Problem 12 - Calculus for Biology and Medicine

Question

In Problems \(1-58\), find the derivative with respect to the independent variable. $$ f(x)=\cot (2-3 x) $$

Step-by-Step Solution

Answer

The derivative is \( 3\csc^2(2 - 3x) \).

Step 1: Recall the Derivative of Cotangent Function

The derivative of the cotangent function is given by the formula \( \frac{d}{dx}[\cot(u)] = -\csc^2(u) \cdot \frac{du}{dx} \), where \( u \) is an expression dependent on \( x \). We will use this formula to find the derivative.

Step 2: Identify the Inner Function

For the function \( f(x) = \cot(2 - 3x) \), the inner function \( u \) is \( u = 2 - 3x \). We next need to find the derivative of this inner function with respect to \( x \).

Step 3: Differentiate the Inner Function

The derivative of \( u = 2 - 3x \) with respect to \( x \) is \( \frac{du}{dx} = \frac{d}{dx}[2 - 3x] = -3 \).

Step 4: Apply the Chain Rule

Using the derivative formula for the cotangent, we substitute \( u = 2 - 3x \) and \( \frac{du}{dx} = -3 \): \[\frac{d}{dx}[\cot(2 - 3x)] = -\csc^2(2 - 3x) \cdot (-3) = 3\csc^2(2 - 3x). \]

Step 5: Write the Final Result

Putting it all together, the derivative of \( f(x) = \cot(2 - 3x) \) with respect to \( x \) is \( 3\csc^2(2 - 3x) \).

Key Concepts

Chain Rule

The Chain Rule is a crucial tool in calculus for finding derivatives of composite functions. Suppose a function is defined as \( f(g(x)) \), where \( f \) is an outer function and \( g \) is an inner function. The Chain Rule states that the derivative of \( f(g(x)) \) with respect to \( x \) is given by the product of the derivative of \( f \) with respect to \( g \), and the derivative of \( g \) with respect to \( x \): \[ \frac{d}{dx}[f(g(x))] = f'(g(x)) \cdot g'(x) \]

• Outer Function: This is the function on the outside; in this case, \( \cot(u) \).
• Inner Function: This is the expression inside the outer function; for this example, \( u = 2 - 3x \).

To differentiate \( \cot(u) \), you apply the derivative formula \( -\csc^2(u) \) and multiply it by the derivative of the inner function \( u \) with respect to \( x \). So, always start by identifying all inner functions and differentiate them one step at a time.

Trigonometric functions

Trigonometric functions are fundamental in calculus and often appear in differentiation exercises. Functions like sine and cosine have their derivatives memorized due to their frequent usage in problems:

• The derivative of \( \sin(x) \) is \( \cos(x) \).
• The derivative of \( \cos(x) \) is \( -\sin(x) \).
• For the cotangent function, the derivative is \( -\csc^2(x) \).

In trigonometric derivative problems, recognizing these basic rules is important. When complex expressions include trigonometric functions, like \( \cot(2 - 3x) \), the chain rule combined with trigonometric derivative formulas simplifies the differentiation task.
Understanding these formulas enables you to differentiate faster and tackle more complex expressions with confidence.

Differentiation techniques

Differentiation is the process of finding a function's derivative, which reveals the rate of change. There are several rules and techniques to keep in mind depending on the type of function:

• Power Rule: Differentiates power functions. For \( x^n \), the derivative is \( n \times x^{n-1} \).
• Product Rule: Useful for the product of two functions \( u(x) \) and \( v(x) \). The derivative is \( u'v + uv' \).
• Quotient Rule: For the quotient of two functions \( u \div v \), use \( (u'v - uv')/v^2 \).
• Chain Rule: Essential for functions within other functions, as explained above.

By mastering these differentiation techniques, you can efficiently tackle any derivative problem. Always identify the type of function involved and apply the appropriate rule step-by-step to ensure accuracy. Differentiation may initially seem complex, but with practice, these techniques become second nature.