Problem 16 - Applied Calculus: For Business, Economics, and the Social and Life Sciences

Question

Differentiate the given function. $$f(x)=\frac{\sin x}{1-\cos x}$$

Step-by-Step Solution

Answer

The derivative is \( f'(x) = -\frac{1}{1 - \text{cos}(x)} \).

Step 1 - Understand the Function

The given function is a quotient of two functions: \[ f(x) = \frac{\text{sin}(x)}{1 - \text{cos}(x)} \]To differentiate this, the Quotient Rule will be needed.

Step 2 - Apply the Quotient Rule

Recall the Quotient Rule: \[ \frac{d}{dx} \bigg( \frac{u}{v} \bigg) = \frac{u'v - uv'}{v^2} \]Here, \( u = \text{sin}(x) \) and \( v = 1 - \text{cos}(x) \).First, find the derivatives \( u' \) and \( v' \).

Step 3 - Differentiate the Numerator

Differentiate \( u = \text{sin}(x) \): \[ u' = \frac{d}{dx} \text{sin}(x) = \text{cos}(x) \]

Step 4 - Differentiate the Denominator

Differentiate \( v = 1 - \text{cos}(x) \): \[ v' = \frac{d}{dx} (1 - \text{cos}(x)) = 0 + \text{sin}(x) = \text{sin}(x) \]

Step 5 - Substitute Into Quotient Rule

Now, substitute \( u, v, u' \) and \( v' \) into the Quotient Rule: \[ \frac{u'}{v} - \frac{uv'}{v^2} = \frac{\text{cos}(x) (1 - \text{cos}(x)) - \text{sin}(x) \text{sin}(x)}{(1 - \text{cos}(x))^2} \]

Step 6 - Simplify the Expression

Simplify the expression in the numerator first: \[ \text{cos}(x)(1 - \text{cos}(x)) - \text{sin}^2(x) = \text{cos}(x) - \text{cos}^2(x) - \text{sin}^2(x) \]Now recall the Pythagorean identity: \[ \text{cos}^2(x) + \text{sin}^2(x) = 1 \]Thus, \[ \text{cos}(x) - 1\]

Step 7 - Finalize the Derivative

After substitution and simplifying, the final expression becomes: \[ f'(x) = \frac{- (1 - \text{cos}(x))}{(1 - \text{cos}(x))^2} \]Further simplify this to obtain: \[ f'(x) = -\frac{1}{1 - \text{cos}(x)} \]

Key Concepts

Quotient Rule

When differentiating a function that is the quotient of two other functions, the Quotient Rule is essential. The rule can be remembered by the phrase 'low d high minus high d low over the square of what's below.' This means, if you have a function \[ f(x) = \frac{u(x)}{v(x)} \] where \( u(x) \) and \( v(x) \) are differentiable, then the derivative \( f'(x) \) is given by \[ f'(x) = \frac{u'(x) v(x) - u(x) v'(x)}{v(x)^2} \] Here's a step-by-step application of this rule:

• Identify the numerator, \( u(x) \), and the denominator, \( v(x) \).
• Differentiate both the numerator and denominator to find \( u'(x) \) and \( v'(x) \).
• Substitute these values into the Quotient Rule formula.
• Simplify the result.

This rule simplifies the process of differentiating complex fractional functions and is crucial in calculus.

Trigonometric Derivatives

Trigonometric functions like sine and cosine have specific derivatives that you should memorize. Knowing these can speed up your work considerably.
Usually, the most common derivatives you will find are:

• \( \frac{d}{dx} \sin(x) = \cos(x) \)
• \( \frac{d}{dx} \cos(x) = -\sin(x) \)

For example, in the problem presented, you had to differentiate \( \sin(x) \) which yielded \( \cos(x) \). Similarly, differentiating \( \cos(x) \) gives \( -\sin(x) \)..
Understanding these basic derivatives will allow you to apply them readily in the larger context of problems requiring more complex derivative rules, such as the Quotient Rule.

Pythagorean Identity

The Pythagorean identity is one of the most fundamental identities in trigonometry and is given by \[ \cos^2(x) + \sin^2(x) = 1 \] This identity is crucial when simplifying expressions involving trigonometric functions.
In the problem provided, you had an opportunity to use this identity while simplifying the final expression. You started with the result: \[ \cos(x)(1 - \cos(x)) - \sin^2(x) \] which simplified further.

By recognizing that \( \sin^2(x) \) can be replaced by \( 1 - \cos^2(x) \) due to the Pythagorean identity, you can make the expression a lot simpler to handle.
Incorporating the Pythagorean identity in trigonometric differentiation problems often helps in reducing the complexity of the expressions you are dealing with.

It is a powerful tool and is widely applied to solve various problems involving trigonometric functions.