Problem 65 - Calculus for Biology and Medicine

Question

Find the derivative of $$ f(x)=\sin \sqrt{x^{2}+1} $$

Step-by-Step Solution

Answer

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

Step 1: Identify the Function

The given function is \( f(x) = \sin(\sqrt{x^2 + 1}) \). It is important to realize that this is a composite function where an inner function is inside an outer function.

Step 2: Determine the Outer Function and Inner Function

The outer function is \( g(u) = \sin(u) \) where \( u = \sqrt{x^2 + 1} \). The inner function is \( h(x) = \sqrt{x^2 + 1} \).

Step 3: Differentiate the Outer Function

The derivative of \( g(u) = \sin(u) \) with respect to \( u \) is \( \cos(u) \).

Step 4: Differentiate the Inner Function

The inner function is \( h(x) = \sqrt{x^2 + 1} = (x^2 + 1)^{1/2} \). Using the power rule and chain rule, the derivative is \( h'(x) = \frac{1}{2}(x^2 + 1)^{-1/2} \cdot 2x = \frac{x}{\sqrt{x^2 + 1}} \).

Step 5: Apply the Chain Rule

The chain rule states that the derivative of a composite function \( f(x) = g(h(x)) \) is \( f'(x) = g'(h(x)) \cdot h'(x) \). Substitute the derivatives from the previous steps: \( f'(x) = \cos(\sqrt{x^2 + 1}) \cdot \frac{x}{\sqrt{x^2 + 1}} \).

Step 6: Simplify the Expression

The expression for the derivative becomes \( f'(x) = \frac{x \cos(\sqrt{x^2 + 1})}{\sqrt{x^2 + 1}} \).

Key Concepts

Understanding the Chain Rule

The chain rule is an essential concept in calculus used to find the derivative of a composite function. Imagine you have a function nestled within another function, much like a series of Russian nesting dolls. The chain rule helps you work through these layers.

Here's a simple analogy: think of the composite function as an assembly line. The chain rule allows you to differentiate each part of the line. It connects the derivatives of the inner and outer functions.

• Start by finding the derivatives of the outer and inner functions separately.
• Next, multiply these derivatives together, following the rule: \(f'(x) = g'(h(x)) \cdot h'(x)\).

In the exercise, the outer function, \(g(u)\), is \(\sin(u)\), and the inner function, \(h(x)\), is \(\sqrt{x^2 + 1}\). The chain rule assists in combining their derivatives to find the derivative of the entire composite function.

Exploring Composite Functions

A composite function is like a function within a function. It's when one function depends on the output of another. Composite functions can seem a bit tricky, but with practice, they become easier.

Consider \(f(x)\) from the exercise: it's \(\sin(\sqrt{x^2 + 1})\). Here’s how to understand this step by step:

• Recognize the outer function: \(\sin(u)\).
• Identify the inner function: \(\sqrt{x^2 + 1}\).
• The inner function is substituted into the outer function, forming the composite.

Realizing the structure of a composite function is crucial for applying the chain rule accurately. It allows you to effectively differentiate complex expressions by breaking them down into simpler parts.

Trigonometric Functions in Calculus

Trigonometric functions like \(\sin(x)\), \(\cos(x)\), and \(\tan(x)\) are fundamental in mathematical analysis, and they frequently appear in calculus. Understanding how to take their derivatives is key.

In calculus, the derivative of \(\sin(u)\) with respect to \(u\) is \(\cos(u)\). This fact is used in the exercise to help find the derivative of the composite function.

• The sine function, \(\sin(u)\), oscillates between -1 and 1.
• When differentiating, the result is \(\cos(u)\), another common trigonometric function.
• This relationship is part of what makes trigonometric functions so powerful in calculus.

These functions also have other derivative rules, such as:

• \(\cos(u)\) differentiates to \(-\sin(u)\).
• \(\tan(u)\) differentiates to \(\sec^2(u)\).

Knowledge of these rules provides a foundation for solving calculus problems involving trigonometric functions effectively.