Problem 38 - Precalculus

Question

Suppose that \(g(x)=f(x) \sec x,\) where \(f(0)=8\) and \(f^{\prime}(0)=-2 .\) Then $$ \begin{aligned} g^{\prime}(0) &=\lim _{h \rightarrow 0} \frac{f(h) \sec h-f(0)}{h}=\lim _{h \rightarrow 0} \frac{8(\sec h-1)}{h} \\ &=\left.8 \cdot \frac{d}{d x}[\sec x]\right|_{x=0}=8 \sec 0 \tan 0=0 \end{aligned} $$

Step-by-Step Solution

Answer

The derivative \(g'(0)\) is \(-2\).

Step 1: Understand the Problem

We need to find the derivative of the function \(g(x) = f(x) \sec x\) at \(x = 0\). We know that \(f(0) = 8\) and \(f'(0) = -2\). Our goal is to find \(g'(0)\).

Step 2: Apply the Product Rule

Since \(g(x) = f(x) \sec x\), we apply the product rule for differentiation: \(g'(x) = f'(x) \sec x + f(x) \frac{d}{dx}(\sec x)\).

Step 3: Differentiate \(\sec x\)

The derivative of \(\sec x\) with respect to \(x\) is \(\sec x \tan x\). So, using this, we have: \(g'(x) = f'(x) \sec x + f(x) \sec x \tan x\).

Step 4: Evaluate at \(x = 0\)

Substitute the known values: \(f'(0) = -2\), \(f(0) = 8\), \(\sec 0 = 1\), and \(\tan 0 = 0\). This gives: \(g'(0) = (-2)(1) + 8(1)(0)\).

Step 5: Simplify the Expression

Calculate the values: \(-2 + 0 = -2\). Thus, \(g'(0) = -2\).

Key Concepts

Product Rule

When dealing with derivatives, especially of functions that are products, the product rule is your friend. Let's break it down for ease of understanding. The product rule states: if you have two functions, say \( u(x) \) and \( v(x) \), then the derivative of their product \( u(x)v(x) \) is given by:

• \( (uv)' = u'v + uv' \)

This rule helps us differentiate products by taking the derivative of each function in turn and summing the results.
In our exercise, \( g(x) = f(x) \sec x \),\( f(x) \) and \( \sec x \) are our functions \( u \) and \( v \) respectively. This means when applying the product rule, we differentiate \( f(x) \) and \( \sec x \) separately:

• \( g'(x) = f'(x) \sec x + f(x) \cdot \frac{d}{dx}(\sec x) \)

By working through these steps, you can find the derivative of a function that is the product of two other functions.

Trigonometric Functions

Trigonometric functions appear frequently in calculus, and understanding their derivatives is key to solving many problems. Today, we focus on one particular trigonometric function: the secant, or \( \sec x \).
The secant function is the reciprocal of cosine: \( \sec x = \frac{1}{\cos x} \). When differentiating \( \sec x \), the result is \( \sec x \tan x \). This might look complex at first glance, but breaking it down shows it’s quite approachable:

• The derivative of \( \sec x \) is found by taking the derivative of \( \frac{1}{\cos x} \), which involves recognizing it as a composition of functions.
• The product \( \sec x \tan x \) emerges from the chain rule and the trigonometric identities.

Grasping how to find derivatives of trig functions like \( \sec x \) will empower you to tackle a variety of derivative-related problems. Always remember that familiarity with trig identities is crucial!

Limits

Limits are fundamental to understanding derivatives, and they play a special role when evaluating derivatives at specific points. A limit essentially helps you understand the behavior of a function as it approaches a certain point.
In calculus, the derivative of a function at a point is defined using a limit:

• \( f'(a) = \lim_{h \to 0} \frac{f(a+h) - f(a)}{h} \)

This idea was applied in our problem when attempting to find \( g'(0) \). The limit expression \( \lim_{h \to 0} \frac{8(\sec h - 1)}{h} \) appears as we evaluate the derivative near 0.
Limit evaluation often requires recognizing when a function behaves simply at a certain point, such as knowing \( \sec 0 = 1 \) and \( \tan 0 = 0 \), which drastically simplifies computations.
Understanding limits allows you to bridge between what a function does near a value and its precise behavior at that exact point.