Problem 66 - Calculus for Biology and Medicine

Question

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

Step-by-Step Solution

Answer

The derivative is \(-\frac{x \sin(\sqrt{x^2 + 1})}{\sqrt{x^2 + 1}}\)."

Step 1: Identify the outer and inner functions

In the function \( f(x) = \cos(\sqrt{x^2 + 1}) \), the outer function is \( \cos(u) \) where \( u = \sqrt{x^2 + 1} \). The inner function is \( \sqrt{x^2 + 1} \). We'll apply the chain rule to differentiate this function.

Step 2: Differentiate the outer function

The derivative of \( \cos(u) \) with respect to \( u \) is \( -\sin(u) \). Therefore, \( \frac{d}{du} \cos(u) = -\sin(\sqrt{x^2 + 1}) \).

Step 3: Differentiate the inner function

To differentiate \( u = \sqrt{x^2 + 1} = (x^2 + 1)^{1/2} \), apply the chain rule. Start by differentiating the power, yielding \( \frac{1}{2}(x^2 + 1)^{-1/2} \times 2x \), which simplifies to \( \frac{x}{\sqrt{x^2 + 1}} \).

Step 4: Apply the chain rule

To find \( \frac{df}{dx} \), multiply the derivative of the outer function by the derivative of the inner function: \[ \frac{df}{dx} = -\sin(\sqrt{x^2 + 1}) \cdot \frac{x}{\sqrt{x^2 + 1}} \].

Step 5: Simplify the expression

The expression simplifies to \[ \frac{df}{dx} = -\frac{x \sin(\sqrt{x^2 + 1})}{\sqrt{x^2 + 1}} \].

Key Concepts

Understanding Derivatives

Derivatives are fundamental in calculus as they measure how a function changes as its input changes. They're like a tool we use to study rates of change or how steep a slope is at any point on a curve. Think of a derivative as answering the question: "How does the function's output move when I tweak its input?"

For real-valued functions, such as polynomials or trigonometric functions, the derivative is simply the slope at a particular point. In simpler terms, it's the rate at which a function is changing at any given moment.

• If a function's derivative is positive, the function is increasing.
• If the derivative is negative, the function is decreasing.
• If the derivative is zero, it typically indicates a flat spot or a peak/valley of the function.

Practically, when we talk about finding a derivative, we apply various derivative rules. One of the most handy tools here is the chain rule, especially when dealing with nested functions.

Exploring Composite Functions

Composite functions are like those magical nesting dolls you open to find more inside. They are functions built from two or more functions, where the output from one function becomes the input to another. The general form is:

• If you have a function \( g(x) \) and another \( f(u) \), a composite function would be \( f(g(x)) \).
• In the given exercise, the function \( f(x) = \cos(\sqrt{x^2 + 1}) \) indicates a layer of functions where one is inside another.

When dealing with composite functions, the derivative can't be found by looking at each piece separately. This is where the chain rule comes into play.

The chain rule helps us manage this complication by stating that the derivative of a composite function is the derivative of the outer function evaluated at the inner function, multiplied by the derivative of the inner function. It's like unpacking each layer of the nested function, one at a time. For this exercise, we identified the 'outer' \( \cos(u) \) and the 'inner' \( \sqrt{x^2+1} \) functions, which makes differentiation much more methodical and manageable.

Trigonometric Functions and Their Derivatives

Trigonometric functions such as \( \sin(x) \), \( \cos(x) \), and \( \tan(x) \) are pivotal in mathematics, especially in calculus due to their various properties and applications. They relate to angles and periodic phenomena, like waves, and have specific derivatives that make them unique.

The crucial rules to remember are:

• The derivative of \( \sin(x) \) is \( \cos(x) \).
• The derivative of \( \cos(x) \) is \( -\sin(x) \), reflecting the cyclic nature of these functions.
• The derivative of \( \tan(x) \) is \( \sec^2(x) \).

These derivatives play a role when they combine with other function types, especially in composite functions. In the exercise presented, we looked specifically at the derivative of \( \cos(u) \), which is \(-\sin(u)\). This was an integral part of using the chain rule effectively.

Understanding how trigonometric derivatives function allows us to approach a wide array of problems involving waves, oscillations, and rotations, offering solutions that are both elegant and comprehensive.