Problem 4 - Calculus Early Transcendentals: Pearson New International Edition

Question

In Problems 1-18, find \(D_{x} y\). $$ y=1-\cos ^{2} x $$

Step-by-Step Solution

Answer

The derivative is \( D_x y = \sin 2x \).

Step 1: Apply Trigonometric Identity

Recall that the identity \( \sin^2 x + \cos^2 x = 1 \) can be rearranged to isolate \( \cos^2 x \). Therefore, \( 1 - \cos^2 x = \sin^2 x \). This allows us to rewrite the function \( y = 1 - \cos^2 x \) as \( y = \sin^2 x \).

Step 2: Differentiate Using Chain Rule

To find the derivative \( D_x y \) of \( y = \sin^2 x \), use the chain rule. Treat \( \sin x \) as the inner function, \( u(x) = \sin x \), and \( u^2 \) as the outer function. The derivative of \( u^2 \) is \( 2u \), and the derivative of \( \sin x \) is \( \cos x \).

Step 3: Apply the Chain Rule

Using the chain rule for \( y = \sin^2 x \), we have:\[D_x y = \frac{dy}{du} \cdot \frac{du}{dx} = 2\sin x \cdot \cos x = \sin 2x\].

Step 4: Present the Final Result

The derivative of \( y = 1 - \cos^2 x \) is \( D_x y = \sin 2x \).

Key Concepts

Trigonometric Identities

Trigonometric identities are mathematical equations that relate different trigonometric functions to one another. One of the most fundamental trigonometric identities is the Pythagorean identity: \( \sin^2 x + \cos^2 x = 1 \). This identity can be rearranged to solve for any of the trigonometric functions involved, providing various useful forms for simplification. In this exercise, we used this identity to rewrite the expression \( 1 - \cos^2 x \) as \( \sin^2 x \). By isolating \( \cos^2 x \), we get \( \cos^2 x = 1 - \sin^2 x \). Substituting back into the original function helps to express it in terms of a single trigonometric function, making differentiation easier later on. Understanding and applying these identities allow us to manipulate and simplify expressions involving trigonometric functions before performing calculus operations.

Chain Rule

The chain rule is a crucial tool in calculus for finding derivatives of composite functions. A composite function involves one function nested inside another, like \( y = (\sin x)^2 \) in the exercise. The chain rule states that the derivative of a composite function \( y = f(g(x)) \) can be found using the formula:

• \( \frac{dy}{dx} = \frac{dy}{du} \cdot \frac{du}{dx} \)

Here, \( y \) is expressed as a composition of an outer function \( f(u) = u^2 \) and an inner function \( g(x) = \sin x \). We first differentiate the outer function with respect to \( u \), resulting in \( 2u \), and then multiply by the derivative of the inner function \( g(x) = \sin x \), which is \( \cos x \). This application yields the final derivative of \( \sin^2 x \) as \( 2\sin x \cdot \cos x \), simplifying to \( \sin 2x \) using a double-angle formula. Practicing the chain rule helps you master differentiation of complex functions efficiently.

Calculus Differentiation

Calculus differentiation involves finding the rate at which a function changes at any given point, and it's a foundational concept in calculus. Differentiation uses rules and methods like the power rule, product rule, quotient rule, and chain rule to find derivatives of various types of functions. In this exercise, we focused on differentiating a function using the chain rule.

When you differentiate, you're essentially finding the slope at any point along a curve represented by a function. For our expression \( y = \sin^2 x \), the chain rule was employed to break down the composite nature of the function. This makes calculus differentiation not just about mechanics, but also about understanding and applying basic identities and rules effectively. Mastery of differentiation leads to insights into growth rates, slopes, and much more across different applications in mathematics and applied sciences. Keep practicing these techniques to become proficient in calculus and its diverse applications.