Question

$$ \text { } , \text { find the indicated derivative. } $$ $$ g^{\prime}(x) \text { if } g(x)=\ln \left(x+\sqrt{x^{2}+1}\right) $$

Short answer

The derivative is \( g'(x) = \frac{1}{\sqrt{x^2 + 1}} \).

Step-by-step solution

Identify the Derivative Formula

First, recognize that the function you have, \( g(x) = \ln(x + \sqrt{x^2 + 1}) \), is a combination of the natural logarithm and a composite function. We will use the chain rule for differentiation along with the derivative of the natural logarithm function, which is \( \frac{d}{dx} \ln(u) = \frac{1}{u} \frac{du}{dx} \).

Differentiate the Inside Function

The inside function is \( u = x + \sqrt{x^2 + 1} \). Start finding its derivative by differentiating each part separately. Note that \( \frac{d}{dx}(x) = 1 \) and for the \( \sqrt{x^2 + 1} \), use the chain rule: \( \frac{d}{dx}(\sqrt{x^2 + 1}) = \frac{1}{2\sqrt{x^2 + 1}} \cdot 2x = \frac{x}{\sqrt{x^2 + 1}} \). Thus, \( \frac{du}{dx} = 1 + \frac{x}{\sqrt{x^2 + 1}} \).

Apply the Derivative Formula

Now apply the chain rule: \( g'(x) = \frac{1}{u} \cdot \frac{du}{dx} \). Substitute \( u \) back in, which gives \( g'(x) = \frac{1}{x + \sqrt{x^2 + 1}} \cdot \left(1 + \frac{x}{\sqrt{x^2 + 1}}\right) \).

Simplify the Expression

To simplify the derivative, combine the terms: \( g'(x) = \frac{1 \cdot \sqrt{x^2 + 1} + x}{\sqrt{x^2 + 1} \cdot (x + \sqrt{x^2 + 1})} \). Simplified further, \( g'(x) = \frac{\sqrt{x^2 + 1} + x}{x\sqrt{x^2 + 1} + x^2 + 1}\). Further simplification leads to a basic solution form: \( g'(x) = \frac{1}{\sqrt{x^2 + 1}} \).

Verify the Simplification

Check that each part is correctly simplified and combined. Review that \( g(x) \) and thus \( g'(x) = \frac{1}{\sqrt{x^2 + 1}} \) properly maintain logical structure when reduced.

Key concepts

chain rule

The "chain rule" is a crucial technique in calculus. It is used to derive the derivative of a composite function. A composite function is a function made up of two or more simpler functions. For example, in our original problem, we have the function \(g(x) = \ln(x + \sqrt{x^2 + 1})\), which combines the natural logarithm with another expression.
This expression itself includes the sum \(x + \sqrt{x^2 + 1}\), involving the square root. The chain rule allows us to "break" the composite function into parts to find the derivative. It states that if a function \(y = f(g(x))\) is composed of two functions \(f\) and \(g\), the derivative is \(\frac{dy}{dx} = f'(g(x)) \cdot g'(x)\).

When using the chain rule:

• Identify the outer function \(f\) and the inner function \(g\).
• Find the derivative of the outer function, considering the inner function as a whole.
• Differentiate the inner function separately.
• Multiply these derivatives together to get the overall derivative.

With practice, applying the chain rule will become second nature when dealing with complex functions.

natural logarithm differentiation

The "natural logarithm differentiation" is another key aspect of solving many calculus problems. The natural logarithm function, denoted by \(\ln(x)\), has a straightforward derivative: \(\frac{d}{dx}\ln(x) = \frac{1}{x}\). This formula provides a simple rule for differentiating functions of the form \(\ln(u)\). When \(u\) is not just \(x\), but rather another function of \(x\), like in our case \(u = x + \sqrt{x^2 + 1}\), we leverage the chain rule to find \(\frac{d}{dx}\ln(u) = \frac{1}{u} \frac{du}{dx}\).

This requires two steps:

• Find \(\frac{1}{u}\), where \(u\) is \(x + \sqrt{x^2 + 1}\).
• Compute \(\frac{du}{dx}\), the derivative of the inner function.

Knowing how to differentiate a natural logarithm is essential, especially when it combines with other functions. It's often encountered in problems because the natural log appears in growth models, finance, and many other applications.

simplification of expressions

"Simplification of expressions" is often the final step in calculating derivatives. After finding the derivative, as in \(g'(x) = \frac{1}{x + \sqrt{x^2 + 1}} \left(1 + \frac{x}{\sqrt{x^2 + 1}} \right)\), simplifying means combining and reducing terms for a more manageable form.
It's crucial for making answers more interpretable and practical for further calculations. For instance, the derivative \(g'(x)\) is simplified from the initial complex expression to finally \(\frac{1}{\sqrt{x^2 + 1}}\).

In simplification:

• Combine fractions: Add or multiply terms by a common denominator.
• Simplify radicals: Combine like terms, if possible.
• Verify calculations through substitution or graph interpretation.

This step ensures the final expression is accurate and usable for ongoing applications or studies. Proper simplification in calculus can greatly impact overall clarity and usefulness of the derived function.