Problem 320
Question
For the following exercises, compute \(d y / d x\) by differentiating \(\ln y\). $$y=\sqrt{x^{2}+1}$$
Step-by-Step Solution
Verified Answer
\( \frac{dy}{dx} = \frac{x}{\sqrt{x^2 + 1}} \)
1Step 1: Express in terms of natural logarithm
Start by expressing the given function in terms of a natural logarithm so that differentiation becomes easier. We have \( y = \sqrt{x^2 + 1} \). Write it as \( \ln y = \ln(\sqrt{x^2 + 1}) \). By using the logarithmic identity, this becomes \( \ln y = \frac{1}{2} \ln(x^2 + 1) \).
2Step 2: Differentiate both sides with respect to x
Differentiate both sides of the equation \( \ln y = \frac{1}{2} \ln(x^2 + 1) \) with respect to \( x \). For the left side, use the chain rule: \( \frac{d}{dx}(\ln y) = \frac{1}{y} \cdot \frac{dy}{dx} \).For the right side, first derive \( \frac{1}{2} \ln(x^2 + 1) \): this becomes \( \frac{1}{2} \cdot \frac{1}{x^2+1} \cdot 2x = \frac{x}{x^2 + 1} \).This gives us the equation: \( \frac{1}{y} \cdot \frac{dy}{dx} = \frac{x}{x^2 + 1} \).
3Step 3: Solve for \( \frac{dy}{dx} \)
To isolate \( \frac{dy}{dx} \), multiply both sides of the equation by \( y \): \( \frac{dy}{dx} = y \cdot \frac{x}{x^2 + 1} \).Since \( y = \sqrt{x^2 + 1} \), substitute back to get:\( \frac{dy}{dx} = \sqrt{x^2 + 1} \cdot \frac{x}{x^2 + 1} \).This simplifies to:\( \frac{dy}{dx} = \frac{x}{\sqrt{x^2 + 1}} \).
Key Concepts
Natural LogarithmChain RuleDerivativesLogarithmic Differentiation
Natural Logarithm
The natural logarithm, denoted as \( \ln \), is the logarithm to the base \( e \), where \( e \approx 2.71828 \). It is a very common function in calculus because it has properties that make differentiation and integration easier.
For the function \( y = \sqrt{x^2 + 1} \), we can use logarithms to transform it into a simpler form. By applying \( \ln \) to both sides, we get \( \ln y = \ln(\sqrt{x^2 + 1}) \). The property that allows this transformation to be simplified is that the logarithm of a square root can be expressed as half the logarithm of the radicand. Thus, \( \ln(\sqrt{x^2 + 1}) = \frac{1}{2} \ln(x^2 + 1) \).
This transformation is particularly useful because it allows us to employ differentiation techniques more conveniently.
For the function \( y = \sqrt{x^2 + 1} \), we can use logarithms to transform it into a simpler form. By applying \( \ln \) to both sides, we get \( \ln y = \ln(\sqrt{x^2 + 1}) \). The property that allows this transformation to be simplified is that the logarithm of a square root can be expressed as half the logarithm of the radicand. Thus, \( \ln(\sqrt{x^2 + 1}) = \frac{1}{2} \ln(x^2 + 1) \).
This transformation is particularly useful because it allows us to employ differentiation techniques more conveniently.
Chain Rule
The chain rule is a fundamental method for finding derivatives. It is used when differentiating composite functions, which means a function inside another function.
In the problem, we have \( \ln y \), where \( y = \sqrt{x^2 + 1} \). To differentiate \( \ln y \) with respect to \( x \), you treat \( y \) as a function of \( x \). The differentiation of \( \ln y \) is performed using the chain rule: \( \frac{d}{dx}(\ln y) = \frac{1}{y} \cdot \frac{dy}{dx} \).
This simplifies the problem because it breaks the differentiation process into manageable parts, allowing you to focus first on differentiating \( \ln y \) and then linking it to the derivative of \( y \) itself.
In the problem, we have \( \ln y \), where \( y = \sqrt{x^2 + 1} \). To differentiate \( \ln y \) with respect to \( x \), you treat \( y \) as a function of \( x \). The differentiation of \( \ln y \) is performed using the chain rule: \( \frac{d}{dx}(\ln y) = \frac{1}{y} \cdot \frac{dy}{dx} \).
This simplifies the problem because it breaks the differentiation process into manageable parts, allowing you to focus first on differentiating \( \ln y \) and then linking it to the derivative of \( y \) itself.
Derivatives
Derivatives represent the rate of change of a function concerning its variables. They are one of the core concepts in calculus.
For this exercise, the derivative needs to be found for both sides of the logarithmic equation: \( \ln y = \frac{1}{2} \ln(x^2 + 1) \). When you differentiate \( \ln y \), you apply the chain rule as mentioned earlier. For \( \frac{1}{2} \ln(x^2 + 1) \), you find the derivative by using the rule for differentiating a logarithm, which is \( \frac{1}{x} \), adjusted for the function inside: \( \frac{d}{dx}\ln(u) = \frac{1}{u} \cdot \frac{du}{dx} \).
In this problem, \( u = x^2 + 1 \) and so \( \frac{du}{dx} = 2x \), leading to the final result \( \frac{x}{x^2 + 1} \) for the right side.
For this exercise, the derivative needs to be found for both sides of the logarithmic equation: \( \ln y = \frac{1}{2} \ln(x^2 + 1) \). When you differentiate \( \ln y \), you apply the chain rule as mentioned earlier. For \( \frac{1}{2} \ln(x^2 + 1) \), you find the derivative by using the rule for differentiating a logarithm, which is \( \frac{1}{x} \), adjusted for the function inside: \( \frac{d}{dx}\ln(u) = \frac{1}{u} \cdot \frac{du}{dx} \).
In this problem, \( u = x^2 + 1 \) and so \( \frac{du}{dx} = 2x \), leading to the final result \( \frac{x}{x^2 + 1} \) for the right side.
Logarithmic Differentiation
Logarithmic differentiation is a technique used when differentiating products, quotients, or powers of functions. By taking logarithms, it simplifies the expression and turns products into sums, quotients into differences, and powers into products again.
The process begins with converting the original function into a logarithmic form. Here, we used \( \ln y = \frac{1}{2} \ln(x^2 + 1) \), which simplifies the differentiation task. Then, by applying the chain rule and simplifying, we found the derivative.
This technique is beneficial when faced with complex multiplication or division within functions, or powers that are hard to handle directly. It transforms these expressions using the properties of logarithms, making them easier to differentiate.
The process begins with converting the original function into a logarithmic form. Here, we used \( \ln y = \frac{1}{2} \ln(x^2 + 1) \), which simplifies the differentiation task. Then, by applying the chain rule and simplifying, we found the derivative.
This technique is beneficial when faced with complex multiplication or division within functions, or powers that are hard to handle directly. It transforms these expressions using the properties of logarithms, making them easier to differentiate.
Other exercises in this chapter
Problem 319
For the following exercises, find the definite or indefinite integral. $$\int \frac{(\ln x)^{2} d x}{x}$$
View solution Problem 320
For the following exercises, compute dy/dx by differentiating ln y. $$y=\sqrt{x^{2}+1}$$
View solution Problem 321
For the following exercises, compute dy/dx by differentiating ln y. $$y=\sqrt{x^{2}+1} \sqrt{x^{2}-1}$$
View solution Problem 321
For the following exercises, compute \(d y / d x\) by differentiating \(\ln y\). $$y=\sqrt{x^{2}+1} \sqrt{x^{2}-1}$$
View solution