Problem 45 - Calculus for Biology and Medicine

Question

In Problems \(1-58\), find the derivative with respect to the independent variable. $$ (x)=\frac{1}{\sin \left(3 x^{2}-1\right)} $$

Step-by-Step Solution

Answer

The derivative is \( -6x \cdot \cot(3x^2 - 1) \cdot \csc(3x^2 - 1) \).

Step 1: Identify the Composite Function

The function given is \( f(x) = \frac{1}{\sin(3x^2 - 1)} \). This is a composite function, which means it is composed of an outer function \( g(u) = \frac{1}{u} \) and an inner function \( u(x) = \sin(3x^2 - 1) \).

Step 2: Differentiate the Outer Function

First, gain the derivative of the outer function \( g(u) = \frac{1}{u} \) with respect to \( u \). The derivative is \( g'(u) = -\frac{1}{u^2} \).

Step 3: Differentiate the Inner Function

Now, we need to differentiate the inner function \( u(x) = \sin(3x^2 - 1) \) with respect to \( x \). First, the derivative of \( \sin(v(x)) \) is \( \cos(v(x)) \cdot v'(x) \). Here, \( v(x) = 3x^2 - 1 \), so its derivative is \( v'(x) = 6x \). Thus, the derivative of \( u(x) \) is \( \cos(3x^2 - 1) \cdot 6x \).

Step 4: Apply the Chain Rule

Using the chain rule to connect the derivatives from Steps 2 and 3. The chain rule states that \( f'(x) = g'(u(x)) \cdot u'(x) \). Therefore, \( f'(x) = \left(-\frac{1}{\sin^2(3x^2 - 1)}\right) \cdot \left(\cos(3x^2 - 1) \cdot 6x \right) \).

Step 5: Simplify the Expression

Simplify the expression from Step 4: \( f'(x) = -6x \cdot \frac{\cos(3x^2 - 1)}{\sin^2(3x^2 - 1)} \). Recognizing that \( \frac{\cos(a)}{\sin^2(a)} = \cot(a) \cdot \csc(a) \), we can further write \( f'(x) = -6x \cdot \cot(3x^2 - 1) \cdot \csc(3x^2 - 1) \).

Key Concepts

Understanding the Chain Rule

The chain rule is a powerful technique when dealing with composite functions in calculus. When you need to differentiate a function that's been composed by layering more than one function together, the chain rule is your go-to method. Think of it like peeling layers off an onion; you handle one layer at a time.
For a composite function, let's say you have a function like \( f(g(x)) \). In this scenario, \( f \) is layered on top of \( g \), and both functions rely on the same variable, \( x \). The chain rule allows us to differentiate this layered or composite function efficiently by expressing the derivative as:

• \( f'(g(x)) \cdot g'(x) \)

This process enables you to break down complex problems into simpler, more manageable parts so you can find their derivatives. Make sure you first find the derivative of the outer function and then follow up with the derivative of the inner function, always multiplying them together based on this rule.

Unpacking Composite Functions

Composite functions occur when one function is applied to the result of another function. For example, in the expression \( f(g(x)) \), \( g(x) \) is the inner function and \( f(u) \) (where \( u = g(x) \)) is the outer function. The expression is considered composite because you’re nesting one function inside another.
In situations such as the one provided, you are often working with functions within functions. Identifying the inner and outer components is crucial. To illustrate, our example function is \( f(x) = \frac{1}{\sin(3x^2 - 1)} \). Here:

• Outer function: \( g(u) = \frac{1}{u} \)
• Inner function: \( u(x) = \sin(3x^2 - 1) \)

Recognizing and breaking down these layers indicates which derivatives you'll need to compute separately before merging them using the chain rule.

Essential Differentiation Techniques

Differentiation techniques are foundational in calculus for finding the rates at which things change. Three primary techniques are most frequently used:
1. **The Power Rule**: Useful for power functions, states the derivative of \( x^n \) is \( nx^{n-1} \).

• Simple yet effective when dealing directly with polynomial functions.

2. **The Product Rule**: Applied when differentiating products of two functions, stated as:

• \((uv)' = u'v + uv'\)

This ensures each function's change is considered.
3. **The Chain Rule**: Discussed previously for composite functions, helps differentiate nested functions.
When approaching problems, select the proper rules based on the function's structure. In our specific example, employing the chain rule is critical as it facilitates breaking down and differentiating the composite function by its inner and outer parts. Differentiation techniques like these form the backbone of calculus analysis and problem-solving, enabling you to tackle even the most complex structures.