Problem 60 - Calculus Early Transcendentals: Pearson New International Edition

Question

Given that \(F(0)=2\) and \(F^{\prime}(0)=-1\), find \(G^{\prime}(0)\) where \(G(x)=\frac{x}{1+\sec F(2 x)}\)

Step-by-Step Solution

Answer

The value of \(G'(0)\) is \(-\frac{1}{(1+\sec(2))^2}\).

Step 1: Understand the Function to Differentiate

We need to find the derivative \(G'(x)\) of the function \(G(x) = \frac{x}{1+\sec F(2x)}\). This involves both a simple quotient rule and the chain rule for differentiation.

Step 2: Apply the Quotient Rule

Using the quotient rule, \(\frac{u}{v}' = \frac{v'u - uv'}{v^2}\), where \(u(x) = x\) and \(v(x) = 1 + \sec F(2x)\). We need to find \(u'\) and \(v'\). Here, \(u'(x) = 1\).

Step 3: Differentiate the Denominator - Part 1

To find \(v'(x)\), which is \((1 + \sec F(2x))'\), begin by differentiating \(\sec F(2x)\) using the chain rule: \(\frac{d}{dx}[\sec(u)] = \sec(u)\tan(u)\frac{du}{dx}\), where \(u = F(2x)\). We need \(F'(2x)\).

Step 4: Differentiate the Denominator - Part 2

We use the chain rule again: \(\frac{d}{dx}[F(2x)] = F'(2x) \cdot 2\). Given \(F'(0) = -1\), \(F'(2x) = F''(0) \cdot 2\), evaluated at \(x = 0\). Note that we'll find \(v'(x) = -\sec(u)\tan(u) \cdot F'(2x)\cdot 2\).

Step 5: Evaluate at x = 0

Substitute any known values where \(x = 0\). Since \(F(0) = 2\), calculate inside \(v(x) = 1 + \sec F(2 \cdot 0) = 1 + \sec(2)\). Then evaluate the derivative at \(x = 0\).

Step 6: Compute G'(0)

Substitution into the quotient rule: \(G'(0) = \frac{v'(0) \cdot u(0) - u'(0) \cdot v(0)}{(v(0))^2}\). After calculating each part, solve for \(G'(0)\).

Key Concepts

Chain Rule

When it comes to differentiating composite functions, the chain rule is your best friend. It helps us find derivatives of functions that are made up of other functions. In this exercise, we are dealing with an expression that involves \(\sec F(2x)\). To differentiate such a function, the chain rule provides a systematic approach.

The chain rule states that if you have a composite function \(g(f(x))\), then its derivative is \(g'(f(x)) \cdot f'(x)\). In this case, it breaks down like this:

• Identify the outer function, here \(\sec(u)\), and the inner function \(u = F(2x)\).
• Find the derivative of the outer function: \(\frac{d}{dx}[\sec(u)] = \sec(u) \tan(u)\).
• Use the given derivative of the inner function: \(\frac{d}{dx}[F(2x)] = F'(2x) \cdot 2\).

This produces the derivative used in our problem: \(-\sec(u) \tan(u) \cdot F'(2x) \cdot 2\). The chain rule allows us to break complexity into manageable parts and apply standard differentiation techniques.

Derivative Evaluation

Derivative evaluation is all about substituting known values into the derived formula to find the result. In our problem, once we have the expressions for \(u'(x)\) and \(v'(x)\), the next step is to plug in any known values when \(x=0\). This ensures an accurate solution.

For example, in this exercise:

• The value of \(F(0) = 2\) helps us find \(v(0)\): \(v(x) = 1 + \sec F(2 \cdot 0) = 1 + \sec(2)\).
• Given that \(F'(0) = -1\), you can determine more about the function at this specific point when combined with other parts of the derivative.

After substituting these into the derivative formula, we get the value of \(G'(0)\). This step is essential to making theoretical work meaningful and helping verify or refute the behavior of the function at specific points.

Function Differentiation

Function differentiation involves breaking down composite functions into simpler parts and employing rules like the quotient and chain rule. In this problem, we are tasked with differentiating \(G(x)=\frac{x}{1+\sec F(2x)}\).

We use the quotient rule, which is essential when differentiating ratios of two functions. It is expressed as:\[\left(\frac{u}{v}\right)' = \frac{v'u - uv'}{v^2}\]Here's how this plays out:

• Identify \(u(x) = x\) with a simple derivative of \(u'(x) = 1\).
• Identify \(v(x) = 1 + \sec F(2x)\) which requires careful differentiation since it involves both a variable term and an external function.

Next, the chain rule helps differentiate \(v(x)\). Finally, substituting all calculated derivatives into the quotient rule provides our function's derivative. Function differentiation is a composite procedure, uniting separate techniques for a cohesive solution.

Calculus Problem Solving

Calculus problem solving often feels like piecing together a puzzle. Each derivative and rule applied helps build towards a solution. For this exercise, the ultimate goal was to find the derivative \(G'(0)\).

Here's a strategic breakdown:

• Recognize the form of the function \(G(x)=\frac{x}{1+\sec F(2x)}\), indicating the use of the quotient rule.
• Use the chain rule to address the interior complexity of the \(\sec F(2x)\) term.
• Calculate derivatives systematically and substitute given values into them.

With accurate derivative values in hand, \(G'(0)\) is derived by substituting these into the quotient rule formula. Each step, from identifying the appropriate rules to evaluating expressions, demands understanding and clarity.
The beauty of calculus solving lies in its logical structure, turning complex problems into comprehensible outcomes.