Problem 27 - Calculus Early Transcendentals: Pearson New International Edition

Question

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

Step-by-Step Solution

Answer

The derivative is \( \frac{dy}{dx} = \frac{2x + \cos x}{2\sqrt{x^2 + \sin x}} \).

Step 1: Understand the Derivative Problem

We need to find the derivative \( \frac{dy}{dx} \) of the function \( y = \sqrt{x^2 + \sin x} \). This involves using the chain rule because the function is a composition of the square root and another function, \( x^2 + \sin x \).

Step 2: Apply the Chain Rule

Let \( u = x^2 + \sin x \). The function becomes \( y = \sqrt{u} \), so according to the chain rule, the derivative \( \frac{dy}{dx} \) is \( \frac{dy}{du} \cdot \frac{du}{dx} \).

Step 3: Differentiate the Outer Function

Differentiate \( y = \sqrt{u} \) with respect to \( u \). We get \( \frac{dy}{du} = \frac{1}{2\sqrt{u}} \). Substitute \( u = x^2 + \sin x \) back, so \( \frac{dy}{du} = \frac{1}{2\sqrt{x^2 + \sin x}} \).

Step 4: Differentiate the Inner Function

Differentiate \( u = x^2 + \sin x \) with respect to \( x \). So we have \( \frac{du}{dx} = 2x + \cos x \).

Step 5: Combine the Results

Multiply \( \frac{dy}{du} \) by \( \frac{du}{dx} \) to get \( \frac{dy}{dx} = \frac{1}{2\sqrt{x^2 + \sin x}} \cdot (2x + \cos x) \). This can be simplified to \( \frac{2x + \cos x}{2\sqrt{x^2 + \sin x}} \).

Step 6: Simplify the Expression

The final derivative is \( \frac{dy}{dx} = \frac{2x + \cos x}{2\sqrt{x^2 + \sin x}} \), which cannot be simplified further.

Key Concepts

Chain Rule

The chain rule is a powerful tool in calculus used to find the derivative of composite functions. In simple terms, a composite function is like a nested function, where one function is inside another. To find the derivative of such a composition, we need to differentiate the outer function and multiply it by the derivative of the inner function.

Here's an analogy: Imagine you are peeling an onion. The outer layer represents the outer function, and the layers beneath represent the inner functions. The chain rule helps us peel back these layers systematically to understand how each affects the overall function.

In the given exercise, the function is \(y = \sqrt{x^2 + \sin x}\). The outer function is the square root, and the inner function is \(x^2 + \sin x\). By letting \(u = x^2 + \sin x\), we apply the chain rule to find \(\frac{dy}{dx}\) as \(\frac{dy}{du} \cdot \frac{du}{dx}\). This step-by-step peeling ensures we correctly capture how the change in \(x\) affects the whole function.

Differentiation

Differentiation is the process of finding the derivative of a function, which measures how a function's value changes as its input changes. It’s like finding the slope of a curve at any given point.

We start by differentiating the outer function, \(y = \sqrt{u}\), with respect to \(u\). The derivative is \(\frac{dy}{du} = \frac{1}{2\sqrt{u}}\). It's essential to remember this step as it gives us the rate of change of \(y\) with respect to \(u\).

Next, we differentiate the inner function, \(u = x^2 + \sin x\), with respect to \(x\), yielding \(\frac{du}{dx} = 2x + \cos x\). This tells us how the inner function changes, which is crucial for the chain rule.

By combining these derivatives, as per the chain rule, we arrive at the solution: \( \frac{dy}{dx} = \frac{2x + \cos x}{2\sqrt{x^2 + \sin x}} \). Differentiation breaks down the problem, enabling us to compute changes effectively and accurately.

Calculus

Calculus is the branch of mathematics that studies how things change. It provides tools like differentiation and integration to analyze dynamic systems. In this context, calculus helps us understand the relationship between a variable and its rate of change.

The exercise showcases how calculus allows us to model a complex function \(y = \sqrt{x^2 + \sin x}\) and investigate how it reacts to changes in \(x\). The derivative \(\frac{dy}{dx}\) is a fundamental concept of calculus, indicating how a small change in \(x\) causes a change in \(y\).

With the chain rule, differentiation becomes manageable, even for functions like the one given that involve both polynomial and trigonometric elements.

Understanding calculus through examples like this deepens our comprehension of the natural world. It explains phenomena ranging from simple motion to complex systems in engineering and economics. It turns abstract concepts into tangible problems, offering solutions that apply to real-world situations.