Problem 40 - Calculus for Biology and Medicine

Question

In Problems \(1-58\), find the derivative with respect to the independent variable. $$ f(x)=\sec x \cos x $$

Step-by-Step Solution

Answer

The derivative of \( f(x) = \sec{x} \cos{x} \) is 0.

Step 1: Identify the Product Rule

The function given is a product of two functions: \( \sec{x} \cdot \cos{x} \). To find the derivative, we apply the product rule. The product rule states that if you have two functions \( u(x) \) and \( v(x) \), their derivative is given by \( u'(x) \cdot v(x) + u(x) \cdot v'(x) \). In this case, let \( u(x) = \sec{x} \) and \( v(x) = \cos{x} \).

Step 2: Find the Derivative of \( u(x) = \sec{x} \)

The derivative of \( \sec{x} \) is \( \sec{x} \tan{x} \). So, \( u'(x) = \sec{x} \tan{x} \).

Step 3: Find the Derivative of \( v(x) = \cos{x} \)

The derivative of \( \cos{x} \) is \( -\sin{x} \). So, \( v'(x) = -\sin{x} \).

Step 4: Apply the Product Rule

Using the product rule, the derivative of the function is: \[ \frac{d}{dx}[\sec{x} \cos{x}] = (\sec{x} \tan{x}) \cdot \cos{x} + \sec{x} \cdot (-\sin{x}). \]

Step 5: Simplify the Expression

Simplify the expression:1. \( (\sec{x} \tan{x}) \cdot \cos{x} = \tan{x} \)2. \( \sec{x} \cdot (-\sin{x}) = -\tan{x} \)Thus, the derivative becomes:\[ \frac{d}{dx}[\sec{x} \cos{x}] = \tan{x} - \tan{x} = 0. \]

Key Concepts

Product Rule

The product rule is a fundamental differentiation technique in calculus used to find the derivative of the product of two functions. If you're dealing with two functions, say \( u(x) \) and \( v(x) \), their derivative is not as straightforward as simply taking the derivative of each separately and multiplying them. The real rule is this:

The product rule states
for functions \( u(x) \) and \( v(x) \), their derivative \( \frac{d}{dx}[u(x) \cdot v(x)] \) is:

• \( u'(x) \cdot v(x) + u(x) \cdot v'(x) \)

In the example of \( f(x) = \sec{x} \cdot \cos{x} \):
- Let \( u(x) = \sec{x} \) and \( v(x) = \cos{x} \).
- Then their derivatives are \( u'(x) = \sec{x} \tan{x} \) and \( v'(x) = -\sin{x} \).
- Plug into the product rule, and you will see how each derivative plays a crucial role in the final equation.

Trigonometric Functions

Trigonometric functions, such as \( \sec{x} \) and \( \cos{x} \), are functions based on angles and are crucial in calculus. When differentiating functions involving them, remember that each trigonometric function has its own unique derivative.

Here are some key derivatives to remember:

• \( \frac{d}{dx} [\sin{x}] = \cos{x} \)
• \( \frac{d}{dx} [\cos{x}] = -\sin{x} \)
• \( \frac{d}{dx} [\tan{x}] = \sec^2{x} \)
• \( \frac{d}{dx} [\sec{x}] = \sec{x} \tan{x} \)

These derivatives are fundamental when applying differentiation techniques, like the product rule, to functions involving trigonometry. Knowing these allows for efficient solving of derivative problems involving angles.

Differentiation Techniques

Differentiation is a process in calculus used to find a function's rate of change with respect to an independent variable. To tackle a wide array of functions, several techniques are employed:

**Basic Differentiation**
This involves taking the derivative of simple functions directly using known rules like the power rule, constant rule, and sum or difference rule.

**Chain Rule**
When a function is composed of another function (like function within function), the chain rule helps differentiate them.

**Product Rule and Quotient Rule**
These are employed when a function is a product or a quotient of two functions. The product rule was detailed above. The quotient rule focuses on division involving derivatives.

**Implicit Differentiation**
If a function is not expressed explicitly as one variable in terms of the other, implicit differentiation is used.

These techniques collectively enable solving not only straightforward derivative problems but also more complex ones involving products, quotients, and compositions of functions.