Problem 27 - Calculus

Question

Compute \(d y / d x\) by differentiating \(\ln y .\) This is LD: $$ y=\sqrt{x^{2}+1} $$

Step-by-Step Solution

Answer

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

Step 1: Differentiate Using Chain Rule

We start by differentiating the given function. Since we have the logarithm of a function, we use the chain rule for differentiation. Given function is \(y = \sqrt{x^2 + 1}\). Taking the natural logarithm of both sides, we get \(\ln y = \ln(\sqrt{x^2 + 1})\).

Step 2: Simplify the Logarithmic Expression

We simplify the expression using logarithmic identities. \(\ln(\sqrt{x^2 + 1}) = \ln((x^2 + 1)^{1/2})\) simplifies to \((1/2) \ln(x^2 + 1)\). Thus, we have \(\ln y = \frac{1}{2} \ln(x^2 + 1)\).

Step 3: Differentiate Both Sides

Differentiate both sides with respect to \(x\). The derivative of \(\ln y\) with respect to \(x\) is \(\frac{1}{y} \frac{dy}{dx}\). The derivative of \((1/2) \ln(x^2 + 1)\) with respect to \(x\) is \((1/2) \cdot \frac{1}{x^2 + 1} \cdot 2x\) due to the chain rule, which simplifies to \(\frac{x}{x^2 + 1}\).

Step 4: Equate and Solve for \(\frac{dy}{dx}\)

Set the derivatives from both sides equal: \(\frac{1}{y} \frac{dy}{dx} = \frac{x}{x^2 + 1}\). Solve for \(\frac{dy}{dx}\) by multiplying both sides by \(y\), resulting in \(\frac{dy}{dx} = y \cdot \frac{x}{x^2 + 1}\).

Step 5: Substitute \(y\) Back

Substitute back the original \(y\) value: \(y = \sqrt{x^2 + 1}\). Thus, \(\frac{dy}{dx} = \sqrt{x^2 + 1} \cdot \frac{x}{x^2 + 1}\), which simplifies to \(\frac{dy}{dx} = \frac{x \sqrt{x^2 + 1}}{x^2 + 1}\).

Key Concepts

Chain Rule in Differentiation

The chain rule is a fundamental technique used in calculus to find the derivative of compositions of functions. In essence, it helps us differentiate functions that are made up of other functions inside them.
For example, when you have a function like \(f(g(x))\), the chain rule allows you to differentiate it by taking the derivative of the outer function \(f\) while keeping the inner function \(g(x)\) intact, and then multiply it by the derivative of the inner function \(g(x)\).
In mathematical terms:

• \(\frac{d}{dx} [f(g(x))] = f'(g(x)) \cdot g'(x)\)

In our original problem, this concept comes into play when differentiating \(\ln(\sqrt{x^2 + 1})\). Here, \(\sqrt{x^2 + 1}\) is the inner function, and we apply the chain rule while differentiating it with respect to \(x\).
Remember, mastering the chain rule will greatly improve your ability to tackle more complex calculus problems that involve nested functions.

Logarithmic Differentiation

Logarithmic differentiation is a clever technique especially useful for differentiating complex functions, such as products, quotients, or powers involving variables.
This method involves three main steps:

• Take the natural logarithm of both sides of an equation \(y = f(x)\).
• Simplify the expression if possible using logarithmic identities.
• Diferentiate both sides of the logarithmic equation.

In our example, we applied logarithmic differentiation when dealing with \(y = \sqrt{x^2 + 1}\). By taking the natural logarithm, we could transform the root into a power that is easier to work with:
\( \ln y = \ln((x^2 + 1)^{1/2}) = \frac{1}{2} \ln(x^2 + 1) \).
This transformation makes differentiating much simpler and highlights the utility of logarithmic differentiation when encountering functions that are otherwise cumbersome to handle directly.

Understanding Derivatives

The derivative of a function is a critical concept in calculus that measures how a function changes with respect to changes in its variables. You can think of it as the slope of the tangent line to the function at any given point.
Derivatives help in:

• Identifying rates of change
• Finding maximum or minimum values (critical points)
• Analyzing the concavity of functions

In our exercise, after applying the chain rule to differentiate \(\ln y\), we set the equation for the derivatives equal and solved for \(\frac{dy}{dx}\).
By using derivatives, we find how the transformation of the function \(y = \sqrt{x^2 + 1}\) changes as \(x\) varies. This results in an expression \(\frac{dy}{dx}\) that signifies the rate of change of \(y\) with respect to \(x\). Understanding derivatives is fundamental for developing intuition about rate and change, an integral part of calculus.

Natural Logarithms and Their Properties

A natural logarithm, denoted \(\ln\), is the logarithm to the base \(e\), where \(e\) is approximately equal to 2.718.
Natural logarithms have many useful properties that make them indispensable in calculus. They often simplify the process of differentiation and integration.Some properties include:

• \(\ln(ab) = \ln(a) + \ln(b)\) - the logarithm of a product is the sum of the logarithms.
• \(\ln\left(\frac{a}{b}\right) = \ln(a) - \ln(b)\) - the logarithm of a quotient is the difference of the logarithms.
• \(\ln(a^b) = b \cdot \ln(a)\) - the logarithm of a power is the exponent times the logarithm of the base.

In our example, using the property of logarithms, we transformed the expression \(\ln(\sqrt{x^2 + 1})\) into \(\frac{1}{2} \ln (x^2 + 1)\), making differentiation straightforward.
The understanding of natural logarithms and their properties greatly enhances problem-solving skills in calculus, making complex differentiations much simpler.