Problem 49 - Calculus for Biology and Medicine

Question

Find the derivative with respect to the independent variable. $$ h(x)=\frac{3}{\tan (2 x)-x} $$

Step-by-Step Solution

Answer

The derivative of the function is \( h'(x) = \frac{-6\sec^2(2x) + 3}{(\tan(2x) - x)^2} \).

Step 1: Identify the Rule to Use

The function given is a quotient: \( h(x) = \frac{3}{\tan(2x) - x} \). To differentiate it, we use the Quotient Rule, which is applicable to functions of the form \( \frac{u}{v} \). The rule states: if \( h(x) = \frac{u}{v} \), then the derivative \( h'(x) = \frac{v\cdot u' - u \cdot v'}{v^2} \). Here, \( u = 3 \) and \( v = \tan(2x) - x \).

Step 2: Derive Each Part of the Function

Determine the derivatives \( u' \) and \( v' \). Since \( u = 3 \), its derivative is \( u' = 0 \). For \( v = \tan(2x) - x \), use the derivatives of \( \tan(x) \) and the linear function. The derivative of \( \tan(2x) \) is \( 2\sec^2(2x) \) using the chain rule, and the derivative of \(-x\) is \(-1\). Thus, \( v' = 2\sec^2(2x) - 1 \).

Step 3: Apply the Quotient Rule

Apply the Quotient Rule: \[ h'(x) = \frac{(\tan(2x) - x) \cdot 0 - 3 \cdot (2\sec^2(2x) - 1)}{(\tan(2x) - x)^2} \]. Since the term \((\tan(2x) - x) \cdot 0\) is zero, the expression simplifies to: \( h'(x) = \frac{-3(2\sec^2(2x) - 1)}{(\tan(2x) - x)^2} \).

Step 4: Simplify the Expression

Simplify the expression for clarity: \( h'(x) = \frac{-6\sec^2(2x) + 3}{(\tan(2x) - x)^2} \). This is the final expression for the derivative of the given function \( h(x) \).

Key Concepts

Quotient Rule

The Quotient Rule is essential when differentiating a function that is expressed as a quotient of two functions, like \(h(x) = \frac{u}{v}\). It provides a systematic way to find the derivative, \(h'(x)\), by applying the formula:

• \(h'(x) = \frac{v \cdot u' - u \cdot v'}{v^2}\).

Here:

• \(u\) is the numerator.
• \(v\) is the denominator.
• \(u', v'\) are the derivatives of \(u\) and \(v\) respectively.

In the given exercise, \(u = 3\), a constant, and \(v = \tan(2x) - x\). Simplifying the calculation is key. By differentiating each part separately and then substituting them into the quotient rule formula, we find the derivative efficiently. Simplifying the expression after applying the rule brings clarity and accuracy.

Chain Rule

The Chain Rule helps us differentiate composite functions, where one function is nested inside another. Its primary purpose is to ensure that we accurately calculate the derivative when variables are dependent on other functions.

• The rule is given by: \(\frac{d}{dx}f(g(x)) = f'(g(x)) \cdot g'(x)\).

When you see a function like \(\tan(2x)\), realize that it involves composing the trigonometric function \(\tan(x)\) with the linear function \(2x\). First, differentiate \(\tan(x)\) to get \(\sec^2(x)\). Then, apply the chain rule:

• \(\frac{d}{dx} \tan(2x) = \sec^2(2x) \cdot 2 = 2\sec^2(2x)\).

This result is crucial for applying the Quotient Rule effectively, ensuring every layer of the function relationship is unpacked.

Trigonometric Functions

Trigonometric functions such as \(\tan(x)\), \(\sin(x)\), and \(\cos(x)\) often appear in derivatives and calculus problems. Their understanding is vital because they frequently serve as building blocks in more complex expressions. For derivatives, particularly:

• \(\frac{d}{dx} \tan(x) = \sec^2(x)\),
• \(\frac{d}{dx} \sin(x) = \cos(x)\),
• \(\frac{d}{dx} \cos(x) = -\sin(x)\).

These foundation derivatives help determine how functions involving trigonometric terms change over an interval. Understanding their behavior allows us to predict the motion implied by these functions, such as wave patterns or angles of rotation.

Function Differentiation

Differentiation is the process of finding the derivative of a function, which represents the function's rate of change or slope at any given point. This foundational concept in calculus is used to solve problems involving rates of change in various fields. It involves:

• Identifying the parts of the function.
• Selecting the appropriate rule (such as power, chain, or quotient).
• Finding derivatives of each part as needed.
• Simplifying the resulting expression.

In the exercise, each function was carefully broken down into manageable parts. The derivative provided insight into how the function behaves for any input \(x\). Differentiation transforms complex problems into smaller tasks, making problem-solving more approachable and concise.