Problem 83 - Calculus Early Transcendentals: Pearson New International Edition

Question

Give a second proof of the Quotient Rule. Write $$D_{x}\left(\frac{f(x)}{g(x)}\right)=D_{x}\left(f(x) \frac{1}{g(x)}\right)$$ and use the Product Rule and the Chain Rule.

Step-by-Step Solution

Answer

The derivative is \( \frac{f'(x)g(x) - f(x)g'(x)}{g(x)^2} \).

Step 1: Rewrite the Quotient as a Product

The function \( \frac{f(x)}{g(x)} \) can be rewritten as \( f(x) \frac{1}{g(x)} \). Hence, finding the derivative of the quotient is equivalent to finding the derivative of this product.

Step 2: Apply the Product Rule

According to the Product Rule, the derivative of the product \( u(x) v(x) \) is \( u'(x)v(x) + u(x)v'(x) \). Here, set \( u(x) = f(x) \) and \( v(x) = \frac{1}{g(x)} \). Thus, the derivative of the product is \( f'(x) \frac{1}{g(x)} + f(x) \frac{d}{dx}\left(\frac{1}{g(x)}\right) \).

Step 3: Apply the Chain Rule to \( \frac{1}{g(x)} \)

To differentiate \( \frac{1}{g(x)} \), use the Chain Rule. Let \( v(x) = g(x)^{-1} \). The derivative \( v'(x) \) is \( -g(x)^{-2}g'(x) \), or equivalently, \( -\frac{g'(x)}{g(x)^2} \). Thus, \( \frac{d}{dx}\left(\frac{1}{g(x)}\right) = -\frac{g'(x)}{g(x)^2} \).

Step 4: Substitute the Derivative from the Chain Rule

Substitute \( \frac{d}{dx}\left(\frac{1}{g(x)}\right) = -\frac{g'(x)}{g(x)^2} \) back into the result from the Product Rule to get \( f'(x) \frac{1}{g(x)} - f(x) \frac{g'(x)}{g(x)^2} \).

Step 5: Simplify the Expression

Combine the two terms into a single fraction: \( \frac{f'(x)g(x) - f(x)g'(x)}{g(x)^2} \). This is the standard form of the Quotient Rule: \( \left(\frac{f}{g}\right)' = \frac{f'g - fg'}{g^2} \).

Key Concepts

Product Rule

The Product Rule is a fundamental concept in calculus used to find the derivative of a product of two functions. When you have two functions, say \( u(x) \) and \( v(x) \), the Product Rule tells us how to differentiate their product: \( (uv)' = u'v + uv' \). This means we take the derivative of the first function and multiply it by the second function, then add the product of the first function and the derivative of the second function.

Let's apply it to the function \( f(x) \cdot \frac{1}{g(x)} \). We set \( u(x) = f(x) \) and \( v(x) = \frac{1}{g(x)} \). Using the Product Rule, we get:

• The derivative of \( u(x) \), \( f'(x) \)
• Multiply it by \( v(x) = \frac{1}{g(x)} \)
• Plus \( f(x) \) times the derivative of \( v(x) \), \( \frac{d}{dx} \left( \frac{1}{g(x)} \right) \)

Combining these provides the necessary components to proceed with solving the derivative of a quotient using the Product Rule.

Chain Rule

The Chain Rule is another key concept in calculus, essential for differentiating compositions of functions. It is used when a function is "inside" another function. Mathematically, if you have \( h(x) = f(g(x)) \), then the Chain Rule states \( h'(x) = f'(g(x)) \cdot g'(x) \).

In our given problem, we encounter \( \frac{1}{g(x)} \). This can be rewritten as \( g(x)^{-1} \). Using the Chain Rule, we want to find the derivative of \( g(x)^{-1} \). First, note that:

• The derivative of \( y = x^{-1} \) is \( y' = -x^{-2} \)
• Applied here, the derivative \( \frac{d}{dx}(g(x)^{-1}) \) becomes \( -g(x)^{-2} \cdot g'(x) \)

This step effectively allows us to evaluate the change in \( \frac{1}{g(x)} \) with respect to \( x \), facilitating the Product Rule application.

Derivative

A derivative signifies the rate of change of a function with respect to its variable, acting as a fundamental concept in calculus. Essentially, it provides a way to determine how a function's output value changes as the input value changes. When we refer to a function \( f(x) \), the derivative, often denoted as \( f'(x) \) or \( \frac{d}{dx}f(x) \), quantifies this rate of change.

In terms of the Quotient Rule, which seeks the derivative of \( \frac{f(x)}{g(x)} \), you must find:

• The derivative of the numerator, \( f'(x) \)
• The application of rules like the Product and Chain Rule to find the derivative of the denominator-related expression.

The end formula \( \frac{f'(x)g(x) - f(x)g'(x)}{g(x)^2} \) effectively captures how \( \frac{f(x)}{g(x)} \) changes with \( x \).

Calculus Proof

Calculus proofs like the one illustrated for the Quotient Rule help in understanding the logic and reasoning behind mathematical formulas. Proofs validate results by deriving them through known premises and logical steps.

The given exercise demonstrates a calculus proof for the Quotient Rule by converting the division into a product. The steps involve:

• Rewriting \( \frac{f(x)}{g(x)} \) as \( f(x) \cdot \frac{1}{g(x)} \)
• Applying the Product Rule to this expression
• Using the Chain Rule to differentiate \( \frac{1}{g(x)} \)

Each of these steps showcases how derivatives of more complex functions can be systematically approached and understood, ultimately leading to the familiar Quotient Rule. This not only proves the rule but deepens comprehension of calculus methods.