Problem 29 - Calculus for Biology and Medicine

Question

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

Step-by-Step Solution

Answer

The derivative is \( f'(x) = \frac{2x \cos(2x^2 - 1)}{\sqrt{\sin(2x^2 - 1)}}.\)

Step 1: Identify the Function Type

The function given is \( f(x) = \sqrt{\sin(2x^2 - 1)} \). This is a composite function, being the square root of another function, which is \( \sin(2x^2 - 1) \). To find the derivative, we will use the chain rule.

Step 2: Apply the Chain Rule

The chain rule states that the derivative of a composite function \( h(g(x)) \) is \( h'(g(x)) \cdot g'(x) \). Let \( u = \sin(2x^2 - 1) \), so \( f(x) = \sqrt{u} = u^{1/2} \). Differentiate \( f \) with respect to \( u \): \( \frac{d}{du}(u^{1/2}) = \frac{1}{2}u^{-1/2} \).

Step 3: Differentiate the Inner Function

Differentiate \( u = \sin(2x^2 - 1) \) using the chain rule: \( \frac{du}{dx} = \cos(2x^2 - 1) \cdot \frac{d}{dx}(2x^2 - 1) = \cos(2x^2 - 1) \cdot 4x \).

Step 4: Combine Derivatives

Use the results from Steps 2 and 3 to find \( f'(x) \): \[ f'(x) = \frac{1}{2}(\sin(2x^2 - 1))^{-1/2} \cdot \cos(2x^2 - 1) \cdot 4x \]. Simplifying gives: \[ f'(x) = \frac{2x \cos(2x^2 - 1)}{\sqrt{\sin(2x^2 - 1)}}. \]

Key Concepts

Understanding the Chain Rule

The chain rule is a fundamental technique in calculus used to differentiate composite functions. It's a tool that allows us to find the derivative of one function that is nested inside another, like taking apart a set of Russian dolls. If you imagine a function as a process, then using the chain rule is like peeling back layers until reaching the core.
To apply the chain rule, first imagine two functions: an outer function, say \( h \), and an inner function, \( g \). The chain rule formula for the derivative of \( h(g(x)) \) is:

• \( h'(g(x)) \cdot g'(x) \)

This means you first take the derivative of the outer function while keeping the inner function unchanged. Then, multiply by the derivative of the inner function. It’s crucial to maintain this order and systematically apply it to each layer of the composition.

What are Composite Functions?

Composite functions occur when one function is applied to the result of another function, like putting on a sweater and then a coat. In our exercise, \( f(x) = \sqrt{\sin(2x^2 - 1)} \), the composite nature is \( \sqrt{\cdot} \) and \( \sin(\cdot) \).Recognizing a composite function involves identifying these layers. The outer function here is the square root function, applied to an inner function, which itself is \( \sin(2x^2 - 1) \). Hence, you have two layers to peel back when differentiating:

• The outer layer: \( \sqrt{u} \) (where \( u = \sin(2x^2 - 1) \))
• The innermost layer: \( 2x^2 - 1 \) inside \( \sin \)

It’s this structure that necessitates using the chain rule for differentiation.

Differentiation Techniques and Their Role

Differentiation techniques are essential tools in calculus, acting like a gardener's scissors trimming out growth along a path. They include formulas and rules to find derivatives effectively. For complex functions like composites, you harness these techniques to unravel each step with precision.In our exercise, differentiation uses multiple methods:

• The chain rule for tackling each function layer in the composite function as described before.
• The power rule to handle the outer square root function, converting it to an exponent for easy differentiation.
• The derivative of trigonometric functions \( \sin \) and \( \cos \) to dissect the inner layer.

Each technique is like a tool in a toolbox, helping to solve different parts of the problem. Understanding when and how to use each one effectively is key to mastering calculus.