Problem 13 - Calculus Early Transcendentals

Question

Use Version I of the Chain Rule to calculate \(\frac{d y}{d x}\). $$y=\sqrt{x^{2}+1}$$

Step-by-Step Solution

Answer

Question: Find the derivative of y with respect to x for the function $$y=\sqrt{x^2+1}$$. Answer: The derivative of y with respect to x for the given function is $$\frac{dy}{dx} = \frac{x}{\sqrt{x^2+1}}$$.

Step 1: Define the functions u and v

Given the function $$y=\sqrt{x^2+1}$$, we can express this as $$y=u(v(x))$$, where $$u(x)=\sqrt{x}$$ and $$v(x)= x^2 + 1$$.

Step 2: Find the derivatives \(\frac{dy}{du}\) and \(\frac{du}{dx}\)

We need to find the derivatives of u and v with respect to x: $$\frac{du}{dv} = \frac{d}{dv} (\sqrt{v}) = \frac{1}{2 \sqrt{v}}$$ $$\frac{dv}{dx} = \frac{d}{dx} (x^2+1) = 2x$$

Step 3: Apply the Chain Rule

To find \(\frac{dy}{dx}\), we will use the chain rule formula: $$\frac{dy}{dx} = \frac{dy}{du} \cdot \frac{du}{dx}$$ Plugging in the derivatives we found in step 2: $$\frac{dy}{dx} = \frac{1}{2 \sqrt{v}} \cdot 2x$$

Step 4: Substitute v(x) and simplify

Replace v with its definition in terms of x to get the final expression for \(\frac{dy}{dx}\): $$\frac{dy}{dx} = \frac{1}{2 \sqrt{x^2+1}} \cdot 2x$$ Simplifying the expression, we get: $$\frac{dy}{dx} = \frac{x}{\sqrt{x^2+1}}$$ So, the derivative of y with respect to x is: $$\frac{dy}{dx} = \frac{x}{\sqrt{x^2+1}}$$

Key Concepts

Calculus

Calculus is a branch of mathematics that deals with continuous change. It is all about understanding how things change, which is crucial for solving problems in physics, engineering, and economics, among other fields. In calculus, we study two main concepts: differentiation and integration.
One of calculus's primary goals is to find the slope of a curve at a particular point on a graph. This involves finding the derivative, which will be discussed in more detail later. The ability to calculate derivatives allows us to understand how a function changes as its input changes.
Calculus also involves integral calculus, which is about finding the area beneath a curve. While differentiation is about finding how fast something is changing, integration gives us a summary of how much has changed over an interval. Together, these concepts form the foundation of calculus and are essential tools in the field.

Derivative

A derivative represents the rate of change of a function concerning one of its variables. If we think of a function as a process that turns input into output, then the derivative tells us how the output changes as we tweak the input slightly. It's like measuring the speed of a car: the derivative is the car’s speed, telling us how fast the position changes over time.
In mathematical terms, if we have a function \(f(x)\), its derivative is sometimes written as \(f'(x)\) or \(\frac{df}{dx}\). The derivative is found by applying rules or formulas, such as the Power Rule, Product Rule, or Chain Rule.
The Chain Rule, used in the original exercise, is especially powerful and important when dealing with composite functions - functions built from two or more functions. It allows us to "chain" the derivatives of functions together, unpacking complicated functions step by step. In the case of \(y = \sqrt{x^2 + 1}\), the Chain Rule was used to differentiate it effectively.

Function Composition

Function composition involves combining two functions so that the output of one function becomes the input of another. Suppose we have two functions, \(f(x)\) and \(g(x)\); their composition is denoted as \((f \circ g)(x) = f(g(x))\). This composition allows for more complex function creation, helping craft new functions that solve diverse problems.
When dealing with derivatives, function composition becomes crucial because it defines scenarios where the Chain Rule must be applied. Consider the function \(y = \sqrt{x^2 + 1}\) from the exercise. It can be viewed as the composition of two simpler functions: one that squares \(x\) and adds 1, and another that takes the square root of its input.
By recognizing function composition in such problems, we can systematically apply the Chain Rule to find derivatives. This structured approach is pivotal in understanding how complex functions behave and paves the way for more advanced calculus topics.