Problem 14 - Calculus for Biology and Medicine

Question

In Problems \(1-58\), find the derivative with respect to the independent variable. $$ f(x)=-3 \csc (3-5 x) $$

Step-by-Step Solution

Answer

The derivative is \(15 \csc(3-5x) \cot(3-5x)\).

Step 1: Understanding the Function

The given function is \( f(x) = -3 \csc(3 - 5x) \). Our task is to find its derivative with respect to \( x \). Recall that the derivative of \( \csc(u) \) where \( u \) is a function of \( x \), involves the chain rule.

Step 2: Applying the Chain Rule

The derivative of \( \csc(u) \) is \( -\csc(u) \cot(u) \). However, since \( u = 3 - 5x \) is a function of \( x \), we need to use the chain rule. The derivative \( \frac{d}{dx} \csc(3 - 5x) \) is \( -\csc(3 - 5x) \cot(3 - 5x) \cdot (-5) \).

Step 3: Differentiating the Outer Function

We now differentiate the entire function \( f(x) = -3 \csc(3 - 5x) \), using the chain rule from step 2. The outside derivative is \( -3 \cdot -\csc(3 - 5x) \cot(3 - 5x) \cdot (-5) \).

Step 4: Simplifying the Derivative

Simplify the expression: \[-3 \cdot - \csc(3 - 5x) \cot(3 - 5x) \cdot (-5) = 15 \csc(3 - 5x) \cot(3 - 5x).\] This is our final derivative.

Key Concepts

Chain Rule

The chain rule is an essential technique in calculus for differentiating composed functions. When you have a function nested inside another, the chain rule is your go-to method. Here's how it works in a nutshell: If you have a function in the form of \( f(g(x)) \), the derivative \( f'(g(x)) \), is found by multiplying the derivative of the outer function by the derivative of the inner function.

• Start with the outer function's derivative and then multiply it by the inner function's derivative.
• It helps to "unwrap" functions that are inside one another by differentiating from the outside in.
• This is especially useful when the argument of your function is itself a non-trivial function of the variable you're differentiating with respect to.

To see it in action, consider differentiating \( f(x) = -3 \csc(3 - 5x) \). Here, we identify \( u = 3 - 5x \) as the "inner function," and \( \csc(u) \) as the "outer function." The chain rule states to first differentiate the outer function, \( -\csc(u) \cot(u) \), and then multiply by the derivative of \( u \), which is "unwrapping" \( -5 \). By this, we seamlessly handle derivatives of compositions of functions.

Functions

Functions are the building blocks of calculus. In essence, a function is a relation that uniquely associates elements of one set (called the domain) with elements of another set (the codomain). Functions take an input, process it, and produce an output.

• The input \( x \) is transformed into an output through the rules of the function.
• Functions are often represented as \( f(x) \), where \( f \) denotes the function and \( x \) is the variable.
• Functions can be simple algebraic expressions, complex trigonometric forms, or even combinations like\( -3 \csc(3 - 5x) \) in our problem.

The importance of understanding functions in calculus lies in the process of differentiation: analyzing how the output of a function changes as its input changes. Recognizing the "outer" and "inner" components in composite functions, like in the exercise \( f(x) = -3 \csc(3 - 5x) \), is particularly crucial when differentiating using the chain rule.

Trigonometric Functions

Trigonometric functions, derived from angles, are fundamental in calculus and describe the relationships between the angles and sides of triangles.

• The functions \( \sin(x), \cos(x), \tan(x), \csc(x), \sec(x), \) and \( \cot(x) \) are the primary trigonometric functions.
• They play a significant role in defining wave patterns, oscillations, and other periodic phenomena.
• The function \( \csc(x) \), or cosecant, is particularly important here. It is the reciprocal of the sine function, \( \csc(x) = \frac{1}{\sin(x)} \).

In differentiation, understanding the derivatives of trigonometric functions is crucial. For instance, in our function \( f(x) = -3 \csc(3 - 5x) \), the derivative of \( \csc(u) \) results in \( -\csc(u) \cot(u) \). Knowing these standard derivatives allows you to apply the chain rule effectively, which involves the trigonometric identities and their relationships to each other. Delving deeper into these functions enables one to tackle a wide variety of calculus problems effectively.