Question

For the following exercises, compute \(d y / d x\) by differentiating \(\ln y\). $$ y=\sqrt{x^{2}+1} $$

Short answer

\( \frac{dy}{dx} = \frac{x}{\sqrt{x^2 + 1}} \)

Step-by-step solution

Identify Original Function

We are given the function \( y = \sqrt{x^2 + 1} \). Our goal is to find \( \frac{dy}{dx} \).

Apply Natural Logarithm

Take the natural logarithm on both sides: \( \ln y = \ln(\sqrt{x^2 + 1}) \). Using the property of logarithms, \( \ln(a^b) = b \cdot \ln a \), this simplifies to \( \ln y = \frac{1}{2} \ln(x^2 + 1) \).

Differentiate Both Sides

Differentiate both sides with respect to \( x \). The derivative of \( \ln y \) is \( \frac{1}{y} \frac{dy}{dx} \) by implicit differentiation. The derivative of \( \frac{1}{2} \ln(x^2 + 1) \) is \( \frac{1}{2} \cdot \frac{1}{x^2 + 1} \cdot 2x \).

Simplify the Right Side of the Equation

The derivative on the right side from the previous step gives us: \( \frac{x}{x^2 + 1} \).

Solve for \( \frac{dy}{dx} \)

Using the result from differentiating, we have: \( \frac{1}{y} \frac{dy}{dx} = \frac{x}{x^2 + 1} \). Multiply both sides by \( y \) to solve for \( \frac{dy}{dx} \): \( \frac{dy}{dx} = y \cdot \frac{x}{x^2 + 1} \). Replace \( y \) with \( \sqrt{x^2 + 1} \): \(\frac{dy}{dx} = \sqrt{x^2 + 1} \cdot \frac{x}{x^2 + 1} = \frac{x}{\sqrt{x^2 + 1}} \).

Key concepts

Natural Logarithm

The natural logarithm, often denoted as \( \ln \), is a fundamental concept in calculus and mathematics, especially when dealing with functions involving exponentiation. Natural logarithms refer to logarithms with base \( e \), where \( e \) is approximately 2.718. This goes all the way back to compounded growth processes in natural systems. Using \( \ln \) makes computations particularly friendly due to its simple derivative properties.

In our exercise, we employ the property of logarithms to simplify the expression \( \ln(\sqrt{x^2 + 1}) \). The logarithmic identity \( \ln(a^b) = b \cdot \ln a \) is extremely useful here. This allows us to transform the logarithm of a square root into a product of a simpler form: \( \frac{1}{2} \ln(x^2 + 1) \).

Such a transformation is crucial when preparing to take derivatives, as it simplifies the function to something directly manageable when invoking the rules of differentiation.

Derivative

The derivative is a fundamental tool in calculus that measures how a function changes as its input changes. Essentially, it tells us the rate of change or the slope of a function at any given point. In our problem, we need the derivative \( \frac{dy}{dx} \) where \( y = \sqrt{x^2 + 1} \).

• By applying implicit differentiation to \( \ln y \), we acknowledge that differentiating means finding \( \frac{d}{dx} (\ln y) = \frac{1}{y} \frac{dy}{dx} \). This concept is anchored in the chain rule, which we will explore further.
• For the expression \( \frac{1}{2} \ln(x^2 + 1) \), we perform the differentiation to get \( \frac{1}{2} \cdot \frac{1}{x^2 + 1} \cdot 2x \), reducing it to \( \frac{x}{x^2 + 1} \).

This derivative simplifies our computation and ultimately allows us to solve for \( \frac{dy}{dx} \). Differentiation, especially within the context of implicit means, provides the backbone of many calculations in calculus, capturing the subtle nuances of function behavior.

Chain Rule

The chain rule is a pivotal principle when dealing with compositions of functions, allowing us to differentiate complex expressions by breaking them down into simpler parts. For our current problem, it's an essential strategy when differentiating the result of applying \( \ln \) to function \( y \).

To use the chain rule, you first identify and differentiate the "outer" function, followed by the "inner" function. The rule instructs us to multiply these derivatives together.

In the context of our exercise:

• The derivative of \( \ln y \) is \( \frac{1}{y} \) (our outer function derivative) multiplied by \( \frac{dy}{dx} \) (derivative of the inner function \( y \)).
• Similarly, for \( \ln(x^2 + 1) \), treating \( x^2 + 1 \) as the inside function, when applying the derivative, we multiply the result \( \frac{1}{x^2 + 1} \) by the derivative of \( 2x \).

The chain rule weaves through our solution, ensuring each differentiation segment correctly adheres to the function composition rules, resulting in \( \frac{x}{x^2 + 1} \) for the transformed expression. Mastering this rule is crucial for tackling the broad spectrum of derivative challenges in calculus.