Problem 39 - Calculus Single Variable

Question

Calculate the derivative of each of the expressions in Exercises 39-44 by applying both the Product and Quotient Rules. $$ x \sin (x) /(x+1) $$

Step-by-Step Solution

Answer

Differentiate using the Product Rule and apply the Quotient Rule for simplification.

Step 1: Identify Functions

To apply the Quotient Rule, identify the functions in the form \( \frac{u}{v} \) where \( u = x \sin(x) \) and \( v = x + 1 \).

Step 2: Apply the Quotient Rule

The Quotient Rule is defined as \( \frac{d}{dx}\left(\frac{u}{v}\right) = \frac{v(\frac{du}{dx}) - u(\frac{dv}{dx})}{v^2} \). Compute each component needed for this formula.

Step 3: Differentiate Function u Using the Product Rule

To differentiate \( u = x \sin(x) \), apply the Product Rule: \( \frac{d}{dx}(fg) = f'g + fg' \). Let \( f = x \) and \( g = \sin(x) \), thus \( f' = 1 \) and \( g' = \cos(x) \). Compute \( \frac{du}{dx} = \cos(x) \times x + \sin(x) \).

Step 4: Differentiate Function v

Differentiate \( v = x + 1 \) simply as \( \frac{dv}{dx} = 1 \).

Step 5: Substitute and Simplify

Substitute the results into the Quotient Rule equation: \[ \frac{d}{dx}\left(\frac{x\sin(x)}{x+1}\right) = \frac{(x+1)(x\cos(x) + \sin(x)) - x\sin(x) \cdot 1}{(x+1)^2} \].Simplify the expression if possible.

Key Concepts

Product Rule

The Product Rule is a crucial differentiation technique used to find the derivative of a product of two functions. Let's say you have two functions, \( f(x) \) and \( g(x) \), and you want to differentiate the product \( f(x)g(x) \). The Product Rule states that:

• \( \frac{d}{dx}[f(x)g(x)] = f'(x)g(x) + f(x)g'(x) \)

This means you take the derivative of the first function \( f(x) \), and multiply it by the second function \( g(x) \). Then, add the first function \( f(x) \) multiplied by the derivative of the second function \( g(x) \).
In our original exercise where \( u = x \sin(x) \), we see that it's a product of two simpler functions, \( f = x \) and \( g = \sin(x) \). Thus:

• \( f'(x) = 1 \)
• \( g'(x) = \cos(x) \)

Plug these into the Product Rule formula to find \( \frac{du}{dx} \):

• \( \frac{du}{dx} = 1 \cdot \sin(x) + x \cdot \cos(x) = \sin(x) + x\cos(x) \)

The Product Rule makes it much easier to differentiate products of functions without manually expanding them.

Quotient Rule

The Quotient Rule helps us find the derivative of the quotient of two functions. When you have a function \( \frac{u(x)}{v(x)} \), use the Quotient Rule. Its formula is:

• \( \frac{d}{dx}\left(\frac{u(x)}{v(x)}\right) = \frac{v(x)u'(x) - u(x)v'(x)}{[v(x)]^2} \)

This equation tells us to differentiate \( u(x) \) and \( v(x) \), then plug them into the formula multiplied by each other as per the rule, and finally, divide by \( v(x) \) squared.
From the step-by-step solution, the functions are \( u = x\sin(x) \) and \( v = x + 1 \), with derivatives:

• \( u'(x) = \sin(x) + x\cos(x) \)
• \( v'(x) = 1 \)

Applying these to the Quotient Rule formula, we derive:

• \( \frac{d}{dx}\left(\frac{x\sin(x)}{x+1}\right) = \frac{(x+1)(\sin(x) + x\cos(x)) - x\sin(x)}{(x+1)^2} \)

The Quotient Rule is invaluable when dealing with fractions of functions in calculus.

Differentiation Techniques

Differentiation is a fundamental concept in calculus focused on finding the rate at which a function changes. Several techniques help simplify this process, especially for more complex functions:

• Power Rule: For any power function \( f(x) = x^n \), the derivative is \( f'(x) = nx^{n-1} \).
• Chain Rule: Used for composite functions, stating \( (f(g(x)))' = f'(g(x)) \cdot g'(x) \).

In the given exercise, the focus is on the Product Rule and Quotient Rule, both intended for functions involving products and quotients, respectively. The Product Rule breaks down complex functions into simpler products and finds derivatives through parts, while the Quotient Rule addresses divisions.
Understanding how and when to apply each technique is key. For example, complicated functions like \( x \sin(x)/(x+1) \) demonstrate how combining rules can tackle tough problems, allowing the student to compute the derivative systematically.
By mastering these techniques, you can solve a vast range of calculus problems efficiently, paving the way for deeper mathematical exploration.