Problem 5 - Calculus for Biology and Medicine

Question

Differentiate the functions with respect to the independent variable. \(f(x)=\sqrt{x^{2}+3}\)

Step-by-Step Solution

Answer

The derivative is \( \frac{x}{\sqrt{x^2 + 3}} \).

Step 1: Recognize the Outer and Inner Functions

The function given is expressed as the square root of an expression. Recognize this outer function as a square root function and the inner function as the expression inside the square root, namely, \( x^2 + 3 \).

Step 2: Apply the Chain Rule

To differentiate the given function, we use the chain rule, which states that \( (f(g(x)))' = f'(g(x)) \cdot g'(x) \). Here, the outer function \( u = \sqrt{u} \) and the inner function \( u = x^2 + 3 \).

Step 3: Differentiate the Outer Function

Differentiate the outer function \( \sqrt{u} \) with respect to \( u \). The derivative is \( \frac{d}{du}(\sqrt{u}) = \frac{1}{2\sqrt{u}} \).

Step 4: Differentiate the Inner Function

Differentiate the inner function \( x^2 + 3 \) with respect to \( x \). The derivative is \( \frac{d}{dx}(x^2 + 3) = 2x \).

Step 5: Combine Results from the Chain Rule

Combine the derivatives from Steps 3 and 4 using the chain rule: \[ \frac{d}{dx}[\sqrt{x^2 + 3}] = \frac{1}{2\sqrt{x^2 + 3}} \cdot 2x = \frac{x}{\sqrt{x^2 + 3}}. \]

Step 6: Simplify the Expression

The expression \( \frac{x}{\sqrt{x^2 + 3}} \) is already simplified, and therefore, this is the derivative of the function \( f(x) = \sqrt{x^2 + 3} \).

Key Concepts

Chain Rule

The chain rule is a fundamental technique in calculus differentiation that helps us find the derivative of composite functions. It is particularly useful when dealing with functions nested within one another, putting them into the form of outer and inner functions.

The rule itself can be written as:

• If you have a function in the form of \( f(g(x)) \), then its derivative is given by \( (f(g(x)))' = f'(g(x)) \cdot g'(x) \).

To apply the chain rule, we need to identify these components first:

• The outer function \( f \), which is acted upon the inner function.
• The inner function \( g(x) \), which is tucked inside the outer function.

The chain rule then guides us to first differentiate the outer function while keeping the inner function intact. Afterward, we differentiate the inner function. Finally, we multiply these derivatives, providing us with a full differentiation of the composite function.

This procedure allows us to take derivatives efficiently, even of functions that initially seem complex due to their composition.

Outer and Inner Functions

When differentiating using the chain rule, correctly identifying the outer and inner functions is an essential first step. In the context of the exercise, where we have \( f(x) = \sqrt{x^2 + 3} \), we see two distinct layers:

• The outer function is \( f(u) = \sqrt{u} \). This is the function that wraps around the inner variable and is what we target first during differentiation.
• The inner function is \( g(x) = x^2 + 3 \), the expression inside the square root.

These functions are distinguished based on their roles in the expression of \( f(x) \).

Why is this distinction crucial? The reason lies in the differentiation process. Each function contributes a layer to the composite function's complexity, meaning we have to treat them as separate and connected entities when applying the chain rule.

Misidentifying which function is outer or inner can lead to incorrect differentiation. Therefore, practice in identifying and separating these layers enhances understanding and helps ensure accuracy in derivative calculations.

Derivative Simplification

After applying the chain rule and obtaining the derivative, simplifying the expression is often necessary for clarity and concise communication of results. Simplification involves combining like terms and reducing complex fractions to their simplest form.

In the exercise, after determining the derivative \( \frac{d}{dx} [\sqrt{x^2 + 3}] = \frac{1}{2\sqrt{x^2 + 3}} \cdot 2x \), it boils down to \( \frac{x}{\sqrt{x^2 + 3}} \).

• Notice that the \( 2 \) in the numerator and denominator cancels out, streamlining the expression.
• Also, ensure that the square root remains in the denominator as placing it in another form could complicate the interpretation.

These clear expressions by simplification help prevent misunderstandings and errors in subsequent computations or evaluations of the function.

Thus, simplification is not merely a cosmetic change but a critical skill in the practice of calculus, particularly when communicating complex mathematical ideas effectively.