Problem 115 - Calculus and its applications

Question

Differentiate. $$ y=\ln \sqrt{5+x^{2}} $$

Step-by-Step Solution

Answer

The derivative is \( \frac{x}{5 + x^2} \).

Step 1: Rewrite the Expression

The first step in differentiating the function is to rewrite the square root in terms of exponents for easier differentiation. The expression given is \( y = \ln \sqrt{5 + x^2} \). We can rewrite this as \( y = \ln (5 + x^2)^{1/2} \).

Step 2: Apply the Logarithm Power Rule

Using the logarithm power rule, \( \ln(a^b) = b \ln(a) \), we can simplify the expression: \( y = \frac{1}{2} \ln(5 + x^2) \).

Step 3: Differentiate Using the Chain Rule

Now, differentiate \( y = \frac{1}{2} \ln(5 + x^2) \) with respect to \( x \). Let \( u = 5 + x^2 \). Then, \( y = \frac{1}{2} \ln(u) \). The derivative \( y' = \frac{1}{2} \cdot \frac{1}{u} \cdot u' \).

Step 4: Differentiate the Inner Function

Determine \( u' \), the derivative of the inner function \( u = 5 + x^2 \). Since \( 5 \) is a constant, \( u' = \frac{d}{dx}(5 + x^2) = 0 + 2x \).

Step 5: Combine the Results

Substitute \( u = 5 + x^2 \) and \( u' = 2x \) back into the expression for \( y' \): \[ y' = \frac{1}{2} \cdot \frac{1}{5 + x^2} \cdot 2x = \frac{x}{5 + x^2}. \]

Step 6: Final Result

The derivative of \( y = \ln \sqrt{5 + x^2} \) with respect to \( x \) is \( \frac{x}{5 + x^2} \).

Key Concepts

Logarithmic Differentiation

Logarithmic differentiation is a useful method when dealing with complex functions, especially those involving products, divisions, or exponents. The key advantage of this technique is the ability it gives us to simplify differentiation by taking the natural logarithm of both sides of an equation.
Here's a simple way to understand it:

• Take the natural logarithm (ln) of both sides of your equation.
• Use properties of logarithms, such as transforming products into sums, transformations of exponents to coefficients, and quotients into differences, to simplify the function.

In the exercise, we start with the equation \( y = \ln \sqrt{5 + x^2} \). By rewriting the square root as an exponent, \( y = \ln(5 + x^2)^{1/2} \), it becomes easier to apply the logarithm power rule.
When solving complex derivatives, logarithmic differentiation offers comprehensive simplification and unleashes the potential of easier manipulation of expressions.

Chain Rule

Understanding the chain rule helps us to efficiently differentiate composite functions. This rule is crucial when functions have layers, like an "inner" function inside an "outer" function.
The core idea is simple:

• Identify the outer function and inner function.
• Differentiate the outer function, keeping the inner function intact.
• Multiply the result by the derivative of the inner function.

In our example, after simplifying the expression to \( y = \frac{1}{2} \ln(5 + x^2) \), the outer function is \( \frac{1}{2} \ln(u) \), and the inner function is \( u = 5 + x^2 \). Differentiating the outer function gives us \( \frac{1}{2} \cdot \frac{1}{u} \), and the inner derivative, \( u' \), yields \( 2x \). Applying the chain rule, the derivative is assembled by multiplying these results, leading to \( \frac{x}{5 + x^2} \).
The chain rule is super powerful in handling the complexity of layered functions and transforming them into much simpler derivatives.

Logarithm Power Rule

The logarithm power rule is a handy and straightforward tool for simplifying expressions involving powers within logarithmic functions. This rule states when you have a logarithmic function with an exponent, you can move the exponent in front of the logarithm as a coefficient.
Here's how it simplifies things:

• Change the form from \( \ln(a^b) \) to \( b \cdot \ln(a) \).
• This transformation makes the differentiation process significantly easier, as seen when applied to simpler forms.

In the given exercise, using the logarithm power rule transformed \( \ln(5 + x^2)^{1/2} \) into \( \frac{1}{2} \ln(5 + x^2) \).
Applying this rule helps break down complicated expressions, making differentiation more straightforward and manageably clear. It's especially beneficial when coupled with other techniques like the chain rule, leading to easier evaluations of derivatives.