Question

In Exercises, find the derivative of the function. $$ y=\ln \left(x \sqrt{4+x^{2}}\right) $$

Short answer

The derivative of the function \(y=\ln \left(x \sqrt{4+x^{2}}\right)\) is \(y' = \frac{(1+x)(1-x)}{4x}\)

Step-by-step solution

Simplify the function using logarithm properties

Rewrite the log function using the properties of the logarithm. Result is: \(y=\ln(x)+\frac{1}{2}\ln(4+x^{2})\)

Find the derivative using the chain rule

Derivative of \(y\) is \(y'=\frac{1}{x} + \frac{1}{2*(4+x^{2})*2x}\) = \( \frac{1}{x} + \frac{x}{4+x^{2}}\)

Simplify the derivative

Obtain a common denominator, and then simplify the equation. \(y' =\frac{1+x^{2}}{x*(4+x^{2})}\)

Factorise and simplify further

Factorize the numerator and simplify the equation, \(y' = \frac{(1+x)(1-x)}{x*(4+x^{2})}\)

Multiply the denominators

Simplify the denominators to finalize the derivative as \(y' = \frac{(1+x)(1-x)}{4x}\)

Key concepts

Logarithm Properties

Understanding the properties of logarithms is essential for simplifying expressions effectively. Logarithms have several key rules that make handling complex expressions more manageable.
First, there's the **product rule**, which states that the logarithm of a product is equal to the sum of the logarithms. In symbols:

• \(\log(ab) = \log(a) + \log(b) \)

In our exercise, the function was first simplified using this rule: \(\ln(x \sqrt{4+x^{2}}) \rightarrow \ln(x) + \frac{1}{2}\ln(4+x^{2}) \).
This approach breaks down a complex expression into simpler parts, making it easier to handle mathematically.

• The second key rule used here is the **power rule**, which expresses that the logarithm of a power is the exponent times the logarithm of the base:
  • \(\log(a^{b}) = b\log(a) \)
  
  This modification allows the logarithm of the square root to be treated as a multiplication, greatly simplifying the derivative calculation process. Understanding and applying these properties helps to reduce complicated logarithmic expressions into simpler, more manageable forms, paving the way for more accurate calculus operations.

Chain Rule

The Chain Rule is a fundamental technique in calculus used to differentiate composite functions. It allows us to find the derivative of a "function within a function."
For our function, \(y = \ln(x \sqrt{4+x^{2}})\), after utilizing the logarithm properties, it was simplified to \(\ln(x) + \frac{1}{2}\ln(4+x^{2})\).
This expression can be seen as a composition of simpler functions, which is where the Chain Rule comes into play.

• The Chain Rule states that if we have a function \(u(v(x))\), then the derivative is:
  • \((\frac{du}{dv} \cdot \frac{dv}{dx})\)

In our scenario, the term \(\frac{1}{2}\ln(4+x^{2})\) requires us to apply the Chain Rule because it involves a nested function inside the logarithm.

• The outer function is the natural logarithm, and the inner function is the quadratic expression \(4+x^{2}\). When differentiated:
  • The derivative of \(\ln(v)\) is \(\frac{1}{v}\), and the derivative of \(4+x^{2}\) is \(2x\).
  
  Using the Chain Rule, we found \(\frac{1}{2}\cdot\frac{2x}{4+x^{2}}\), which simplifies to \(\frac{x}{4+x^{2}}\).

Mastering the Chain Rule is key when dealing with functions recorded in complex forms. It reveals how intricate expressions can be managed incrementally by addressing each layer of function relationships systematically.

Common Denominator

Finding a common denominator is crucial for combining fractions into a single expression. This step helps to simplify expressions and solve equations more smoothly.
In our calculus problem, once the derivative \(y' = \frac{1}{x} + \frac{x}{4+x^{2}}\) was obtained, it needed to be combined into a single fraction to make further simplification feasible.
A common denominator is vital for this task:

• **Common Denominator:** The smallest shared multiple of the denominators of each fraction.
  • Here, the denominators are \(x\) and \(4+x^{2}\). So, the common denominator is their product, \(x(4+x^{2})\).
  The fractions are then rewritten as single fractions with this new common denominator.
  By manipulating the numerators to align with this shared base, you can sum the fractions:
  • This yields, \(y' = \frac{1+x^{2}}{x(4+x^{2})}\), officially consolidated into one fraction for simpler manipulation.
  
  This technique is extremely useful in various mathematical contexts, particularly for simplifying algebraic expressions and solving complex equations. Using a common denominator rightly allows you to streamline calculations and gain clearer insights into the behavior of functions.

Factorization

Factorization simplifies expressions by expressing them as products of simpler factors. It's an essential skill in algebra, especially when you want to simplify expressions, solve equations, or split numerators and denominators.
In our derivative problem, the simplified form of the derivative was \(y' = \frac{1+x^{2}}{x(4+x^{2})}\). Further simplification using factorization helped break down the numerator:

• **Factorization:** Expressing a polynomial or number as a product of its factors.
  • In this context, it means identifying factors \( (1+x)(1-x) \) within the numerator \(1+x^{2}\).

By recognizing \(1+x^{2}\) as a difference of squares \( (1+x)(1-x) \), you can simplify the expression further.
Thus, the factorized derivative becomes:

• \(y' = \frac{(1+x)(1-x)}{x(4+x^{2})}\). Ultimately, further algebraic manipulation yielded the final answer:
  • \(y' = \frac{(1+x)(1-x)}{4x} \).

Understanding and implementing factorization aids significantly in reducing expressions to simpler components, making complex calculus problems more approachable. It's a powerful tool for solving and simplifying computations across mathematics effectively.