Problem 14 - Calculus for Biology and Medicine

Question

Differentiate the functions with respect to the independent variable. \(f(x)=\frac{\sqrt{x^{2}-1}}{2+\sqrt{x^{2}+1}}\)

Step-by-Step Solution

Answer

Differentiate using the quotient rule; simplify if needed.

Step 1: Identify the Differentiation Rule

The function given is a quotient of two functions: the numerator is \(\sqrt{x^2 - 1}\) and the denominator is \(2 + \sqrt{x^2 + 1}\). We need to employ the quotient rule for differentiation, which states: \(\left(\frac{u}{v}\right)' = \frac{v \cdot u' - u \cdot v'}{v^2}\), where \(u(x) = \sqrt{x^2 - 1}\) and \(v(x) = 2 + \sqrt{x^2 + 1}\).

Step 2: Differentiate the Numerator

First, differentiate \(u(x) = \sqrt{x^2 - 1}\). Using the chain rule, we find \(u'(x) = \frac{1}{2\sqrt{x^2 - 1}} \cdot (2x) = \frac{x}{\sqrt{x^2 - 1}}\).

Step 3: Differentiate the Denominator

Now, differentiate \(v(x) = 2 + \sqrt{x^2 + 1}\). Applying the chain rule, we get \(v'(x) = \frac{1}{2\sqrt{x^2 + 1}} \cdot (2x) = \frac{x}{\sqrt{x^2 + 1}}\).

Step 4: Apply the Quotient Rule

Using the quotient rule, substitute \(u(x)\), \(u'(x)\), \(v(x)\), and \(v'(x)\) into the formula: \[f'(x) = \frac{(2 + \sqrt{x^2 + 1}) \cdot \frac{x}{\sqrt{x^2 - 1}} - \sqrt{x^2 - 1} \cdot \frac{x}{\sqrt{x^2 + 1}}}{(2 + \sqrt{x^2 + 1})^2}.\]

Step 5: Simplify the Expression

To simplify, notice the common factor of \(x\) in the numerator. This can be factored out to obtain: \[f'(x) = \frac{x \left((2 + \sqrt{x^2 + 1}) \cdot \sqrt{x^2 + 1} - \sqrt{x^2 - 1} \cdot (\sqrt{x^2 + 1})\right)}{\sqrt{x^2 + 1} \cdot (2 + \sqrt{x^2 + 1})^2}.\] Further simplification and rationalization might be needed depending on the specific question demands.

Key Concepts

Quotient Rule in Differentiation

When dealing with differentiation of a function presented as a ratio of two other functions, the quotient rule becomes incredibly useful. This rule helps us find the derivative of a quotient

• It is written as \(\left( \frac{u}{v} \right)' = \frac{v \cdot u' - u \cdot v'}{v^2} \).
• Here, \(u(x)\) is the function in the numerator and \(v(x)\) is the function in the denominator.

Consider the equation:
\(f(x) = \frac{\sqrt{x^2 - 1}}{2 + \sqrt{x^2 + 1}}\).
We identify \(u(x) = \sqrt{x^2 - 1}\) and \(v(x) = 2 + \sqrt{x^2 + 1}\).

Applying the quotient rule allows you to systematically differentiate the function

• First, differentiate the numerator.
• Then, differentiate the denominator.
• Substitute these into the quotient rule formula.

This organized approach ensures accurate computation of derivatives for complex functions.

Chain Rule Simplified

Differentiating functions inside other functions can be tricky, which is where the chain rule comes into play. It simplifies the differentiation process for composite functions.

The chain rule states:

• If you have a function \(f(g(x))\), then its derivative is \(f'(g(x)) \cdot g'(x)\).

In our original exercise, both the numerator and denominator involve the square root of an expression.

To differentiate \(u(x) = \sqrt{x^2 - 1}\):

• First, find the outer function's derivative, \(\frac{1}{2\sqrt{x^2 - 1}}\).
• Then multiply by the inner function's derivative (of \(x^2 - 1\), which is \(2x\)).
• The result is \(\frac{x}{\sqrt{x^2 - 1}}\).

By following a similar approach for the denominator, such rules make complex problems much more manageable.

Understanding Functions

A function is a mathematical relationship where each input is related to exactly one output. In calculus, functions can often represent real-world phenomena.

In the context of differentiation

• We often examine how output values (dependent variable) change in relation to input values (independent variable).

Reflecting on the problem:

• Consider \(f(x) = \frac{\sqrt{x^2 - 1}}{2 + \sqrt{x^2 + 1}}\). Here, \(f(x)\) is a function of \(x\).
• This means each specific \(x\) gives a unique \(f(x)\) value.

Differentiating this function with respect to \(x\) helps determine how changes in \(x\) affect \(f(x)\).

Such understanding is not just theoretical. It provides insight into how things work and change, forming the backbone of mathematical analysis.

Independent Variable Explained

In calculus and mathematics in general, the independent variable is the input of a function. Its change affects the function's output.

For the function \(f(x)=\frac{\sqrt{x^2-1}}{2+\sqrt{x^2+1}}\):

• \(x\) serves as the independent variable.
• When \(x\) changes, each component of the function is recalculated.
• This change affects the overall value of \(f(x)\).

A solid understanding of the independent variable is crucial for solving any differentiation problem as it emphasizes the variable's essential role in the function's behavior.

Through differentiation, we explore how minute changes in \(x\) affect \(f(x)\), providing deeper insights into the shape and nature of graphs, as well as real-world applications of these principles.