Problem 36 - Precalculus

Question

Determine whether the statement is true or false. Explain your answer. If \(g(x)=f(x) \sin x,\) then \(g^{\prime}(0)=f(0)\)

Step-by-Step Solution

Answer

True; using the product rule, we find \(g'(0) = f(0)\).

Step 1: Recall the Product Rule for Differentiation

The product rule states that if you have two functions multiplied together, their derivative is the first function times the derivative of the second plus the second function times the derivative of the first. Mathematically, if \( g(x) = u(x)v(x) \), then \( g'(x) = u'(x)v(x) + u(x)v'(x) \).

Step 2: Identify the Functions u(x) and v(x)

In this problem, set \( u(x) = f(x) \) and \( v(x) = \sin x \). We will differentiate \( g(x) = f(x) \sin x \) using these functions.

Step 3: Differentiate u(x) and v(x)

Differentiate \( u(x) = f(x) \), which gives \( u'(x) = f'(x) \). Differentiate \( v(x) = \sin x \), which gives \( v'(x) = \cos x \).

Step 4: Apply the Product Rule

Using the product rule, the derivative \( g'(x) \) is given by \( g'(x) = f'(x) \sin x + f(x) \cos x \).

Step 5: Evaluate the Derivative at x=0

Substitute \( x = 0 \) into the derivative. We have \( g'(0) = f'(0) \sin 0 + f(0) \cos 0 = f'(0) \times 0 + f(0) \times 1 \).

Step 6: Simplify the Derivative at x=0

This simplifies to \( g'(0) = 0 + f(0) = f(0) \). Hence, \( g'(0) = f(0) \).

Step 7: Conclusion

The statement is true, as we have shown that \( g'(0) = f(0) \) based on the evaluated derivative and simplification.

Key Concepts

Differentiation

Differentiation is a central concept in calculus that focuses on finding the rate at which a function changes at any given point. This concept gives us the derivative, which is essentially the slope of the tangent line at any point on the function's graph. The differentiation process allows us to examine how functions behave as their input values change. In calculus, it's essential to know some basic rules, like the product rule, chain rule, and quotient rule, which guide us in finding derivatives of more complex functions.
When addressing composite or combined functions, such as products of two functions like in this exercise, we utilize the product rule for differentiation. Given two functions, say, \( u(x) \) and \( v(x) \), their product's derivative \( g(x) = u(x) v(x) \) is calculated as \( g'(x) = u'(x) v(x) + u(x) v'(x) \). This means the derivative of the first function times the second plus the first function times the derivative of the second.
This principle is helpful when you have a function expressed as the product of two separate functions, like \( f(x) \sin x \), and you need to determine how that product function changes at a specific point.

Trigonometric Functions

Trigonometric functions are fundamental in mathematics, especially in calculus and analysis. Functions like \( \sin x \) and \( \cos x \) are periodic and oscillate between -1 and 1, making them crucial for describing phenomena that involve waves or cyclical patterns.
When dealing with trigonometric differentiation, it's essential to remember how these functions behave under differentiation:

• The derivative of \( \sin x \) is \( \cos x \).
• The derivative of \( \cos x \) is \( -\sin x \).

These derivatives play a critical role in numerous calculus problems, such as finding the rate of change of waveforms or in engineering calculations involving mechanical vibrations. In the exercise, when taking the derivative of \( g(x) = f(x) \sin x \) using the product rule, the specific derivative \( \sin x \rightarrow \cos x \) was used to find how the function changes around the point \( x = 0 \). Understanding these transformations makes solving trigonometric-based differentiation problems much more straightforward.

Calculus Problem-Solving

Solving calculus problems often involves identifying the type of function and then applying the appropriate differentiation rules. The product rule, as seen in the exercise, is just one tool that can be used to address problems involving multiplication of two functions.
When tackling calculus exercises:

• First, identify the individual components or functions within the problem.
• Next, choose the correct differentiation rule or technique suited to the problem.
• Calculate the derivatives step-by-step, as needed.
• Finally, evaluate the result, considering the initial conditions or specific point, like \( x = 0 \) in this case.

This systematic approach ensures accuracy and efficiency. In the specific problem at hand, after identifying \( u(x) = f(x) \) and \( v(x) = \sin x \), the steps involved using these functions with the product rule. By evaluating at \( x = 0 \), it was shown that \( g'(0) = f(0) \), confirming the statement's true nature. This kind of structured problem-solving not only leads to the correct answer but also builds a deeper understanding of how calculus principles apply to various functions and circumstances.