2 x\) requiring the derivative of \(\tan x\) multiplied by \(2\tan x\).

Applying these principles enabled us to solve the derivative of the inner function \(u = \tan^2 x + \sin^2 x\), leading to the final expression for \(\frac{du}{dx}\), which we then used in the chain rule.

It's critical to understand how to differentiate each trigonometric function accurately, as this knowledge is required for almost all calculus problems involving waves and cyclic behavior.