Problem 61 - Calculus Early Transcendentals: Pearson New International Edition

Question

Given that \(f(1)=2, f^{\prime}(1)=-1, g(1)=0\) and \(g^{\prime}(1)=1\), find \(F^{\prime}(1)\) where \(F(x)=f(x) \cos g(x)\).

Step-by-Step Solution

Answer

\(F'(1) = -1\).

Step 1: Identify the Formula

To find the derivative of the function \( F(x) = f(x) \cos g(x) \), we need to use the product rule for differentiation, which states \( (uv)' = u'v + uv' \). In this case, \( u = f(x) \) and \( v = \cos g(x) \).

Step 2: Apply the Product Rule

Using the product rule, we differentiate \( F(x) = u \cdot v = f(x) \cdot \cos g(x) \) to get: \[ F'(x) = f'(x) \cdot \cos g(x) + f(x) \cdot (-\sin g(x)) \cdot g'(x) \].

Step 3: Evaluate at x=1

Substitute \( x = 1 \) into the derivative found in the previous step:\[ F'(1) = f'(1) \cdot \cos g(1) + f(1) \cdot (-\sin g(1)) \cdot g'(1) \].

Step 4: Substitute Known Values

Replace the derivatives and function values at \( x=1 \): \( f(1)=2, f'(1)=-1, g(1)=0, g'(1)=1 \). Also, calculate \( \cos 0 = 1 \) and \( \sin 0 = 0 \). Thus, \[ F'(1) = -1 \cdot 1 + 2 \cdot (-0) \cdot 1 \].

Step 5: Simplify the Expression

Simplify the expression obtained:\[ F'(1) = -1 + 0 = -1 \].

Key Concepts

Product Rule

In calculus, the product rule is a crucial method used for finding the derivative of a product of two functions. Let's simplify this further.
Suppose we have two functions, say \( u \) and \( v \). If we wish to differentiate the product of these functions, i.e., \( u(x) \, v(x) \), the product rule comes into play:

• According to the product rule, the derivative of \( u(x)v(x) \) is given by \((uv)' = u'v + uv'\).
• This means that we first differentiate \( u \) and keep \( v \) as it is, then multiply \( u \) as it is and differentiate \( v \).

In our exercise:

• We applied this rule to \( F(x) = f(x) \cos g(x) \), where \( u=f(x) \) and \( v=\cos g(x) \).
• Each component of the rule gives us important terms that contribute to finding the final derivative, helping in evaluating the function at a specific point.

Derivative Evaluation

Derivative evaluation is the process of calculating the derivative of a function at a specific point.
This process helps in understanding the rate of change of a function in relation to one of its variables.
For the function \( F(x) = f(x) \cos g(x) \):

• Using the product rule, we derived its derivative as \( F'(x) = f'(x) \cos g(x) + f(x) \cdot (-\sin g(x)) \cdot g'(x) \).
• Next, we evaluate this derivative at \( x=1 \) using known values: \( f(1)=2 \), \( f'(1)=-1 \), \( g(1)=0 \), and \( g'(1)=1 \).

Calculating these:

• \( \cos g(1) = \cos 0 = 1 \)
• \( \sin g(1) = \sin 0 = 0 \).

Substituting all these into the derived formula gives the final result at that particular point.

Trigonometric Functions

Trigonometric functions, such as sine and cosine, are essential in many areas of calculus.
They allow us to describe periodic phenomena and calculate various derivatives effectively.
When dealing with derivatives of trigonometric functions:

• The derivative of \( \cos(x) \) is \( -\sin(x) \), and this is crucial in evaluating complex expressions involving \( \cos(x) \).
• In our problem, \( \cos(g(x)) \) and \( \sin(g(x)) \) help break down the function \( F(x) = f(x) \cos g(x) \) into simpler parts that can be differentiated.

To apply these correctly:

• Remembering the fundamental derivatives such as \( \cos 0 = 1 \) and \( \sin 0 = 0 \) simplifies the computation significantly when evaluating at specific points.
• This understanding bridges the gap between abstract trigonometric identities and their practical application in differentiation processes.