Problem 65 - Calculus for Biology and Medicine

Question

Find the derivatives of the following functions: $$ f(x)=\sin \sqrt{x^{2}+1} $$

Step-by-Step Solution

Answer

The derivative of the function is \( f'(x) = \frac{x \cos(\sqrt{x^2 + 1})}{\sqrt{x^2 + 1}} \).

Step 1: Identify the Outer and Inner Functions

The given function is a composition of two functions: \[ f(x) = \sin(\sqrt{x^2 + 1}) \] Here, the outer function is \( \sin(u) \) and the inner function is \( u = \sqrt{x^2 + 1} \). We will need to use the chain rule to find the derivative.

Step 2: Differentiate the Outer Function

Differentiate the outer function \( \sin(u) \) with respect to \( u \): \[ \frac{d}{du} \sin(u) = \cos(u) \] Therefore, the derivative of the outer function is \( \cos(\sqrt{x^2 + 1}) \).

Step 3: Differentiate the Inner Function

The inner function is \( u = \sqrt{x^2 + 1} \) which can be rewritten as \( u = (x^2 + 1)^{1/2} \). Use the chain rule again to differentiate: \[ \frac{d}{dx}(x^2 + 1)^{1/2} = \frac{1}{2}(x^2 + 1)^{-1/2} \cdot 2x = \frac{x}{\sqrt{x^2 + 1}} \] Thus, the derivative of \( u \) with respect to \( x \) is \( \frac{x}{\sqrt{x^2 + 1}} \).

Step 4: Apply the Chain Rule

By the chain rule, the derivative of \( f(x) \) is the derivative of the outer function times the derivative of the inner function: \[ f'(x) = \cos(\sqrt{x^2 + 1}) \cdot \frac{x}{\sqrt{x^2 + 1}} \] This simplifies to: \[ f'(x) = \frac{x \cos(\sqrt{x^2 + 1})}{\sqrt{x^2 + 1}} \]

Step 5: Simplify the Expression

The expression for \( f'(x) \) is already quite simplified. However, ensure it is clear: \[ f'(x) = \frac{x \cos(\sqrt{x^2 + 1})}{\sqrt{x^2 + 1}} \] This is the final derivative of the function \( f(x) \).

Key Concepts

Understanding the Chain Rule

In calculus, the chain rule is a fundamental method used for differentiating composite functions. To put it simply, if you have one function nested inside another, you can use the chain rule to find the derivative. Consider a situation where you have two functions: an inner function and an outer function. The chain rule involves two major steps:

• First, differentiate the outer function while keeping the inner function intact.
• Then, multiply this result by the derivative of the inner function itself.

When applied together, these steps allow you to tackle derivatives of complex expressions that aren't straightforward polynomials. This is especially useful when you encounter compositions involving functions like trigonometric or radical expressions.

Composition of Functions Explained

A composition of functions is where one function is applied to the result of another function. It's like stacking processes on top of each other: imagine you have a washing machine that first sorts clothes (inner function) and then washes them (outer function). The composition is doing both together. In the exercise, we dealt with the function composition \( \sin(\sqrt{x^2 + 1}) \). Here, \( \sqrt{x^2 + 1} \) is the inner function and \( \sin(u) \) where \( u = \sqrt{x^2 + 1} \), is the outer function. Recognizing each part of the composition is critical for applying the correct differentiation rules and using the chain rule effectively. Once you break down a composite function, differentiation becomes much more manageable, allowing you to understand the problem step by step.

The Role of Trigonometric Functions in Derivatives

Trigonometric functions, such as sine and cosine, are essential in calculus, appearing in many differentiation problems. Differentiating these functions requires specific rules:

• The derivative of \( \sin(u) \) is \( \cos(u) \),
• While the derivative of \( \cos(u) \) is \( -\sin(u) \).

In the given exercise, the trigonometric function \( \sin(\sqrt{x^2 + 1}) \) was part of the composite function. The differentiation involved recognizing \( \sin \) as the outer function and finding its derivative as \( \cos \). Understanding these rules is important not only for calculus exercises but also for real-world applications where waveforms or periodic functions are involved. Trigonometric derivatives often appear in physics and engineering scenarios, highlighting the practical utility of mastering these concepts.