Problem 41 - Calculus for Biology and Medicine

Question

Find the derivative with respect to the independent variable. $$ f(x)=\sin x \sec x $$

Step-by-Step Solution

Answer

The derivative is \( f'(x) = 1 + \sin x \tan x \).

Step 1: Identify the Functions

We need to find the derivative of the function \( f(x) = \sin x \sec x \). The function is a product of \( \sin x \) and \( \sec x \).

Step 2: Recall Product Rule

The derivative of a product of two functions \( u(x) \) and \( v(x) \) is given by:\[(u \cdot v)' = u'v + uv'\]Here, \( u(x) = \sin x \) and \( v(x) = \sec x \).

Step 3: Calculate Derivatives

Calculate the derivatives of each function:- The derivative of \( \sin x \) is \( \cos x \).- The derivative of \( \sec x \) is \( \sec x \tan x \).

Step 4: Apply Product Rule

Using the product rule, we have:\[(f(x))' = (\sin x)' \cdot \sec x + \sin x \cdot (\sec x)'\]Substitute the derivatives found in Step 3:\[(f(x))' = \cos x \cdot \sec x + \sin x \cdot \sec x \tan x\]

Step 5: Simplify the Expression

Rewrite the expression obtained:\[(f(x))' = \cos x \cdot \sec x + \sin x \cdot \sec x \tan x = 1 + \sin x \tan x\]This follows because \( \sec x = \frac{1}{\cos x} \) so \( \cos x \cdot \sec x = 1 \).

Key Concepts

Understanding the Product Rule for Derivatives

When dealing with calculus derivatives, one of the most common scenarios is finding the derivative of a product of two functions. This is where the **product rule** comes into play. To visualize this, imagine two functions denoted as \( u(x) \) and \( v(x) \). The product rule states that the derivative of the product \( u(x) \cdot v(x) \) is given by:

• \((u \cdot v)' = u'v + uv'\)

This can be broken down into very straightforward steps:

• First, differentiate the first function \( u(x) \) to get \( u'(x) \).
• Then, multiply it by the second function \( v(x) \).
• Next, keep the first function \( u(x) \) intact and differentiate the second function \( v(x) \) to get \( v'(x) \).
• Multiply this result by the original first function \( u(x) \).

Combine these two results, and you have your derivative! Applying this rule simplifies the differentiation process for products of functions, making it an essential tool in calculus.

Exploring Trigonometric Derivatives

Trigonometric functions have specific derivatives that are pivotal in calculus. Understanding these derivatives can greatly facilitate solving problems like finding the slope or rate of change implied by a given function.

• The derivative of \( \sin x \) is \( \cos x \). This means wherever you see \( \sin x \) in a differentiation problem, you immediately replace it with \( \cos x \).
• For \( \sec x \), the derivative is \( \sec x \tan x \). Knowing these basic transformations assists in quickly computing derivatives involving trigonometric functions.

To apply these, always ensure you first differentiate the trigonometric function individually as done in the product rule application. Whether dealing with basic or complex functions, the derivatives of these fundamental trigonometric functions remain constant, forming the backbone for more intricate calculus scenarios.

Tips for Function Simplification

After obtaining a derivative, simplifying the expression is an important final step to ensure clarity and utility. Simplifying trigonometric expressions involves certain mathematical identities that help reduce functions to a more manageable form.

• Remember that \( \sec x = \frac{1}{\cos x} \). Therefore, any expression like \( \cos x \cdot \sec x \) simplifies to \( 1 \).
• Additional trigonometric relationships, such as \( \tan x = \frac{\sin x}{\cos x} \), can also be handy. Using them can untangle complex expressions after differentiation.

The goal of simplification is to make the expression as elegant and as simple as possible. Simplifying results increases readability and decreases computational complexity in subsequent calculations. Whether you solve problems analytically or numerically, always strive to leave expressions in their simplest forms.