Problem 54 - Calculus for Biology and Medicine

Question

Find the derivative with respect to the independent variable. $$ f(x)=\frac{1+\cos (3 x)}{2 x^{3}-x} $$

Step-by-Step Solution

Answer

The derivative is \( \frac{-3\sin(3x)(2x^3 - x) - (1 + \cos(3x))(6x^2 - 1)}{(2x^3 - x)^2} \).

Step 1: Identify the Function Type

The function given is a ratio of two functions of the form \( \frac{u(x)}{v(x)} \), where \( u(x) = 1 + \cos(3x) \) and \( v(x) = 2x^3 - x \). The appropriate method to find the derivative is the quotient rule.

Step 2: Recall the Quotient Rule

The quotient rule states that if \( y = \frac{u}{v} \), then the derivative \( \frac{dy}{dx} = \frac{v \cdot u' - u \cdot v'}{v^2} \). We will use this formula to find the derivative of the given function.

Step 3: Calculate \( u'(x) \) and \( v'(x) \)

First, find the derivative of the numerator, \( u(x) = 1 + \cos(3x) \). The derivative is \( u'(x) = -3\sin(3x) \), using the chain rule. Next, find the derivative of the denominator, \( v(x) = 2x^3 - x \). This gives \( v'(x) = 6x^2 - 1 \).

Step 4: Substitute into the Quotient Rule Formula

Substitute \( u(x), v(x), u'(x), \) and \( v'(x) \) into the quotient rule formula: \[\frac{dy}{dx} = \frac{(2x^3 - x)(-3\sin(3x)) - (1 + \cos(3x))(6x^2 - 1)}{(2x^3 - x)^2}\]

Step 5: Simplify the Expression

Expand and simplify the expression obtained in Step 4:\[\frac{dy}{dx} = \frac{-3\sin(3x)(2x^3 - x) - (1 + \cos(3x))(6x^2 - 1)}{(2x^3 - x)^2}\]Further simplify by expanding the numerators and combining like terms, keeping the expression for the derivative as neat as possible.

Key Concepts

Quotient Rule

When dealing with the derivative of a quotient of two functions, we use the quotient rule. This rule is crucial whenever the problem at hand involves a fraction where both the numerator and the denominator are functions of the same variable. In our exercise, we have the function defined as
\[ f(x) = \frac{u(x)}{v(x)} = \frac{1 + \cos(3x)}{2x^3 - x} \]
To find the derivative of this function, the quotient rule is applied, which is remembered using the formula:

• Given \( y = \frac{u}{v} \), then \( \frac{dy}{dx} = \frac{v \cdot u' - u \cdot v'}{v^2} \).

The formula might seem complex at first, but with practice, it becomes a straightforward tool to manage these types of functions. The numerator of the derivative involves the subtractive combination of the derivatives of the top and bottom functions, weighted by each other. This setup allows for the derivation to account for the changing rates of both parts of the fraction simultaneously.

Chain Rule

The chain rule is another vital tool in differentiation, especially useful when dealing with composite functions. When a function is defined in terms of another function, or when a variable affects another inside a function, the chain rule comes into play.
In our current solution, the chain rule is used to differentiate the term \( \cos(3x) \).
Here's how it applies:

• Consider the derivative of \( u(x) = 1 + \cos(3x) \).
• The chain rule helps differentiate \( \cos(3x) \) as a composite function.
• By the chain rule, \( \frac{d}{dx}[\cos(3x)] = -\sin(3x) \cdot 3 \), reflecting both the derivative of \( \cos \) as \( -\sin \) and the inner function \( 3x \) having a derivative of \( 3 \).
• Thus, the derivative \( u'(x) = -3\sin(3x) \), showcasing the power of the chain rule.

Remember: Always consider the inner function. By identifying it, you will know exactly where the chain rule can be used effectively.

Trigonometric Functions

Trigonometric functions like sine and cosine often appear in calculus problems, especially involving angles or periodic behavior. These functions have unique properties and derivatives, which play a crucial role in solving such problems.
For the trigonometric part of our given function:

• The term \( \cos(3x) \) is pivotal. We know from trigonometric derivatives that \( \frac{d}{dx}[\cos(x)] = -\sin(x) \).
• Using the chain rule, differentiating \( \cos(3x) \) results in \( -3\sin(3x) \), as previously discussed.

Understanding how trigonometric functions behave during differentiation is an essential part of mastering calculus. These functions often require both the basic derivative properties and additional rules—like the chain rule—to be successfully differentiated.

Differentiation Techniques

Differentiation is the cornerstone of calculus, allowing us to understand how functions change. Different techniques are employed depending on the structure of the function at hand.
In solving the derivative of composite and quotient functions, several skills are combined:

• Quotient Rule: Used for dividing two functions. Essential for managing fractions.
• Chain Rule: Helps when a function is inside another. Crucial for breaking down complex expressions.
• Basic Derivatives: Recognizing and using the standard derivatives, like for \( x^n \) and trigonometric functions.

The combination of these techniques allows us to solve complex expressions systematically. Each method has its place and knowing when and how to apply them becomes clear with practice. Understanding these techniques empowers us to approach any differentiation problem with confidence.