Problem 45 - Calculus Early Transcendentals: Pearson New International Edition

Question

If \(f(0)=4, f^{\prime}(0)=-1, g(0)=-3\), and \(g^{\prime}(0)=5\), find (a) \((f \cdot g)^{\prime}(0)\) (b) \((f+g)^{\prime}(0)\) (c) \((f / g)^{\prime}(0)\)

Step-by-Step Solution

Answer

(a) 23, (b) 4, (c) \(-\frac{17}{9}\).

Step 1: Understand the Problem

We need to find the derivatives of functions involving products, sums, and quotients of two functions \(f\) and \(g\) at the point \(x = 0\). We are given their values and derivatives at this point.

Step 2: Apply Product Rule for Part (a)

The product rule for derivatives states that for two functions \(f\) and \(g\), the derivative of their product \((f \cdot g)^{\prime}(x) = f^{\prime}(x) \cdot g(x) + f(x) \cdot g^{\prime}(x)\). At \(x = 0\), it becomes \((f \cdot g)^{\prime}(0) = f^{\prime}(0) \cdot g(0) + f(0) \cdot g^{\prime}(0)\). Substitute the given values: \(-1 \cdot (-3) + 4 \cdot 5 = 3 + 20 = 23\).

Step 3: Apply Sum Rule for Part (b)

The derivative of the sum of two functions is the sum of their derivatives: \((f + g)^{\prime}(x) = f^{\prime}(x) + g^{\prime}(x)\). Therefore, \((f + g)^{\prime}(0) = f^{\prime}(0) + g^{\prime}(0) = -1 + 5 = 4\).

Step 4: Apply Quotient Rule for Part (c)

The quotient rule states \((\frac{f}{g})^{\prime}(x) = \frac{f^{\prime}(x)g(x) - f(x)g^{\prime}(x)}{(g(x))^2}\). At \(x = 0\), substitute the given values: \((\frac{f}{g})^{\prime}(0) = \frac{-1 \cdot (-3) - 4 \cdot 5}{(-3)^2} = \frac{3 - 20}{9} = \frac{-17}{9}\).

Key Concepts

Product Rule

The product rule is an essential tool in calculus for differentiating the product of two functions. When you have functions like \(f\) and \(g\), and you need to find the derivative of their product \((f \cdot g)\), this rule comes into play. Here’s how it works:
To find the derivative of \((f \cdot g)\), apply the formula:

• \((f \cdot g)^{\prime}(x) = f^{\prime}(x) \cdot g(x) + f(x) \cdot g^{\prime}(x)\)

Essentially, you differentiate 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.
In our given problem, we’ve used the product rule at \(x = 0\):

• \(f^{\prime}(0) \cdot g(0) + f(0) \cdot g^{\prime}(0) = -1 \cdot (-3) + 4 \cdot 5 = 23\)

This clearly shows how each part contributes to the final result.
Understanding and mastering the product rule helps greatly, as it’s frequently used in complex derivatives.

Sum Rule

The sum rule is one of the simplest rules to apply in calculus derivatives. It states that if you want to find the derivative of a sum of functions, you simply add their derivatives together.The formula is straightforward:

• \((f + g)^{\prime}(x) = f^{\prime}(x) + g^{\prime}(x)\)

Using this rule simplifies calculations and helps tackle more complex functions with ease.
For our specific exercise, applying the sum rule at \(x = 0\) was simple:

• \(f^{\prime}(0) + g^{\prime}(0) = -1 + 5 = 4\)

By understanding and using the sum rule, students can quickly and efficiently differentiate combined functions.
Always remember, the sum rule makes derivative calculations for sums straightforward and is a fundamental skill in calculus.

Quotient Rule

The quotient rule is applied when you need to differentiate a quotient of two functions. It can be a bit trickier than the other rules, but it's incredibly useful. Let’s break it down.The quotient rule formula is:

• \(\left(\frac{f}{g}\right)^{\prime}(x) = \frac{f^{\prime}(x)g(x) - f(x)g^{\prime}(x)}{(g(x))^2}\)

This formula tells us exactly how to deal with derivatives involving division.
To apply this in our problem at \(x = 0\):

• \(\left(\frac{f}{g}\right)^{\prime}(0) = \frac{-1 \cdot (-3) - 4 \cdot 5}{(-3)^2} = \frac{3 - 20}{9} = \frac{-17}{9}\)

It’s important to remember that the denominator of the derivative is squared, which prevents any division by zero.
The quotient rule is essential for calculus students as it extends the capacity to handle complex rational expressions.