Problem 38 - Calculus Early Transcendentals

Question

Determine whether the statement is true or false. Explain your answer. 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 statement is false; the correct value is \(-2\).

Step 1: Clarify the derivatives

First, we need to find the derivative of the function \( g(x) = f(x) \sec x \). The derivative is found using the product rule since \( g(x) \) is a product of two functions: \( f(x) \) and \( \sec x \).

Step 2: Apply the Product Rule

Using the product rule, \( g'(x) = f'(x) \sec x + f(x) \sec x \tan x \).

Step 3: Evaluate at x = 0

Substitute \( x = 0 \) into the derivative. We know that \( f(0) = 8 \) and \( f'(0) = -2 \). So, \( g'(0) = (-2)(\sec 0) + (8)(\sec 0 \tan 0) \).

Step 4: Simplify the expression

Calculate the trigonometric values: \( \sec(0) = 1 \) and \( \tan(0) = 0 \). Therefore, \( g'(0) = (-2)(1) + (8)(1)(0) = -2 \).

Step 5: Determine the validity of the statement

The given expression for \( g'(0) \) incorrectly calculates the limit to be 0. However, using the product rule, we found \( g'(0) = -2 \). Thus, the statement is false.

Key Concepts

Calculus

Calculus is a significant branch of mathematics that studies continuous change. It is divided into two main parts: differential calculus and integral calculus. Differential calculus focuses on the concept of a derivative. In simpler terms, calculus and its different tools can help in understanding and describing how things change and how quantities interact with each other.

When you break it down, calculus helps to tackle dynamic problems. For instance, it helps find rates of change like speed or the slope of a curve. A student in calculus might often analyze functions to see how they behave. Understanding calculus forms the essential foundation for higher-level mathematics and sciences, enabling more complex theories and models.

In calculus, concepts like limits and infinitesimal changes are fundamental. By learning about how small changes affect the overall system, we can model real-world phenomena that are in constant motion or flux. So, whether you're dealing with physics, economics, or biology, calculus is an invaluable tool.

Derivatives

In calculus, a derivative represents the rate of change of a function with respect to a variable. It's like measuring how fast or slow a given quantity changes over time. The concept of derivatives can be understood through simple comparisons, like if we consider speed as the derivative of the distance concerning time.

Let's explore further with the example from the exercise. We have a function \( g(x) = f(x) \sec x \). To find its derivative \( g'(x) \), we use the product rule given that it is the product of two individual functions: \( f(x) \) and \( \sec x \).

The product rule helps us differentiate products of two or more functions and is expressed as:

• \( (uv)' = u'v + uv' \)

Here, \( u = f(x) \) and \( v = \sec x \). So, the derivative \( g'(x) \) is found using the product rule as:

• \( g'(x) = f'(x) \sec x + f(x) \sec x \tan x \)

This crucial calculus tool facilitates solving complex derivations by breaking them into simpler parts.

Trigonometric Functions

Trigonometric functions are mathematical functions related to angles and ratios of triangle sides. They play an essential role in calculus and playfully connect algebraic concepts with geometric interpretations. Functions like sine, cosine, and tangent, along with their reciprocals such as cosecant, secant, and cotangent, help model periodic phenomena in the real world.

In our particular exercise, we're dealing with \( \sec x \) and \( \tan x \). Understanding these functions is crucial when they become parts of bigger calculus problems. For instance, \( \sec x \) is the reciprocal of \( \cos x \).

When evaluating the function \( g'(x) = f'(x) \sec x + f(x) \sec x \tan x \) at \( x=0 \), we need to substitute the values of \( \sec(0) \) and \( \tan(0) \). These values are:

• \( \sec(0) = 1 \)
• \( \tan(0) = 0 \)

This substitution yielded simplified and accurate calculations for the problem solution. Trigonometric functions, with their cyclic natures and unique properties, often turn up in calculus tasks demanding a blend of algebraic skill and geometric understanding.