Problem 11 - Calculus for Biology and Medicine

Question

In Problems 1-28, differentiate the functions with respect to the independent variable. $$ f(x)=\frac{3 x+1}{\sqrt{2 x^{2}-1}} $$

Step-by-Step Solution

Answer

The derivative is \( f'(x) = \frac{-3 - 2x}{(2x^2 - 1)\sqrt{2x^2 - 1}} \).

Step 1: Identify the rules to use

To differentiate the function \( f(x) = \frac{3x+1}{\sqrt{2x^2-1}} \), we need to apply the quotient rule of differentiation. The quotient rule states that for functions \( u(x) \) and \( v(x) \), the derivative \( \left(\frac{u}{v}\right)'=\frac{u'v-uv'}{v^2} \). Identify \( u(x) = 3x + 1 \) and \( v(x) = \sqrt{2x^2 - 1} \).

Step 2: Find derivatives of u(x) and v(x)

Calculate the derivative of the numerator and denominator separately. For \( u(x) = 3x + 1 \), the derivative is \( u'(x) = 3 \). For \( v(x) = \sqrt{2x^2 - 1} \), rewrite as \( (2x^2 - 1)^{1/2} \) and use the chain rule to get \( v'(x) = \frac{1}{2}(2x^2 - 1)^{-1/2} \cdot 4x = \frac{2x}{\sqrt{2x^2 - 1}} \).

Step 3: Apply the quotient rule

Substitute the derivatives into the quotient rule formula. Thus, \( f'(x) = \frac{u'v - uv'}{v^2} \) becomes \( f'(x) = \frac{3 \cdot \sqrt{2x^2 - 1} - (3x + 1) \cdot \frac{2x}{\sqrt{2x^2 - 1}}}{(\sqrt{2x^2 - 1})^2} \).

Step 4: Simplify the expression

The denominator simplifies to \( 2x^2 - 1 \), and the numerator simplifies by factoring out \( \sqrt{2x^2 - 1} \). So we have \( f'(x) = \frac{3\sqrt{2x^2 - 1} - \frac{2x(3x + 1)}{\sqrt{2x^2 - 1}}}{2x^2 - 1} = \frac{3(2x^2 - 1) - 2x(3x + 1)}{(2x^2 - 1)\sqrt{2x^2 - 1}} \).

Step 5: Finalize the derivative

Simplify the numerator further: \( 3(2x^2 - 1) - 2x(3x + 1) = 6x^2 - 3 - (6x^2 + 2x) = -3 - 2x \). So the derivative is \( f'(x) = \frac{-3 - 2x}{(2x^2 - 1)\sqrt{2x^2 - 1}} \).

Key Concepts

Understanding the Quotient Rule

The **quotient rule** is a critical differentiation technique used to find the derivative of a function that is the division of two other functions. It's particularly helpful when dealing with rational functions, those that can be expressed as one function divided by another. The rule is expressed as follows:\[\left( \frac{u}{v} \right)' = \frac{u'v - uv'}{v^2}\]Here, \( u(x) \) is the numerator function, and \( v(x) \) is the denominator function. The derivative \( u'(x) \) is the rate of change of the numerator, and \( v'(x) \) is the rate of change of the denominator.
**Key Steps Include:**

• Identify the numerator \( u(x) \) and the denominator \( v(x) \) of the function.
• Find the derivatives \( u'(x) \) and \( v'(x) \).
• Substitute these into the quotient rule formula.
• Simplify the expression to find the final derivative.

The beauty of the quotient rule lies in its ability to systematically break down and differentiate complex rational expressions, as in our example function \( f(x) = \frac{3x+1}{\sqrt{2x^{2}-1}} \), effectively.

Demystifying the Chain Rule

The **chain rule** is a powerful tool in calculus for differentiating composite functions. A composite function is essentially a function inside another function. The chain rule allows us to differentiate such functions effectively by considering their inner and outer components separately.
The rule states:\[(g(f(x)))' = g'(f(x)) \cdot f'(x)\]**Breaking Down the Steps:**

• Identify the "outer function" \( g(u) \) and the "inner function" \( f(x) \).
• Calculate \( f'(x) \), the derivative of the inner function.
• Calculate \( g'(u) \), treated as a function of \( u = f(x) \).
• Combine them to express the derivative of the composite function.

In the example problem, when differentiating \( \sqrt{2x^2 - 1} \), we see it as \( (2x^2 - 1)^{1/2} \), thus needing the chain rule. Here, the outer function is \( g(u) = u^{1/2} \) and inner \( f(x) = 2x^2 - 1 \). By linking their derivatives, we efficiently find the derivative of this composite structure.

The Art of Derivative Calculation

Derivative calculation is a cornerstone of calculus. It involves finding how a function changes as its input changes. Essentially, it captures the rate at which one quantity changes relative to another.
**Important Elements to Master:**

• Understanding different rules like the product rule, quotient rule, and chain rule.
• Recognizing when to apply each rule based on the type of function you're dealing with.
• Correctly simplifying and interpreting results to ensure accurate representations of function behavior.

In this exercise, we apply different rules in stages to handle complex expressions. Initially, we use the quotient rule since the function is a division of two parts. Within the application of the quotient rule, we further leverage the chain rule for differentiating the square root in the denominator. The final step is simplification, ensuring clarity of the resulting derivative. Mastering these steps equips learners to tackle a wide array of calculus problems efficiently.