Problem 35 - Precalculus

Question

Determine whether the statement is true or false. Explain your answer. If \(g(x)=f(x) \sin x,\) then \(g^{\prime}(x)=f^{\prime}(x) \cos x\)

Step-by-Step Solution

Answer

The statement is false; the derivative includes an additional term.

Step 1: Understand the Given Functions

We have a function defined as \( g(x) = f(x) \sin x \). Our goal is to differentiate this function correctly and verify if the given derivative \( g'(x) = f'(x) \cos x \) is true.

Step 2: Apply the Product Rule for Differentiation

To differentiate \( g(x) = f(x) \sin x \), we will use the product rule. The product rule states that if \( h(x) = u(x) v(x) \), then \( h'(x) = u'(x)v(x) + u(x)v'(x) \). Here \( u(x) = f(x) \) and \( v(x) = \sin x \).

Step 3: Differentiate Each Component

Differentiate \( u(x) = f(x) \) giving \( u'(x) = f'(x) \). Differentiate \( v(x) = \sin x \) giving \( v'(x) = \cos x \).

Step 4: Substitute into Product Rule Formula

Substitute \( u'(x) = f'(x) \), \( v(x) = \sin x \), \( u(x) = f(x) \), and \( v'(x) = \cos x \) into the product rule formula: \[ g'(x) = f'(x) \sin x + f(x) \cos x \]

Step 5: Compare the Results

The derived expression \( g'(x) = f'(x) \sin x + f(x) \cos x \) must be compared against the suggested derivative \( g'(x) = f'(x) \cos x \). Clearly, these are not equivalent because there is an extra term \( f'(x) \sin x \) in the correct derivative.

Step 6: Conclusion

The given statement that \( g'(x) = f'(x) \cos x \) is false, since the correct derivative obtained from the product rule is \( g'(x) = f'(x) \sin x + f(x) \cos x \).

Key Concepts

Product Rule

When differentiating a product of two functions, the product rule comes into play. It provides the correct way to differentiate expressions where two functions are multiplied together. When you have a function like \( h(x) = u(x) v(x) \), the derivative \( h'(x) \) is not just the derivative of each part multiplied. Instead, you apply the formula:

• \( h'(x) = u'(x) v(x) + u(x) v'(x) \)

This formula essentially requires two steps: taking the derivative of the first function times the second function untouched and adding it to the first function untouched times the derivative of the second function.
This method ensures that no part of the function's complexity is overlooked, leading to an accurate derivative. For the exercise at hand, we applied the product rule where \( u(x) = f(x) \) and \( v(x) = \sin x \). These components were differentiated separately and assembled, following the rule perfectly.

Differentiation

Differentiation is the process of finding the derivative of a function. The derivative represents the rate at which a function is changing at any given point, much like the slope of a tangent line to a curve. In mathematics, differentiation is crucial because it helps in understanding how functions behave.
For the expression \( g(x) = f(x) \sin x \), differentiation requires us to precisely follow tools like the product rule to deconstruct the function into simpler parts.

• Differentiate \( u(x) = f(x) \) to get \( u'(x) = f'(x) \).
• Differentiate \( v(x) = \sin x \) to get \( v'(x) = \cos x \).

Combining these results through the rules of differentiation gives the complete derivative. Mastering differentiation skills is indispensable for solving calculus-related problems.

Trigonometric Functions

Trigonometric functions, such as \( \sin x \), \( \cos x \), and so on, are common in calculus and have specific differentiation rules. They describe angles and various phenomena in periodic motions like waves. Knowing how to differentiate trigonometric functions is essential for dealing with functions where these elements appear.
In our exercise, \( \sin x \) is one of the components of the product \( g(x) = f(x) \sin x \). Differentiating \( \sin x \) yields \( \cos x \), a basic yet impactful relationship that frequently emerges in calculus problems.

• Derivative of \( \sin x \) is \( \cos x \).

This differentiation forms a critical part of solving problems using trigonometric functions as a component of larger expressions. Understanding these basic identities and their derivatives makes handling complex calculus operations more manageable.