Problem 36 - Calculus Early Transcendentals

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

The statement is true; \(g'(0) = f(0)\).

Step 1: Understand the Functions Involved

We are given a function \(g(x) = f(x) \sin x\). This indicates that \(g(x)\) is the product of two functions: \(f(x)\) and \(\sin x\).

Step 2: Use Product Rule for Differentiation

To find the derivative \(g'(x)\), we apply the product rule. The product rule states that if \(h(x) = u(x) \, v(x)\), then \(h'(x) = u'(x) \, v(x) + u(x) \, v'(x)\). Let \(u(x) = f(x)\) and \(v(x) = \sin x\).

Step 3: Compute Derivatives

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

Step 4: Apply the Product Rule

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

Step 5: Evaluate the Derivative at x = 0

Substitute \(x = 0\) into \(g'(x)\):\[g'(0) = f'(0) \sin 0 + f(0) \cos 0\]Since \(\sin 0 = 0\) and \(\cos 0 = 1\), the expression simplifies to:\[g'(0) = f(0)\]

Step 6: Conclusion

The expression \(g'(0) = f(0)\) is correctly obtained from the derivative calculation, confirming the statement is true.

Key Concepts

Product Rule

The product rule is a fundamental technique used in calculus for finding the derivative of a product of two functions. When you have a function like \( h(x) = u(x) \, v(x) \), and you want to differentiate it, the product rule helps in this process. It states that \( h'(x) = u'(x) \, v(x) + u(x) \, v'(x) \). This means we take the derivative of the first function and multiply it by the second function as is, and then add it to the derivative of the second function multiplied by the first function as is.
Consider a simpler analogy: If you have something like "apples times bananas", and you want to change apples, you multiply the change in apples (\(u'(x)\)) with the bananas as they are. And then, you add the change in bananas (\(v'(x)\)) multiplied by the apples as they are.
This is exactly what is happening when we apply it to \( g(x) = f(x) \sin x \). We see \( g'(x) \) include terms from both \( f(x) \) and \( \sin x \) derivatives combined in a smart way.

Derivatives

Derivatives are a core concept in calculus, representing the rate of change of a function. When we compute a derivative, we are finding out how a function behaves when its input changes. In simpler terms, it's like asking how fast water is flowing out of a tap at a given moment.
In the exercise, we have two functions to differentiate: \( f(x) \) and \( \sin x \). Each requires a different approach:

• For \( f(x) \), the derivative is written as \( f'(x) \). We don't have explicit information on \( f(x) \), so we leave it general as \( f'(x) \).
• For \( \sin x \), a well-known trigonometric function, the derivative is precisely \( \cos x \). This stems from the basic rules of derivatives for sine and cosine functions.

Computing these derivatives is essential for applying the product rule correctly.

Trigonometric Functions

Trigonometric functions like sine and cosine are everywhere in mathematics. They describe relationships in triangular geometry and map angles to ratios. Two important trigonometric functions involved in our problem are \( \sin x \) and \( \cos x \).
Understanding their derivatives is crucial:

• The derivative of \( \sin x \) is \( \cos x \), which you can think of as describing how the sine function changes with respect to its angle.
• Similarly, the derivative of \( \cos x \) is \(-\sin x \). This information comes in handy when dealing with combined functions like \( f(x) \sin x \) as in our exercise.

These derivatives tell us how the trigonometric function behaves as its input changes, crucial information for successfully differentiating the product in our example.